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Statistics By Jim
Making statistics intuitive
Z Test: Uses, Formula & Examples
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What is a Z Test?
Use a Z test when you need to compare group means. Use the 1sample analysis to determine whether a population mean is different from a hypothesized value. Or use the 2sample version to determine whether two population means differ.
A Z test is a form of inferential statistics . It uses samples to draw conclusions about populations.
For example, use Z tests to assess the following:
 One sample : Do students in an honors program have an average IQ score different than a hypothesized value of 100?
 Two sample : Do two IQ boosting programs have different mean scores?
In this post, learn about when to use a Z test vs T test. Then we’ll review the Z test’s hypotheses, assumptions, interpretation, and formula. Finally, we’ll use the formula in a worked example.
Related post : Difference between Descriptive and Inferential Statistics
Z test vs T test
Z tests and t tests are similar. They both assess the means of one or two groups, have similar assumptions, and allow you to draw the same conclusions about population means.
However, there is one critical difference.
Z tests require you to know the population standard deviation, while t tests use a sample estimate of the standard deviation. Learn more about Population Parameters vs. Sample Statistics .
In practice, analysts rarely use Z tests because it’s rare that they’ll know the population standard deviation. It’s even rarer that they’ll know it and yet need to assess an unknown population mean!
A Z test is often the first hypothesis test students learn because its results are easier to calculate by hand and it builds on the standard normal distribution that they probably already understand. Additionally, students don’t need to know about the degrees of freedom .
Z and T test results converge as the sample size approaches infinity. Indeed, for sample sizes greater than 30, the differences between the two analyses become small.
William Sealy Gosset developed the t test specifically to account for the additional uncertainty associated with smaller samples. Conversely, Z tests are too sensitive to mean differences in smaller samples and can produce statistically significant results incorrectly (i.e., false positives).
When to use a T Test vs Z Test
Let’s put a button on it.
When you know the population standard deviation, use a Z test.
When you have a sample estimate of the standard deviation, which will be the vast majority of the time, the best statistical practice is to use a t test regardless of the sample size.
However, the difference between the two analyses becomes trivial when the sample size exceeds 30.
Learn more about a TTest Overview: How to Use & Examples and How TTests Work .
Z Test Hypotheses
This analysis uses sample data to evaluate hypotheses that refer to population means (µ). The hypotheses depend on whether you’re assessing one or two samples.
OneSample Z Test Hypotheses
 Null hypothesis (H 0 ): The population mean equals a hypothesized value (µ = µ 0 ).
 Alternative hypothesis (H A ): The population mean DOES NOT equal a hypothesized value (µ ≠ µ 0 ).
When the pvalue is less or equal to your significance level (e.g., 0.05), reject the null hypothesis. The difference between your sample mean and the hypothesized value is statistically significant. Your sample data support the notion that the population mean does not equal the hypothesized value.
Related posts : Null Hypothesis: Definition, Rejecting & Examples and Understanding Significance Levels
TwoSample Z Test Hypotheses
 Null hypothesis (H 0 ): Two population means are equal (µ 1 = µ 2 ).
 Alternative hypothesis (H A ): Two population means are not equal (µ 1 ≠ µ 2 ).
Again, when the pvalue is less than or equal to your significance level, reject the null hypothesis. The difference between the two means is statistically significant. Your sample data support the idea that the two population means are different.
These hypotheses are for twosided analyses. You can use onesided, directional hypotheses instead. Learn more in my post, OneTailed and TwoTailed Hypothesis Tests Explained .
Related posts : How to Interpret P Values and Statistical Significance
Z Test Assumptions
For reliable results, your data should satisfy the following assumptions:
You have a random sample
Drawing a random sample from your target population helps ensure that the sample represents the population. Representative samples are crucial for accurately inferring population properties. The Z test results won’t be valid if your data do not reflect the population.
Related posts : Random Sampling and Representative Samples
Continuous data
Z tests require continuous data . Continuous variables can assume any numeric value, and the scale can be divided meaningfully into smaller increments, such as fractional and decimal values. For example, weight, height, and temperature are continuous.
Other analyses can assess additional data types. For more information, read Comparing Hypothesis Tests for Continuous, Binary, and Count Data .
Your sample data follow a normal distribution, or you have a large sample size
All Z tests assume your data follow a normal distribution . However, due to the central limit theorem, you can ignore this assumption when your sample is large enough.
The following sample size guidelines indicate when normality becomes less of a concern:
 OneSample : 20 or more observations.
 TwoSample : At least 15 in each group.
Related posts : Central Limit Theorem and Skewed Distributions
Independent samples
For the twosample analysis, the groups must contain different sets of items. This analysis compares two distinct samples.
Related post : Independent and Dependent Samples
Population standard deviation is known
As I mention in the Z test vs T test section, use a Z test when you know the population standard deviation. However, when n > 30, the difference between the analyses becomes trivial.
Related post : Standard Deviations
Z Test Formula
These Z test formulas allow you to calculate the test statistic. Use the Z statistic to determine statistical significance by comparing it to the appropriate critical values and use it to find pvalues.
The correct formula depends on whether you’re performing a one or twosample analysis. Both formulas require sample means (x̅) and sample sizes (n) from your sample. Additionally, you specify the population standard deviation (σ) or variance (σ 2 ), which does not come from your sample.
I present a worked example using the Z test formula at the end of this post.
Learn more about ZScores and Test Statistics .
One Sample Z Test Formula
The one sample Z test formula is a ratio.
The numerator is the difference between your sample mean and a hypothesized value for the population mean (µ 0 ). This value is often a strawman argument that you hope to disprove.
The denominator is the standard error of the mean. It represents the uncertainty in how well the sample mean estimates the population mean.
Learn more about the Standard Error of the Mean .
Two Sample Z Test Formula
The two sample Z test formula is also a ratio.
The numerator is the difference between your two sample means.
The denominator calculates the pooled standard error of the mean by combining both samples. In this Z test formula, enter the population variances (σ 2 ) for each sample.
Z Test Critical Values
As I mentioned in the Z vs T test section, a Z test does not use degrees of freedom. It evaluates Zscores in the context of the standard normal distribution. Unlike the tdistribution , the standard normal distribution doesn’t change shape as the sample size changes. Consequently, the critical values don’t change with the sample size.
To find the critical value for a Z test, you need to know the significance level and whether it is one or twotailed.
0.01  TwoTailed  ±2.576 
0.01  Left Tail  –2.326 
0.01  Right Tail  +2.326 
0.05  TwoTailed  ±1.960 
0.05  Left Tail  +1.650 
0.05  Right Tail  –1.650 
Learn more about Critical Values: Definition, Finding & Calculator .
Z Test Worked Example
Let’s close this post by calculating the results for a Z test by hand!
Suppose we randomly sampled subjects from an honors program. We want to determine whether their mean IQ score differs from the general population. The general population’s IQ scores are defined as having a mean of 100 and a standard deviation of 15.
We’ll determine whether the difference between our sample mean and the hypothesized population mean of 100 is statistically significant.
Specifically, we’ll use a twotailed analysis with a significance level of 0.05. Looking at the table above, you’ll see that this Z test has critical values of ± 1.960. Our results are statistically significant if our Z statistic is below –1.960 or above +1.960.
The hypotheses are the following:
 Null (H 0 ): µ = 100
 Alternative (H A ): µ ≠ 100
Entering Our Results into the Formula
Here are the values from our study that we need to enter into the Z test formula:
 IQ score sample mean (x̅): 107
 Sample size (n): 25
 Hypothesized population mean (µ 0 ): 100
 Population standard deviation (σ): 15
The Zscore is 2.333. This value is greater than the critical value of 1.960, making the results statistically significant. Below is a graphical representation of our Z test results showing how the Z statistic falls within the critical region.
We can reject the null and conclude that the mean IQ score for the population of honors students does not equal 100. Based on the sample mean of 107, we know their mean IQ score is higher.
Now let’s find the pvalue. We could use technology to do that, such as an online calculator. However, let’s go old school and use a Z table.
To find the pvalue that corresponds to a Zscore from a twotailed analysis, we need to find the negative value of our Zscore (even when it’s positive) and double it.
In the truncated Ztable below, I highlight the cell corresponding to a Zscore of 2.33.
The cell value of 0.00990 represents the area or probability to the left of the Zscore 2.33. We need to double it to include the area > +2.33 to obtain the pvalue for a twotailed analysis.
Pvalue = 0.00990 * 2 = 0.0198
That pvalue is an approximation because it uses a Zscore of 2.33 rather than 2.333. Using an online calculator, the pvalue for our Z test is a more precise 0.0196. This pvalue is less than our significance level of 0.05, which reconfirms the statistically significant results.
See my full Ztable , which explains how to use it to solve other types of problems.
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Ztest : Formula, Types, Examples
Ztest is especially useful when you have a large sample size and know the population’s standard deviation. Different tests are used in statistics to compare distinct samples or groups and make conclusions about populations. These tests, also referred to as statistical tests, concentrate on examining the probability or possibility of acquiring the observed data under particular premises or hypotheses. They offer a framework for evaluating the evidence for or against a given hypothesis.
Table of Content
What is ZTest?
Ztest formula, when to use ztest, hypothesis testing, steps to perform ztest, type of ztest, practice problems.
Ztest is a statistical test that is used to determine whether the mean of a sample is significantly different from a known population mean when the population standard deviation is known. It is particularly useful when the sample size is large (>30).
Ztest can also be defined as a statistical method that is used to determine whether the distribution of the test statistics can be approximated using the normal distribution or not. It is the method to determine whether two sample means are approximately the same or different when their variance is known and the sample size is large (should be >= 30).
The Ztest compares the difference between the sample mean and the population means by considering the standard deviation of the sampling distribution. The resulting Zscore represents the number of standard deviations that the sample mean deviates from the population mean. This ZScore is also known as ZStatistics, and can be formulated as:
[Tex]\text{ZScore} = \frac{\bar{x}\mu}{\sigma} [/Tex]
 [Tex]\bar{x} [/Tex] : mean of the sample.
 [Tex]\mu [/Tex] : mean of the population.
 [Tex]\sigma [/Tex] : Standard deviation of the population.
ztest assumes that the test statistic (zscore) follows a standard normal distribution.
The average family annual income in India is 200k, with a standard deviation of 5k, and the average family annual income in Delhi is 300k.
Then ZScore for Delhi will be.
[Tex]\begin{aligned} \text{ZScore}&=\frac{\bar{x}\mu}{\sigma} \\&=\frac{300200}{5} \\&=20 \end{aligned} [/Tex]
This indicates that the average family’s annual income in Delhi is 20 standard deviations above the mean of the population (India).
 The sample size should be greater than 30. Otherwise, we should use the ttest.
 Samples should be drawn at random from the population.
 The standard deviation of the population should be known.
 Samples that are drawn from the population should be independent of each other.
 The data should be normally distributed , however, for a large sample size, it is assumed to have a normal distribution because central limit theorem
A hypothesis is an educated guess/claim about a particular property of an object. Hypothesis testing is a way to validate the claim of an experiment.
 Null Hypothesis: The null hypothesis is a statement that the value of a population parameter (such as proportion, mean, or standard deviation) is equal to some claimed value. We either reject or fail to reject the null hypothesis. The null hypothesis is denoted by H 0 .
 Alternate Hypothesis: The alternative hypothesis is the statement that the parameter has a value that is different from the claimed value. It is denoted by H A .
 Level of significance: It means the degree of significance in which we accept or reject the null hypothesis. Since in most of the experiments 100% accuracy is not possible for accepting or rejecting a hypothesis, we, therefore, select a level of significance. It is denoted by alpha (∝).
 First, identify the null and alternate hypotheses.
 Determine the level of significance (∝).
 Find the critical value of z in the ztest using
 n: sample size.
 Now compare with the hypothesis and decide whether to reject or not reject the null hypothesis
Lefttailed Test
In this test, our region of rejection is located to the extreme left of the distribution. Here our null hypothesis is that the claimed value is less than or equal to the mean population value.
Righttailed Test
In this test, our region of rejection is located to the extreme right of the distribution. Here our null hypothesis is that the claimed value is less than or equal to the mean population value.
OneTailed Test
A school claimed that the students who study that are more intelligent than the average school. On calculating the IQ scores of 50 students, the average turns out to be 110. The mean of the population IQ is 100 and the standard deviation is 15. State whether the claim of the principal is right or not at a 5% significance level.
 First, we define the null hypothesis and the alternate hypothesis. Our null hypothesis will be: [Tex]H_0 : \mu = 100 [/Tex] and our alternate hypothesis. [Tex]H_A : \mu > 100 [/Tex]
 State the level of significance. Here, our level of significance is given in this question ( [Tex]\alpha [/Tex] =0.05), if not given then we take ∝=0.05 in general.
 Now, we compute the ZScore: X = 110 Mean = 100 Standard Deviation = 15 Number of samples = 50 [Tex]\begin{aligned} \text{ZScore}&=\frac{\bar{x}\mu}{\sigma/\sqrt{n}} \\&=\frac{110100}{15/\sqrt{50}} \\&=\frac{10}{2.12} \\&=4.71 \end{aligned} [/Tex]
 Now, we look up to the ztable. For the value of ∝=0.05, the zscore for the righttailed test is 1.645.
 Here 4.71 >1.645, so we reject the null hypothesis.
 If the ztest statistics are less than the zscore, then we will not reject the null hypothesis.
Code Implementations of OneTailed ZTest
# Import the necessary libraries import numpy as np import scipy.stats as stats # Given information sample_mean = 110 population_mean = 100 population_std = 15 sample_size = 50 alpha = 0.05 # compute the zscore z_score = ( sample_mean  population_mean ) / ( population_std / np . sqrt ( 50 )) print ( 'ZScore :' , z_score ) # Approach 1: Using Critical ZScore # Critical ZScore z_critical = stats . norm . ppf ( 1  alpha ) print ( 'Critical ZScore :' , z_critical ) # Hypothesis if z_score > z_critical : print ( "Reject Null Hypothesis" ) else : print ( "Fail to Reject Null Hypothesis" ) # Approach 2: Using Pvalue # PValue : Probability of getting less than a Zscore p_value = 1  stats . norm . cdf ( z_score ) print ( 'pvalue :' , p_value ) # Hypothesis if p_value < alpha : print ( "Reject Null Hypothesis" ) else : print ( "Fail to Reject Null Hypothesis" )
ZScore : 4.714045207910317Critical ZScore : 1.6448536269514722Reject Null Hypothesispvalue : 1.2142337364462463e06Reject Null Hypothesis
Twotailed test
In this test, our region of rejection is located to both extremes of the distribution. Here our null hypothesis is that the claimed value is equal to the mean population value.
Below is an example of performing the ztest:
Twosampled ztest
In this test, we have provided 2 normally distributed and independent populations, and we have drawn samples at random from both populations. Here, we consider u 1 and u 2 to be the population mean, and X 1 and X 2 to be the observed sample mean. Here, our null hypothesis could be like this:
[Tex]H_{0} : \mu_{1} \mu_{2} = 0 [/Tex]
and alternative hypothesis
[Tex]H_{1} : \mu_{1} – \mu_{2} \ne 0 [/Tex]
and the formula for calculating the ztest score:
[Tex]Z = \frac{\left ( \overline{X_{1}} – \overline{X_{2}} \right ) – \left ( \mu_{1} – \mu_{2} \right )}{\sqrt{\frac{\sigma_{1}^2}{n_{1}} + \frac{\sigma_{2}^2}{n_{2}}}} [/Tex]
where [Tex]\sigma_1 [/Tex] and [Tex]\sigma_2 [/Tex] are the standard deviation and n 1 and n 2 are the sample size of population corresponding to u 1 and u 2 .
There are two groups of students preparing for a competition: Group A and Group B. Group A has studied offline classes, while Group B has studied online classes. After the examination, the score of each student comes. Now we want to determine whether the online or offline classes are better.
Group A: Sample size = 50, Sample mean = 75, Sample standard deviation = 10 Group B: Sample size = 60, Sample mean = 80, Sample standard deviation = 12
Assuming a 5% significance level, perform a twosample ztest to determine if there is a significant difference between the online and offline classes.
Step 1: Null & Alternate Hypothesis
 Null Hypothesis: There is no significant difference between the mean score between the online and offline classes [Tex] \mu_1 \mu_2 = 0 [/Tex]
 Alternate Hypothesis: There is a significant difference in the mean scores between the online and offline classes. [Tex] \mu_1 \mu_2 \neq 0 [/Tex]
Step 2: Significance Label
 Significance Label: 5% [Tex]\alpha = 0.05 [/Tex]
Step 3: ZScore
[Tex]\begin{aligned} \text{Zscore} &= \frac{(x_1x_2)(\mu_1 \mu_2)} {\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_1}}} \\ &= \frac{(7580)0} {\sqrt{\frac{10^2}{50}+\frac{12^2}{60}}} \\ &= \frac{5} {\sqrt{2+2.4}} \\ &= \frac{5} {2.0976} \\&=2.384 \end{aligned} [/Tex]
Step 4: Check to Critical ZScore value in the ZTable for apha/2 = 0.025
 Critical ZScore = 1.96
Step 5: Compare with the absolute ZScore value
 absolute(ZScore) > Critical ZScore
 Reject the null hypothesis. There is a significant difference between the online and offline classes.
Code Implementations on Twosampled Ztest
import numpy as np import scipy.stats as stats # Group A (Offline Classes) n1 = 50 x1 = 75 s1 = 10 # Group B (Online Classes) n2 = 60 x2 = 80 s2 = 12 # Null Hypothesis = mu_1mu_2 = 0 # Hypothesized difference (under the null hypothesis) D = 0 # Set the significance level alpha = 0.05 # Calculate the test statistic (zscore) z_score = (( x1  x2 )  D ) / np . sqrt (( s1 ** 2 / n1 ) + ( s2 ** 2 / n2 )) print ( 'ZScore:' , np . abs ( z_score )) # Calculate the critical value z_critical = stats . norm . ppf ( 1  alpha / 2 ) print ( 'Critical ZScore:' , z_critical ) # Compare the test statistic with the critical value if np . abs ( z_score ) > z_critical : print ( """Reject the null hypothesis. There is a significant difference between the online and offline classes.""" ) else : print ( """Fail to reject the null hypothesis. There is not enough evidence to suggest a significant difference between the online and offline classes.""" ) # Approach 2: Using Pvalue # PValue : Probability of getting less than a Zscore p_value = 2 * ( 1  stats . norm . cdf ( np . abs ( z_score ))) print ( 'PValue :' , p_value ) # Compare the pvalue with the significance level if p_value < alpha : print ( """Reject the null hypothesis. There is a significant difference between the online and offline classes.""" ) else : print ( """Fail to reject the null hypothesis. There is not enough evidence to suggest significant difference between the online and offline classes.""" )
ZScore: 2.3836564731139807 Critical ZScore: 1.959963984540054 Reject the null hypothesis. There is a significant difference between the online and offline classes. PValue : 0.01714159544079563 Reject the null hypothesis. There is a significant difference between the online and offline classes.
Solved examples :
Example 1: Onesample Ztest
Problem: A company claims that the average battery life of their new smartphone is 12 hours. A consumer group tests 100 phones and finds the average battery life to be 11.8 hours with a population standard deviation of 0.5 hours. At a 5% significance level, is there evidence to refute the company’s claim?
Solution: Step 1: State the hypotheses H₀: μ = 12 (null hypothesis) H₁: μ ≠ 12 (alternative hypothesis) Step 2: Calculate the Zscore Z = (x̄ – μ) / (σ / √n) = (11.8 – 12) / (0.5 / √100) = 0.2 / 0.05 = 4 Step 3: Find the critical value (twotailed test at 5% significance) Z₀.₀₂₅ = ±1.96 Step 4: Compare Zscore with critical value 4 > 1.96, so we reject the null hypothesis. Conclusion: There is sufficient evidence to refute the company’s claim about battery life.
Problem: A researcher wants to compare the effectiveness of two different medications for reducing blood pressure. Medication A is tested on 50 patients, resulting in a mean reduction of 15 mmHg with a standard deviation of 3 mmHg. Medication B is tested on 60 patients, resulting in a mean reduction of 13 mmHg with a standard deviation of 4 mmHg. At a 1% significance level, is there a significant difference between the two medications?
Step 1: State the hypotheses H₀: μ₁ – μ₂ = 0 (null hypothesis) H₁: μ₁ – μ₂ ≠ 0 (alternative hypothesis) Step 2: Calculate the Zscore Z = (x̄₁ – x̄₂) / √((σ₁²/n₁) + (σ₂²/n₂)) = (15 – 13) / √((3²/50) + (4²/60)) = 2 / √(0.18 + 0.2667) = 2 / 0.6455 = 3.10 Step 3: Find the critical value (twotailed test at 1% significance) Z₀.₀₀₅ = ±2.576 Step 4: Compare Zscore with critical value 3.10 > 2.576, so we reject the null hypothesis. Conclusion: There is a significant difference between the effectiveness of the two medications at the 1% significance level.
Problem 3 : A polling company claims that 60% of voters support a new policy. In a sample of 1000 voters, 570 support the policy. At a 5% significance level, is there evidence to support the company’s claim?
Step 1: State the hypotheses H₀: p = 0.60 (null hypothesis) H₁: p ≠ 0.60 (alternative hypothesis) Step 2: Calculate the Zscore p̂ = 570/1000 = 0.57 (sample proportion) Z = (p̂ – p) / √(p(1p)/n) = (0.57 – 0.60) / √(0.60(10.60)/1000) = 0.03 / √(0.24/1000) = 0.03 / 0.0155 = 1.94 Step 3: Find the critical value (twotailed test at 5% significance) Z₀.₀₂₅ = ±1.96 Step 4: Compare Zscore with critical value 1.94 < 1.96, so we fail to reject the null hypothesis. Conclusion: There is not enough evidence to refute the polling company’s claim at the 5% significance level.
Problem 4 : A manufacturer claims that their light bulbs last an average of 1000 hours. A sample of 100 bulbs has a mean life of 985 hours. The population standard deviation is known to be 50 hours. At a 5% significance level, is there evidence to reject the manufacturer’s claim?
Solution: H₀: μ = 1000 H₁: μ ≠ 1000 Z = (x̄ – μ) / (σ / √n) = (985 – 1000) / (50 / √100) = 15 / 5 = 3 Critical value (α = 0.05, twotailed): ±1.96 3 > 1.96, so reject H₀. Conclusion: There is sufficient evidence to reject the manufacturer’s claim at the 5% significance level.
Example 5 : Two factories produce semiconductors. Factory A’s chips have a mean resistance of 100 ohms with a standard deviation of 5 ohms. Factory B’s chips have a mean resistance of 98 ohms with a standard deviation of 4 ohms. Samples of 50 chips from each factory are tested. At a 1% significance level, is there a difference in mean resistance between the two factories?
H₀: μA – μB = 0 H₁: μA – μB ≠ 0 Z = (x̄A – x̄B) / √((σA²/nA) + (σB²/nB)) = (100 – 98) / √((5²/50) + (4²/50)) = 2 / √(0.5 + 0.32) = 2 / 0.872 = 2.29 Critical value (α = 0.01, twotailed): ±2.576 2.29 < 2.576, so fail to reject H₀. Conclusion: There is not enough evidence to conclude a difference in mean resistance at the 1% significance level.
Problem 6 : A political analyst claims that 40% of voters in a certain district support a new tax policy. In a random sample of 500 voters, 220 support the policy. At a 5% significance level, is there evidence to reject the analyst’s claim?
H₀: p = 0.40 H₁: p ≠ 0.40 p̂ = 220/500 = 0.44 Z = (p̂ – p) / √(p(1p)/n) = (0.44 – 0.40) / √(0.40(10.40)/500) = 0.04 / 0.0219 = 1.83 Critical value (α = 0.05, twotailed): ±1.96 1.83 < 1.96, so fail to reject H₀. Conclusion: There is not enough evidence to reject the analyst’s claim at the 5% significance level.
Problem 7 : Two advertising methods are compared. Method A results in 150 sales out of 1000 contacts. Method B results in 180 sales out of 1200 contacts. At a 5% significance level, is there a difference in the effectiveness of the two methods?
H₀: pA – pB = 0 H₁: pA – pB ≠ 0 p̂A = 150/1000 = 0.15 p̂B = 180/1200 = 0.15 p̂ = (150 + 180) / (1000 + 1200) = 0.15 Z = (p̂A – p̂B) / √(p̂(1p̂)(1/nA + 1/nB)) = (0.15 – 0.15) / √(0.15(10.15)(1/1000 + 1/1200)) = 0 / 0.0149 = 0 Critical value (α = 0.05, twotailed): ±1.96 0 < 1.96, so fail to reject H₀. Conclusion: There is no significant difference in the effectiveness of the two advertising methods at the 5% significance level.
Problem 8 : A new treatment for a disease is tested in two cities. In City A, 120 out of 400 patients recover. In City B, 140 out of 500 patients recover. At a 5% significance level, is there a difference in the recovery rates between the two cities?
H₀: pA – pB = 0 H₁: pA – pB ≠ 0 p̂A = 120/400 = 0.30 p̂B = 140/500 = 0.28 p̂ = (120 + 140) / (400 + 500) = 0.2889 Z = (p̂A – p̂B) / √(p̂(1p̂)(1/nA + 1/nB)) = (0.30 – 0.28) / √(0.2889(10.2889)(1/400 + 1/500)) = 0.02 / 0.0316 = 0.633 Critical value (α = 0.05, twotailed): ±1.96 0.633 < 1.96, so fail to reject H₀. Conclusion: There is not enough evidence to conclude a difference in recovery rates between the two cities at the 5% significance level.
Problem 9 : Two advertising methods are compared. Method A results in 150 sales out of 1000 contacts. Method B results in 180 sales out of 1200 contacts. At a 5% significance level, is there a difference in the effectiveness of the two methods?
Problem 10 : A company claims that their product weighs 500 grams on average. A sample of 64 products has a mean weight of 498 grams. The population standard deviation is known to be 8 grams. At a 1% significance level, is there evidence to reject the company’s claim?
H₀: μ = 500 H₁: μ ≠ 500 Z = (x̄ – μ) / (σ / √n) = (498 – 500) / (8 / √64) = 2 / 1 = 2 Critical value (α = 0.01, twotailed): ±2.576 2 < 2.576, so fail to reject H₀. Conclusion: There is not enough evidence to reject the company’s claim at the 1% significance level.
1).A cereal company claims that their boxes contain an average of 350 grams of cereal. A consumer group tests 100 boxes and finds a mean weight of 345 grams with a known population standard deviation of 15 grams. At a 5% significance level, is there evidence to refute the company’s claim?
2).A study compares the effect of two different diets on cholesterol levels. Diet A is tested on 50 people, resulting in a mean reduction of 25 mg/dL with a standard deviation of 8 mg/dL. Diet B is tested on 60 people, resulting in a mean reduction of 22 mg/dL with a standard deviation of 7 mg/dL. At a 1% significance level, is there a significant difference between the two diets?
3).A politician claims that 60% of voters in her district support her reelection. In a random sample of 1000 voters, 570 support her. At a 5% significance level, is there evidence to reject the politician’s claim?
4).Two different teaching methods are compared. Method A results in 80 students passing out of 120 students. Method B results in 90 students passing out of 150 students. At a 5% significance level, is there a difference in the effectiveness of the two methods?
5).A company claims that their new energysaving light bulbs last an average of 10,000 hours. A sample of 64 bulbs has a mean life of 9,800 hours. The population standard deviation is known to be 500 hours. At a 1% significance level, is there evidence to reject the company’s claim?
6).The mean salary of employees in a large corporation is said to be $75,000 per year. A union representative suspects this is too high and surveys 100 randomly selected employees, finding a mean salary of $72,500. The population standard deviation is known to be $8,000. At a 5% significance level, is there evidence to support the union representative’s suspicion?
7).Two factories produce computer chips. Factory A’s chips have a mean processing speed of 3.2 GHz with a standard deviation of 0.2 GHz. Factory B’s chips have a mean processing speed of 3.3 GHz with a standard deviation of 0.25 GHz. Samples of 100 chips from each factory are tested. At a 5% significance level, is there a difference in mean processing speed between the two factories?
8).A new vaccine is claimed to be 90% effective. In a clinical trial with 500 participants, 440 develop immunity. At a 1% significance level, is there evidence to reject the claim about the vaccine’s effectiveness?
9).Two different advertising campaigns are tested. Campaign A results in 250 sales out of 2000 views. Campaign B results in 300 sales out of 2500 views. At a 5% significance level, is there a difference in the effectiveness of the two campaigns?
10).A quality control manager claims that the defect rate in a production line is 5%. In a sample of 1000 items, 65 are found to be defective. At a 5% significance level, is there evidence to suggest that the actual defect rate is different from the claimed 5%?
Type 1 error and Type II error
 Type I error: Type 1 error has occurred when we reject the null hypothesis, even when the hypothesis is true. This error is denoted by alpha.
 Type II error: Type II error occurred when we didn’t reject the null hypothesis, even when the hypothesis is false. This error is denoted by beta.
Null Hypothesis is TRUE  Null Hypothesis is FALSE  

Reject Null Hypothesis  Type I Error (False Positive)  Correct decision 
Fail to Reject the Null Hypothesis  Correct decision  Type II error (False Negative) 
Ztests are used to determine whether there is a statistically significant difference between a sample statistic and a population parameter, or between two population parameters.Ztests are statistical tools used to determine if there’s a significant difference between a sample statistic and a population parameter, or between two population parameters. They’re applicable when dealing with large sample sizes (typically n > 30) and known population standard deviations. Ztests can be used for analyzing means or proportions in both onesample and twosample scenarios. The process involves stating hypotheses, calculating a Zscore, comparing it to a critical value based on the chosen significance level (often 5% or 1%), and then making a decision to reject or fail to reject the null hypothesis.
What is the main limitation of the ztest?
The limitation of ZTests is that we don’t usually know the population standard deviation. What we do is: When we don’t know the population’s variability, we assume that the sample’s variability is a good basis for estimating the population’s variability.
What is the minimum sample for ztest?
A ztest can only be used if the population standard deviation is known and the sample size is 30 data points or larger. Otherwise, a ttest should be employed.
What is the application of ztest?
It is also used to determine if there is a significant difference between the mean of two independent samples. The ztest can also be used to compare the population proportion to an assumed proportion or to determine the difference between the population proportion of two samples.
What is the theory of the ztest?
The z test is a commonly used hypothesis test in inferential statistics that allows us to compare two populations using the mean values of samples from those populations, or to compare the mean of one population to a hypothesized value, when what we are interested in comparing is a continuous variable.
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Z test is a statistical test that is conducted on data that approximately follows a normal distribution. The z test can be performed on one sample, two samples, or on proportions for hypothesis testing. It checks if the means of two large samples are different or not when the population variance is known.
A z test can further be classified into lefttailed, righttailed, and twotailed hypothesis tests depending upon the parameters of the data. In this article, we will learn more about the z test, its formula, the z test statistic, and how to perform the test for different types of data using examples.
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What is Z Test?
A z test is a test that is used to check if the means of two populations are different or not provided the data follows a normal distribution. For this purpose, the null hypothesis and the alternative hypothesis must be set up and the value of the z test statistic must be calculated. The decision criterion is based on the z critical value.
Z Test Definition
A z test is conducted on a population that follows a normal distribution with independent data points and has a sample size that is greater than or equal to 30. It is used to check whether the means of two populations are equal to each other when the population variance is known. The null hypothesis of a z test can be rejected if the z test statistic is statistically significant when compared with the critical value.
Z Test Formula
The z test formula compares the z statistic with the z critical value to test whether there is a difference in the means of two populations. In hypothesis testing , the z critical value divides the distribution graph into the acceptance and the rejection regions. If the test statistic falls in the rejection region then the null hypothesis can be rejected otherwise it cannot be rejected. The z test formula to set up the required hypothesis tests for a one sample and a twosample z test are given below.
OneSample Z Test
A onesample z test is used to check if there is a difference between the sample mean and the population mean when the population standard deviation is known. The formula for the z test statistic is given as follows:
z = \(\frac{\overline{x}\mu}{\frac{\sigma}{\sqrt{n}}}\). \(\overline{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation and n is the sample size.
The algorithm to set a one sample z test based on the z test statistic is given as follows:
Left Tailed Test:
Null Hypothesis: \(H_{0}\) : \(\mu = \mu_{0}\)
Alternate Hypothesis: \(H_{1}\) : \(\mu < \mu_{0}\)
Decision Criteria: If the z statistic < z critical value then reject the null hypothesis.
Right Tailed Test:
Alternate Hypothesis: \(H_{1}\) : \(\mu > \mu_{0}\)
Decision Criteria: If the z statistic > z critical value then reject the null hypothesis.
Two Tailed Test:
Alternate Hypothesis: \(H_{1}\) : \(\mu \neq \mu_{0}\)
Two Sample Z Test
A two sample z test is used to check if there is a difference between the means of two samples. The z test statistic formula is given as follows:
z = \(\frac{(\overline{x_{1}}\overline{x_{2}})(\mu_{1}\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\). \(\overline{x_{1}}\), \(\mu_{1}\), \(\sigma_{1}^{2}\) are the sample mean, population mean and population variance respectively for the first sample. \(\overline{x_{2}}\), \(\mu_{2}\), \(\sigma_{2}^{2}\) are the sample mean, population mean and population variance respectively for the second sample.
The twosample z test can be set up in the same way as the onesample test. However, this test will be used to compare the means of the two samples. For example, the null hypothesis is given as \(H_{0}\) : \(\mu_{1} = \mu_{2}\).
Z Test for Proportions
A z test for proportions is used to check the difference in proportions. A z test can either be used for one proportion or two proportions. The formulas are given as follows.
One Proportion Z Test
A one proportion z test is used when there are two groups and compares the value of an observed proportion to a theoretical one. The z test statistic for a one proportion z test is given as follows:
z = \(\frac{pp_{0}}{\sqrt{\frac{p_{0}(1p_{0})}{n}}}\). Here, p is the observed value of the proportion, \(p_{0}\) is the theoretical proportion value and n is the sample size.
The null hypothesis is that the two proportions are the same while the alternative hypothesis is that they are not the same.
Two Proportion Z Test
A two proportion z test is conducted on two proportions to check if they are the same or not. The test statistic formula is given as follows:
z =\(\frac{p_{1}p_{2}0}{\sqrt{p(1p)\left ( \frac{1}{n_{1}} +\frac{1}{n_{2}}\right )}}\)
where p = \(\frac{x_{1}+x_{2}}{n_{1}+n_{2}}\)
\(p_{1}\) is the proportion of sample 1 with sample size \(n_{1}\) and \(x_{1}\) number of trials.
\(p_{2}\) is the proportion of sample 2 with sample size \(n_{2}\) and \(x_{2}\) number of trials.
How to Calculate Z Test Statistic?
The most important step in calculating the z test statistic is to interpret the problem correctly. It is necessary to determine which tailed test needs to be conducted and what type of test does the z statistic belong to. Suppose a teacher claims that his section's students will score higher than his colleague's section. The mean score is 22.1 for 60 students belonging to his section with a standard deviation of 4.8. For his colleague's section, the mean score is 18.8 for 40 students and the standard deviation is 8.1. Test his claim at \(\alpha\) = 0.05. The steps to calculate the z test statistic are as follows:
 Identify the type of test. In this example, the means of two populations have to be compared in one direction thus, the test is a righttailed twosample z test.
 Set up the hypotheses. \(H_{0}\): \(\mu_{1} = \mu_{2}\), \(H_{1}\): \(\mu_{1} > \mu_{2}\).
 Find the critical value at the given alpha level using the z table. The critical value is 1.645.
 Determine the z test statistic using the appropriate formula. This is given by z = \(\frac{(\overline{x_{1}}\overline{x_{2}})(\mu_{1}\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\). Substitute values in this equation. \(\overline{x_{1}}\) = 22.1, \(\sigma_{1}\) = 4.8, \(n_{1}\) = 60, \(\overline{x_{2}}\) = 18.8, \(\sigma_{2}\) = 8.1, \(n_{2}\) = 40 and \(\mu_{1}  \mu_{2} = 0\). Thus, z = 2.32
 Compare the critical value and test statistic to arrive at a conclusion. As 2.32 > 1.645 thus, the null hypothesis can be rejected. It can be concluded that there is enough evidence to support the teacher's claim that the scores of students are better in his class.
Z Test vs TTest
Both z test and ttest are univariate tests used on the means of two datasets. The differences between both tests are outlined in the table given below:
Z Test  TTest 

A z test is a statistical test that is used to check if the means of two data sets are different when the population variance is known.  A is used to check if the means of two data sets are different when the population variance is not known. 
The sample size is greater than or equal to 30.  The sample size is lesser than 30. 
The follows a normal distribution.  The data follows a studentt distribution. 
The onesample z test statistic is given by \(\frac{\overline{x}\mu}{\frac{\sigma}{\sqrt{n}}}\)  The t test statistic is given as \(\frac{\overline{x}\mu}{\frac{s}{\sqrt{n}}}\) where s is the sample standard deviation 
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Important Notes on Z Test
 Z test is a statistical test that is conducted on normally distributed data to check if there is a difference in means of two data sets.
 The sample size should be greater than 30 and the population variance must be known to perform a z test.
 The onesample z test checks if there is a difference in the sample and population mean,
 The two sample z test checks if the means of two different groups are equal.
Examples on Z Test
Example 1: A teacher claims that the mean score of students in his class is greater than 82 with a standard deviation of 20. If a sample of 81 students was selected with a mean score of 90 then check if there is enough evidence to support this claim at a 0.05 significance level.
Solution: As the sample size is 81 and population standard deviation is known, this is an example of a righttailed onesample z test.
\(H_{0}\) : \(\mu = 82\)
\(H_{1}\) : \(\mu > 82\)
From the z table the critical value at \(\alpha\) = 1.645
z = \(\frac{\overline{x}\mu}{\frac{\sigma}{\sqrt{n}}}\)
\(\overline{x}\) = 90, \(\mu\) = 82, n = 81, \(\sigma\) = 20
As 3.6 > 1.645 thus, the null hypothesis is rejected and it is concluded that there is enough evidence to support the teacher's claim.
Answer: Reject the null hypothesis
Example 2: An online medicine shop claims that the mean delivery time for medicines is less than 120 minutes with a standard deviation of 30 minutes. Is there enough evidence to support this claim at a 0.05 significance level if 49 orders were examined with a mean of 100 minutes?
Solution: As the sample size is 49 and population standard deviation is known, this is an example of a lefttailed onesample z test.
\(H_{0}\) : \(\mu = 120\)
\(H_{1}\) : \(\mu < 120\)
From the z table the critical value at \(\alpha\) = 1.645. A negative sign is used as this is a left tailed test.
\(\overline{x}\) = 100, \(\mu\) = 120, n = 49, \(\sigma\) = 30
As 4.66 < 1.645 thus, the null hypothesis is rejected and it is concluded that there is enough evidence to support the medicine shop's claim.
Example 3: A company wants to improve the quality of products by reducing defects and monitoring the efficiency of assembly lines. In assembly line A, there were 18 defects reported out of 200 samples while in line B, 25 defects out of 600 samples were noted. Is there a difference in the procedures at a 0.05 alpha level?
Solution: This is an example of a twotailed two proportion z test.
\(H_{0}\): The two proportions are the same.
\(H_{1}\): The two proportions are not the same.
As this is a twotailed test the alpha level needs to be divided by 2 to get 0.025.
Using this, the critical value from the z table is 1.96.
\(n_{1}\) = 200, \(n_{2}\) = 600
\(p_{1}\) = 18 / 200 = 0.09
\(p_{2}\) = 25 / 600 = 0.0416
p = (18 + 25) / (200 + 600) = 0.0537
z =\(\frac{p_{1}p_{2}0}{\sqrt{p(1p)\left ( \frac{1}{n_{1}} +\frac{1}{n_{2}}\right )}}\) = 2.62
As 2.62 > 1.96 thus, the null hypothesis is rejected and it is concluded that there is a significant difference between the two lines.
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FAQs on Z Test
What is a z test in statistics.
A z test in statistics is conducted on data that is normally distributed to test if the means of two datasets are equal. It can be performed when the sample size is greater than 30 and the population variance is known.
What is a OneSample Z Test?
A onesample z test is used when the population standard deviation is known, to compare the sample mean and the population mean. The z test statistic is given by the formula \(\frac{\overline{x}\mu}{\frac{\sigma}{\sqrt{n}}}\).
What is the TwoSample Z Test Formula?
The two sample z test is used when the means of two populations have to be compared. The z test formula is given as \(\frac{(\overline{x_{1}}\overline{x_{2}})(\mu_{1}\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).
What is a One Proportion Z test?
A one proportion z test is used to check if the value of the observed proportion is different from the value of the theoretical proportion. The z statistic is given by \(\frac{pp_{0}}{\sqrt{\frac{p_{0}(1p_{0})}{n}}}\).
What is a Two Proportion Z Test?
When the proportions of two samples have to be compared then the two proportion z test is used. The formula is given by \(\frac{p_{1}p_{2}0}{\sqrt{p(1p)\left ( \frac{1}{n_{1}} +\frac{1}{n_{2}}\right )}}\).
How Do You Find the Z Test?
The steps to perform the z test are as follows:
 Set up the null and alternative hypotheses.
 Find the critical value using the alpha level and z table.
 Calculate the z statistic.
 Compare the critical value and the test statistic to decide whether to reject or not to reject the null hypothesis.
What is the Difference Between the Z Test and the TTest?
A z test is used on large samples n ≥ 30 and normally distributed data while a ttest is used on small samples (n < 30) following a student t distribution . Both tests are used to check if the means of two datasets are the same.
ZTest for Statistical Hypothesis Testing Explained
The Ztest is a statistical hypothesis test that determines where the distribution of the statistic we are measuring, like the mean, is part of the normal distribution.
The Ztest is a statistical hypothesis test used to determine where the distribution of the test statistic we are measuring, like the mean , is part of the normal distribution .
There are multiple types of Ztests, however, we’ll focus on the easiest and most well known one, the one sample mean test. This is used to determine if the difference between the mean of a sample and the mean of a population is statistically significant.
What Is a ZTest?
A Ztest is a type of statistical hypothesis test where the teststatistic follows a normal distribution.
The name Ztest comes from the Zscore of the normal distribution. This is a measure of how many standard deviations away a raw score or sample statistics is from the populations’ mean.
Ztests are the most common statistical tests conducted in fields such as healthcare and data science . Therefore, it’s an essential concept to understand.
Requirements for a ZTest
In order to conduct a Ztest, your statistics need to meet a few requirements, including:
 A Sample size that’s greater than 30. This is because we want to ensure our sample mean comes from a distribution that is normal. As stated by the c entral limit theorem , any distribution can be approximated as normally distributed if it contains more than 30 data points.
 The standard deviation and mean of the population is known .
 The sample data is collected/acquired randomly .
More on Data Science: What Is Bootstrapping Statistics?
ZTest Steps
There are four steps to complete a Ztest. Let’s examine each one.
4 Steps to a ZTest
 State the null hypothesis.
 State the alternate hypothesis.
 Choose your critical value.
 Calculate your Ztest statistics.
1. State the Null Hypothesis
The first step in a Ztest is to state the null hypothesis, H_0 . This what you believe to be true from the population, which could be the mean of the population, μ_0 :
2. State the Alternate Hypothesis
Next, state the alternate hypothesis, H_1 . This is what you observe from your sample. If the sample mean is different from the population’s mean, then we say the mean is not equal to μ_0:
3. Choose Your Critical Value
Then, choose your critical value, α , which determines whether you accept or reject the null hypothesis. Typically for a Ztest we would use a statistical significance of 5 percent which is z = +/ 1.96 standard deviations from the population’s mean in the normal distribution:
This critical value is based on confidence intervals.
4. Calculate Your ZTest Statistic
Compute the Ztest Statistic using the sample mean, μ_1 , the population mean, μ_0 , the number of data points in the sample, n and the population’s standard deviation, σ :
If the test statistic is greater (or lower depending on the test we are conducting) than the critical value, then the alternate hypothesis is true because the sample’s mean is statistically significant enough from the population mean.
Another way to think about this is if the sample mean is so far away from the population mean, the alternate hypothesis has to be true or the sample is a complete anomaly.
More on Data Science: Basic Probability Theory and Statistics Terms to Know
ZTest Example
Let’s go through an example to fully understand the onesample mean Ztest.
A school says that its pupils are, on average, smarter than other schools. It takes a sample of 50 students whose average IQ measures to be 110. The population, or the rest of the schools, has an average IQ of 100 and standard deviation of 20. Is the school’s claim correct?
The null and alternate hypotheses are:
Where we are saying that our sample, the school, has a higher mean IQ than the population mean.
Now, this is what’s called a rightsided, onetailed test as our sample mean is greater than the population’s mean. So, choosing a critical value of 5 percent, which equals a Zscore of 1.96 , we can only reject the null hypothesis if our Ztest statistic is greater than 1.96.
If the school claimed its students’ IQs were an average of 90, then we would use a lefttailed test, as shown in the figure above. We would then only reject the null hypothesis if our Ztest statistic is less than 1.96.
Computing our Ztest statistic, we see:
Therefore, we have sufficient evidence to reject the null hypothesis, and the school’s claim is right.
Hope you enjoyed this article on Ztests. In this post, we only addressed the most simple case, the onesample mean test. However, there are other types of tests, but they all follow the same process just with some small nuances.
Recent Data Science Articles
Z Test: Definition & Two Proportion ZTest
What is a z test.
For example, if someone said they had found a new drug that cures cancer, you would want to be sure it was probably true. A hypothesis test will tell you if it’s probably true, or probably not true. A Z test, is used when your data is approximately normally distributed (i.e. the data has the shape of a bell curve when you graph it).
When you can run a Z Test.
Several different types of tests are used in statistics (i.e. f test , chi square test , t test ). You would use a Z test if:
 Your sample size is greater than 30 . Otherwise, use a t test .
 Data points should be independent from each other. In other words, one data point isn’t related or doesn’t affect another data point.
 Your data should be normally distributed . However, for large sample sizes (over 30) this doesn’t always matter.
 Your data should be randomly selected from a population, where each item has an equal chance of being selected.
 Sample sizes should be equal if at all possible.
How do I run a Z Test?
Running a Z test on your data requires five steps:
 State the null hypothesis and alternate hypothesis .
 Choose an alpha level .
 Find the critical value of z in a z table .
 Calculate the z test statistic (see below).
 Compare the test statistic to the critical z value and decide if you should support or reject the null hypothesis .
You could perform all these steps by hand. For example, you could find a critical value by hand , or calculate a z value by hand . For a step by step example, watch the following video: Watch the video for an example:
Can’t see the video? Click here to watch it on YouTube. You could also use technology, for example:
 Two sample z test in Excel .
 Find a critical z value on the TI 83 .
 Find a critical value on the TI 89 (lefttail) .
Two Proportion ZTest
Watch the video to see a two proportion ztest:
Can’t see the video? Click here to watch it on YouTube.
A Two Proportion ZTest (or Zinterval) allows you to calculate the true difference in proportions of two independent groups to a given confidence interval .
There are a few familiar conditions that need to be met for the Two Proportion ZInterval to be valid.
 The groups must be independent. Subjects can be in one group or the other, but not both – like teens and adults.
 The data must be selected randomly and independently from a homogenous population. A survey is a common example.
 The population should be at least ten times bigger than the sample size. If the population is teenagers for example, there should be at least ten times as many total teenagers as the number of teenagers being surveyed.
 The null hypothesis (H 0 ) for the test is that the proportions are the same.
 The alternate hypothesis (H 1 ) is that the proportions are not the same.
Example question: let’s say you’re testing two flu drugs A and B. Drug A works on 41 people out of a sample of 195. Drug B works on 351 people in a sample of 605. Are the two drugs comparable? Use a 5% alpha level .
Step 1: Find the two proportions:
 P 1 = 41/195 = 0.21 (that’s 21%)
 P 2 = 351/605 = 0.58 (that’s 58%).
Set these numbers aside for a moment.
Step 2: Find the overall sample proportion . The numerator will be the total number of “positive” results for the two samples and the denominator is the total number of people in the two samples.
 p = (41 + 351) / (195 + 605) = 0.49.
Set this number aside for a moment.
Solving the formula, we get: Z = 8.99
We need to find out if the zscore falls into the “ rejection region .”
Step 5: Compare the calculated zscore from Step 3 with the table zscore from Step 4. If the calculated zscore is larger, you can reject the null hypothesis.
8.99 > 1.96, so we can reject the null hypothesis .
Example 2: Suppose that in a survey of 700 women and 700 men, 35% of women and 30% of men indicated that they support a particular presidential candidate. Let’s say we wanted to find the true difference in proportions of these two groups to a 95% confidence interval .
At first glance the survey indicates that women support the candidate more than men by about 5% . However, for this statistical inference to be valid we need to construct a range of values to a given confidence interval.
To do this, we use the formula for Two Proportion ZInterval:
Plugging in values we find the true difference in proportions to be
Based on the results of the survey, we are 95% confident that the difference in proportions of women and men that support the presidential candidate is between about 0 % and 10% .
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Introduction to Statistics and Data Analysis
Chapter 6 hypothesis testing: the ztest.
We’ve all had the experience of standing at a crosswalk waiting staring at a pedestrian traffic light showing the little red man. You’re waiting for the little green man so you can cross. After a little while you’re still waiting and there aren’t any cars around. You might think ‘this light is really taking a long time’, but you continue waiting. Minutes pass and there’s still no little green man. At some point you come to the conclusion that the light is broken and you’ll never see that little green man. You cross on the little red man when it’s clear.
You may not have known this but you just conducted a hypothesis test. When you arrived at the crosswalk, you assumed that the light was functioning properly, although you will always entertain the possibility that it’s broken. In terms of hypothesis testing, your ‘null hypothesis’ is that the light is working and your ‘alternative hypothesis’ is that it’s broken. As time passes, it seems less and less likely that light is working properly. Eventually, the probability of the light working given how long you’ve been waiting becomes so low that you reject the null hypothesis in favor of the alternative hypothesis.
This sort of reasoning is the backbone of hypothesis testing and inferential statistics. It’s also the point in the course where we turn the corner from descriptive statistics to inferential statistics. Rather than describing our data in terms of means and plots, we will now start using our data to make inferences, or generalizations, about the population that our samples are drawn from. In this course we’ll focus on standard hypothesis testing where we set up a null hypothesis and determine the probability of our observed data under the assumption that the null hypothesis is true (the much maligned pvalue). If this probability is small enough, then we conclude that our data suggests that the null hypothesis is false, so we reject it.
In this chapter, we’ll introduce hypothesis testing with examples from a ‘ztest’, when we’re comparing a single mean to what we’d expect from a population with known mean and standard deviation. In this case, we can convert our observed mean into a zscore for the standard normal distribution. Hence the name ztest.
It’s time to introduce the hypothesis test flow chart . It’s pretty self explanatory, even if you’re not familiar with all of these hypothesis tests. The ztest is (1) based on means, (2) with only one mean, and (3) where we know \(\sigma\) , the standard deviation of the population. Here’s how to find the ztest in the flow chart:
6.1 Women’s height example
Let’s work with the example from the end of the last chapter where we started with the fact that the heights of US women has a mean of 63 and a standard deviation of 2.5 inches. We calculated that the average height of the 122 women in Psych 315 is 64.7 inches. We then used the central limit theorem and calculated the probability of a random sample 122 heights from this population having a mean of 64.7 or greater is 2.4868996^{14}. This is a very, very small number.
Here’s how we do it using R:
Let’s think of our sample as a random sample of UW psychology students, which is a reasonable assumption since all psychology students have to take a statistics class. What does this sample say about the psychology students that are women at UW compared to the US population? It could be that these psychology students at UW have the same mean and standard deviation as the US population, but our sample just happens to have an unusual number of tall women, but we calculated that the probability of this happening is really low. Instead, it makes more sense that the population that we’re drawing from has a mean that’s greater than the US population mean. Notice that we’re making a conclusion about the whole population of women psychology students based on our one sample.
Using the terminology of hypothesis testing, we first assumed the null hypothesis that UW women psych students have the same mean (and standard deviation) as the US population. The null hypothesis is written as:
\[ H_{0}: \mu = 63 \] In this example, our alternative hypothesis is that the mean of our population is larger than the mean of null hypothesis population. We write this as:
\[ H_{A}: \mu > 63 \]
Next, after obtaining a random sample and calculate the mean, we calculate the probability of drawing a mean this large (or larger) from the null hypothesis distribution.
If this probability is low enough, we reject the null hypothesis in favor of the alternative hypothesis. When our probability allows us to reject the null hypothesis, we say that our observed results are ‘statistically significant’.
In statistics terms, we never say we ‘accept that alternative hypothesis’ as true. All we can say is that we don’t think the null hypothesis is true. I know it’s subtle, but in science can never prove that a hypothesis is true or not. There’s always the possibility that we just happened to grab an unusual sample from the null hypothesis distribution.
6.2 The hated p<.05
The probability that we obtain our observed mean or greater given that the null hypothesis is true is called the pvalue. How improbable is improbable enough to reject the null hypothesis? The pvalue for our example above on women’s heights is astronomically low, so it’s clear that we should reject \(H_{0}\) .
The pvalue that’s on the border of rejection is called the alpha ( \(\alpha\) ) value. We reject \(H_{0}\) when our pvalue is less than \(\alpha\) .
You probably know that the most common value of alpha is \(\alpha = .05\) .
The first publication of this value dates back to Sir Ronald Fisher, in his seminal 1925 book Statistical Methods for Research Workers where he states:
“It is convenient to take this point as a limit in judging whether a deviation is considered significant or not. Deviations exceeding twice the standard deviation are thus formally regarded as significant.” (p. 47)
If you read the chapter on the normal distribution, then you should know that 95% of the area under the normal distribution lies within \(\pm\) two standard deviations of the mean. So the probability of obtaining a sample that exceeds two standard deviations from the mean (in either direction) is .05.
6.3 IQ example
Let’s do an example using IQ scores. IQ scores are normalized to have a mean of 100 and a standard deviation of 15 points. Because they’re normalized, they are a rare example of a population which has a known mean and standard deviation. In the next chapter we’ll discuss the ttest, which is used in the more common situation when we don’t know the population standard deviation.
Suppose you have the suspicion that graduate students have higher IQ’s than the general population. You have enough time to go and measure the IQ’s of 25 randomly sampled grad students and obtain a mean of 105. Is this difference between our this observed mean and 100 statistically significant using an alpha value of \(\alpha = 0.05\) ?
Here the null hypothesis is:
\[ H_{0}: \mu = 100\]
And the alternative hypothesis is:
\[ H_{A}: \mu > 100 \]
We know that the parameters for the null hypothesis are:
\[ \mu = 100 \] and \[ \sigma = 15 \]
From this, we can calculate the probability of observing our mean of 105 or higher using the central limit theorem and what we know about the normal distribution:
\[ \sigma_{\bar{x}} = \frac{\sigma_{x}}{\sqrt{n}} = \frac{15}{\sqrt{25}} = 3 \] From this, we can calculate the probability of our observed mean using R’s ‘pnorm’ function. Here’s how to do the whole thing in R.
Since our pvalue of 0.0478 is (just barely) less than our chosen value of \(\alpha = 0.05\) as our criterion, we reject \(H_{0}\) for this (contrived) example and conclude that our observed mean of 105 is significantly greater than 100, so our study suggests that the average graduate student has a higher IQ than the overall population.
You should feel uncomfortable making such a hard, binary decision for such a borderline case. After all, if we had chosen our second favorite value of alpha, \(\alpha = .01\) , we would have failed to reject \(H_{0}\) . This discomfort is a primary reason why statisticians are moving away from this discrete decision making process. Later on we’ll discuss where things are going, including reporting effect sizes, and using confidence intervals.
6.4 Alpha values vs. critical values
Using R’s qnorm function, we can find the zscore for which only 5% of the area lies above:
So the probability of a randomly sampled zscore exceeding 1.644854 is less than 5%. It follows that if we convert our observed mean into zscore values, we will reject \(H_{0}\) if and only if our zscore is greater than 1.644854. This value is called the ‘critical value’ because it lies on the boundary between rejecting and failing to reject \(H_{0}\) .
In our last example, the zscore for our observed mean is:
\[ z = \frac{X\mu}{\frac{\sigma}{\sqrt{n}}} = \frac{105  100}{3} = 1.67 \] Our zscore is just barely greater than the critical value of 1.644854, which makes sense because our pvalue is just barely less than 0.05.
Sometimes you’ll see textbooks will compare critical values to observed scores for the decision making process in hypothesis testing. This dates back to days were computers were less available and we had to rely on tables instead. There wasn’t enough space in a book to hold complete tables which prohibited the ability to look up a pvalue for any observed value. Instead only critical values for specific values of alpha were included. If you look at really old papers, you’ll see statistics reported as \(p<.05\) or \(p<.01\) instead of actual pvalues for this reason.
It may help to visualize the relationship between pvalues, alpha values and critical values like this:
The red shaded region is the upper 5% of the standard normal distribution which starts at the critical value of z=1.644854. This is sometimes called the ‘rejection region’. The blue vertical line is drawn at our observed value of z=1.67. You can see that the red line falls just inside the rejection region, so we Reject \(H_{0}\) !
6.5 One vs. twotailed tests
Recall that our alternative hypothesis was to reject if our mean IQ was significantly greater than the null hypothesis mean: \(H_{A}: \mu > 100\) . This implies that the situation where \(\mu < 100\) is never even in consideration, which is weird. In science, we’re trying to understand the true state of the world. Although we have a hunch that grad student IQ’s are higher than average, there is always the possibility that they are lower than average. If our sample came up with an IQ well below 100, we’d simply fail to reject \(H_{0}\) and move on. This feels like throwing out important information.
The test we just ran is called a ‘onetailed’ test because we only reject \(H_{0}\) if our results fall in one of the two tails of the population distribution.
Instead, it might make more sense to reject \(H_{0}\) if we get either an unusually large or small score. This means we need two critical values  one above and one below zero. At first thought you might think we just duplicate our critical value from a onetailed test to the other side. But will double the area of the rejection region. That’s not a good thing because if \(H_{0}\) is true, there’s actually a \(2\alpha\) probability that we’ll draw a score in the rejection region.
Instead, we divide the area into two tails, each containing an area of \(\frac{\alpha}{2}\) . For \(\alpha\) = 0.05, we can find the critical value of z with qnorm:
So with a twotailed test at \(\alpha = 0.05\) we reject \(H_{0}\) if our observed zscore is either above z = 1.96 or less than 1.96. This is that value around 2 that Sir Ronald Fischer was talking about!
Here’s what the critical regions and observed value of z looks like for our example with a twotailed test:
You can see that splitting the area of \(\alpha = 0.05\) into two halves increased the critical value in the positive direction from 1.64 to 1.96, making it harder to reject \(H_{0}\) . For our example, this changes our decision: our observed value of z = 1.67 no longer falls into the rejection region, so now we fail to reject \(H_{0}\) .
If we now fail to reject \(H_{0}\) , what about the pvalue? Remember, for a onetailed test, p = \(\alpha\) if our observed zscore lands right on the critical value of z. The same is true for a twotailed test. But the zscore moved so that the area above that score is \(\frac{\alpha}{2}\) . So for a twotailed test, in order to have a pvalue of \(\alpha\) when our zscore lands right on the critical value, we need to double pvalue hat we’d get for a onetailed test.
For our example, the pvalue for the one tailed test was \(p=0.0478\) . So if we use a twotailed test, our pvalue is \((2)(0.0478) = 0.0956\) . This value is greater than \(\alpha\) = 0.05, which makes sense because we just showed above that we fail to reject \(H_{0}\) with a two tailed test.
Which is the right test, onetailed or twotailed? Ideally, as scientists, we should be agnostic about the results of our experiment. But in reality, we all know that the results are more interesting if they are statistically significant. So you can imagine that for this example, given a choice between one and twotailed, we’d choose a onetailed test so that we can reject \(H_{0}\) .
There are two problems with this. First, we should never adjust our choice of hypothesis test after we observe the data. That would be an example of ‘phacking’, a topic we’ll discuss later. Second, most statisticians these days strongly recommend against onetailed tests. The only reason for a onetailed test is if there is no logical or physical possibility for a population mean to fall below the null hypothesis mean.
10 Chapter 10: Hypothesis Testing with Z
Setting up the hypotheses.
When setting up the hypotheses with z, the parameter is associated with a sample mean (in the previous chapter examples the parameters for the null used 0). Using z is an occasion in which the null hypothesis is a value other than 0. For example, if we are working with mothers in the U.S. whose children are at risk of low birth weight, we can use 7.47 pounds, the average birth weight in the US, as our null value and test for differences against that. For now, we will focus on testing a value of a single mean against what we expect from the population.
Using birthweight as an example, our null hypothesis takes the form: H 0 : μ = 7.47 Notice that we are testing the value for μ, the population parameter, NOT the sample statistic ̅X (or M). We are referring to the data right now in raw form (we have not standardized it using z yet). Again, using inferential statistics, we are interested in understanding the population, drawing from our sample observations. For the research question, we have a mean value from the sample to use, we have specific data is – it is observed and used as a comparison for a set point.
As mentioned earlier, the alternative hypothesis is simply the reverse of the null hypothesis, and there are three options, depending on where we expect the difference to lie. We will set the criteria for rejecting the null hypothesis based on the directionality (greater than, less than, or not equal to) of the alternative.
If we expect our obtained sample mean to be above or below the null hypothesis value (knowing which direction), we set a directional hypothesis. O ur alternative hypothesis takes the form based on the research question itself. In our example with birthweight, this could be presented as H A : μ > 7.47 or H A : μ < 7.47.
Note that we should only use a directional hypothesis if we have a good reason, based on prior observations or research, to suspect a particular direction. When we do not know the direction, such as when we are entering a new area of research, we use a nondirectional alternative hypothesis. In our birthweight example, this could be set as H A : μ ≠ 7.47
In working with data for this course we will need to set a critical value of the test statistic for alpha (α) for use of test statistic tables in the back of the book. This is determining the critical rejection region that has a set critical value based on α.
Determining Critical Value from α
We set alpha (α) before collecting data in order to determine whether or not we should reject the null hypothesis. We set this value beforehand to avoid biasing ourselves by viewing our results and then determining what criteria we should use.
When a research hypothesis predicts an effect but does not predict a direction for the effect, it is called a nondirectional hypothesis . To test the significance of a nondirectional hypothesis, we have to consider the possibility that the sample could be extreme at either tail of the comparison distribution. We call this a twotailed test .
Figure 1. showing a 2tail test for nondirectional hypothesis for z for area C is the critical rejection region.
When a research hypothesis predicts a direction for the effect, it is called a directional hypothesis . To test the significance of a directional hypothesis, we have to consider the possibility that the sample could be extreme at onetail of the comparison distribution. We call this a onetailed test .
Figure 2. showing a 1tail test for a directional hypothesis (predicting an increase) for z for area C is the critical rejection region.
Determining Cutoff Scores with TwoTailed Tests
Typically we specify an α level before analyzing the data. If the data analysis results in a probability value below the α level, then the null hypothesis is rejected; if it is not, then the null hypothesis is not rejected. In other words, if our data produce values that meet or exceed this threshold, then we have sufficient evidence to reject the null hypothesis ; if not, we fail to reject the null (we never “accept” the null). According to this perspective, if a result is significant, then it does not matter how significant it is. Moreover, if it is not significant, then it does not matter how close to being significant it is. Therefore, if the 0.05 level is being used, then probability values of 0.049 and 0.001 are treated identically. Similarly, probability values of 0.06 and 0.34 are treated identically. Note we will discuss ways to address effect size (which is related to this challenge of NHST).
When setting the probability value, there is a special complication in a twotailed test. We have to divide the significance percentage between the two tails. For example, with a 5% significance level, we reject the null hypothesis only if the sample is so extreme that it is in either the top 2.5% or the bottom 2.5% of the comparison distribution. This keeps the overall level of significance at a total of 5%. A onetailed test does have such an extreme value but with a onetailed test only one side of the distribution is considered.
Figure 3. Critical value differences in one and twotail tests. Photo Credit
Let’s re view th e set critical values for Z.
We discussed zscores and probability in chapter 8. If we revisit the zscore for 5% and 1%, we can identify the critical regions for the critical rejection areas from the unit standard normal table.
 A twotailed test at the 5% level has a critical boundary Z score of +1.96 and 1.96
 A onetailed test at the 5% level has a critical boundary Z score of +1.64 or 1.64
 A twotailed test at the 1% level has a critical boundary Z score of +2.58 and 2.58
 A onetailed test at the 1% level has a critical boundary Z score of +2.33 or 2.33.
Review: Critical values, pvalues, and significance level
There are two criteria we use to assess whether our data meet the thresholds established by our chosen significance level, and they both have to do with our discussions of probability and distributions. Recall that probability refers to the likelihood of an event, given some situation or set of conditions. In hypothesis testing, that situation is the assumption that the null hypothesis value is the correct value, or that there is no effec t. The value laid out in H 0 is our condition under which we interpret our results. To reject this assumption, and thereby reject the null hypothesis, we need results that would be very unlikely if the null was true.
Now recall that values of z which fall in the tails of the standard normal distribution represent unlikely values. That is, the proportion of the area under the curve as or more extreme than z is very small as we get into the tails of the distribution. Our significance level corresponds to the area under the tail that is exactly equal to α: if we use our normal criterion of α = .05, then 5% of the area under the curve becomes what we call the rejection region (also called the critical region) of the distribution. This is illustrated in Figure 4.
Figure 4: The rejection region for a onetailed test
The shaded rejection region takes us 5% of the area under the curve. Any result which falls in that region is sufficient evidence to reject the null hypothesis.
The rejection region is bounded by a specific zvalue, as is any area under the curve. In hypothesis testing, the value corresponding to a specific rejection region is called the critical value, z crit (“zcrit”) or z* (hence the other name “critical region”). Finding the critical value works exactly the same as finding the zscore corresponding to any area under the curve like we did in Unit 1. If we go to the normal table, we will find that the zscore corresponding to 5% of the area under the curve is equal to 1.645 (z = 1.64 corresponds to 0.0405 and z = 1.65 corresponds to 0.0495, so .05 is exactly in between them) if we go to the right and 1.645 if we go to the left. The direction must be determined by your alternative hypothesis, and drawing then shading the distribution is helpful for keeping directionality straight.
Suppose, however, that we want to do a nondirectional test. We need to put the critical region in both tails, but we don’t want to increase the overall size of the rejection region (for reasons we will see later). To do this, we simply split it in half so that an equal proportion of the area under the curve falls in each tail’s rejection region. For α = .05, this means 2.5% of the area is in each tail, which, based on the ztable, corresponds to critical values of z* = ±1.96. This is shown in Figure 5.
Figure 5: Twotailed rejection region
Thus, any zscore falling outside ±1.96 (greater than 1.96 in absolute value) falls in the rejection region. When we use zscores in this way, the obtained value of z (sometimes called zobtained) is something known as a test statistic, which is simply an inferential statistic used to test a null hypothesis.
Calculate the test statistic: Z
Now that we understand setting up the hypothesis and determining the outcome, let’s examine hypothesis testing with z! The next step is to carry out the study and get the actual results for our sample. Central to hypothesis test is comparison of the population and sample means. To make our calculation and determine where the sample is in the hypothesized distribution we calculate the Z for the sample data.
Make a decision
To decide whether to reject the null hypothesis, we compare our sample’s Z score to the Z score that marks our critical boundary. If our sample Z score falls inside the rejection region of the comparison distribution (is greater than the zscore critical boundary) we reject the null hypothesis.
The formula for our z statistic has not changed:
To formally test our hypothesis, we compare our obtained zstatistic to our critical zvalue. If z obt > z crit , that means it falls in the rejection region (to see why, draw a line for z = 2.5 on Figure 1 or Figure 2) and so we reject H 0 . If z obt < z crit , we fail to reject. Remember that as z gets larger, the corresponding area under the curve beyond z gets smaller. Thus, the proportion, or pvalue, will be smaller than the area for α, and if the area is smaller, the probability gets smaller. Specifically, the probability of obtaining that result, or a more extreme result, under the condition that the null hypothesis is true gets smaller.
Conversely, if we fail to reject, we know that the proportion will be larger than α because the zstatistic will not be as far into the tail. This is illustrated for a one tailed test in Figure 6.
Figure 6. Relation between α, z obt , and p
When the null hypothesis is rejected, the effect is said to be statistically significant . Do not confuse statistical significance with practical significance. A small effect can be highly significant if the sample size is large enough.
Why does the word “significant” in the phrase “statistically significant” mean something so different from other uses of the word? Interestingly, this is because the meaning of “significant” in everyday language has changed. It turns out that when the procedures for hypothesis testing were developed, something was “significant” if it signified something. Thus, finding that an effect is statistically significant signifies that the effect is real and not due to chance. Over the years, the meaning of “significant” changed, leading to the potential misinterpretation.
Review: Steps of the Hypothesis Testing Process
The process of testing hypotheses follows a simple fourstep procedure. This process will be what we use for the remained of the textbook and course, and though the hypothesis and statistics we use will change, this process will not.
Step 1: State the Hypotheses
Your hypotheses are the first thing you need to lay out. Otherwise, there is nothing to test! You have to state the null hypothesis (which is what we test) and the alternative hypothesis (which is what we expect). These should be stated mathematically as they were presented above AND in words, explaining in normal English what each one means in terms of the research question.
Step 2: Find the Critical Values
Next, we formally lay out the criteria we will use to test our hypotheses. There are two pieces of information that inform our critical values: α, which determines how much of the area under the curve composes our rejection region, and the directionality of the test, which determines where the region will be.
Step 3: Compute the Test Statistic
Once we have our hypotheses and the standards we use to test them, we can collect data and calculate our test statistic, in this case z . This step is where the vast majority of differences in future chapters will arise: different tests used for different data are calculated in different ways, but the way we use and interpret them remains the same.
Step 4: Make the Decision
Finally, once we have our obtained test statistic, we can compare it to our critical value and decide whether we should reject or fail to reject the null hypothesis. When we do this, we must interpret the decision in relation to our research question, stating what we concluded, what we based our conclusion on, and the specific statistics we obtained.
Example: Movie Popcorn
Let’s see how hypothesis testing works in action by working through an example. Say that a movie theater owner likes to keep a very close eye on how much popcorn goes into each bag sold, so he knows that the average bag has 8 cups of popcorn and that this varies a little bit, about half a cup. That is, the known population mean is μ = 8.00 and the known population standard deviation is σ =0.50. The owner wants to make sure that the newest employee is filling bags correctly, so over the course of a week he randomly assesses 25 bags filled by the employee to test for a difference (n = 25). He doesn’t want bags overfilled or under filled, so he looks for differences in both directions. This scenario has all of the information we need to begin our hypothesis testing procedure.
Our manager is looking for a difference in the mean cups of popcorn bags compared to the population mean of 8. We will need both a null and an alternative hypothesis written both mathematically and in words. We’ll always start with the null hypothesis:
H 0 : There is no difference in the cups of popcorn bags from this employee H 0 : μ = 8.00
Notice that we phrase the hypothesis in terms of the population parameter μ, which in this case would be the true average cups of bags filled by the new employee.
Our assumption of no difference, the null hypothesis, is that this mean is exactly
the same as the known population mean value we want it to match, 8.00. Now let’s do the alternative:
H A : There is a difference in the cups of popcorn bags from this employee H A : μ ≠ 8.00
In this case, we don’t know if the bags will be too full or not full enough, so we do a twotailed alternative hypothesis that there is a difference.
Our critical values are based on two things: the directionality of the test and the level of significance. We decided in step 1 that a twotailed test is the appropriate directionality. We were given no information about the level of significance, so we assume that α = 0.05 is what we will use. As stated earlier in the chapter, the critical values for a twotailed ztest at α = 0.05 are z* = ±1.96. This will be the criteria we use to test our hypothesis. We can now draw out our distribution so we can visualize the rejection region and make sure it makes sense
Figure 7: Rejection region for z* = ±1.96
Step 3: Calculate the Test Statistic
Now we come to our formal calculations. Let’s say that the manager collects data and finds that the average cups of this employee’s popcorn bags is ̅X = 7.75 cups. We can now plug this value, along with the values presented in the original problem, into our equation for z:
So our test statistic is z = 2.50, which we can draw onto our rejection region distribution:
Figure 8: Test statistic location
Looking at Figure 5, we can see that our obtained zstatistic falls in the rejection region. We can also directly compare it to our critical value: in terms of absolute value, 2.50 > 1.96, so we reject the null hypothesis. We can now write our conclusion:
When we write our conclusion, we write out the words to communicate what it actually means, but we also include the average sample size we calculated (the exact location doesn’t matter, just somewhere that flows naturally and makes sense) and the zstatistic and pvalue. We don’t know the exact pvalue, but we do know that because we rejected the null, it must be less than α.
Effect Size
When we reject the null hypothesis, we are stating that the difference we found was statistically significant, but we have mentioned several times that this tells us nothing about practical significance. To get an idea of the actual size of what we found, we can compute a new statistic called an effect size. Effect sizes give us an idea of how large, important, or meaningful a statistically significant effect is.
For mean differences like we calculated here, our effect size is Cohen’s d :
Effect sizes are incredibly useful and provide important information and clarification that overcomes some of the weakness of hypothesis testing. Whenever you find a significant result, you should always calculate an effect size
d  Interpretation 

0.0 – 0.2  negligible 
0.2 – 0.5  small 
0.5 – 0.8  medium 
0.8 –  large 
Table 1. Interpretation of Cohen’s d
Example: Office Temperature
Let’s do another example to solidify our understanding. Let’s say that the office building you work in is supposed to be kept at 74 degree Fahrenheit but is allowed
to vary by 1 degree in either direction. You suspect that, as a cost saving measure, the temperature was secretly set higher. You set up a formal way to test your hypothesis.
You start by laying out the null hypothesis:
H 0 : There is no difference in the average building temperature H 0 : μ = 74
Next you state the alternative hypothesis. You have reason to suspect a specific direction of change, so you make a onetailed test:
H A : The average building temperature is higher than claimed H A : μ > 74
Now that you have everything set up, you spend one week collecting temperature data:
Day  Temp 
Monday  77 
Tuesday  76 
Wednesday  74 
Thursday  78 
Friday  78 
You calculate the average of these scores to be 𝑋̅ = 76.6 degrees. You use this to calculate the test statistic, using μ = 74 (the supposed average temperature), σ = 1.00 (how much the temperature should vary), and n = 5 (how many data points you collected):
z = 76.60 − 74.00 = 2.60 = 5.78
1.00/√5 0.45
This value falls so far into the tail that it cannot even be plotted on the distribution!
Figure 7: Obtained zstatistic
You compare your obtained zstatistic, z = 5.77, to the critical value, z* = 1.645, and find that z > z*. Therefore you reject the null hypothesis, concluding: Based on 5 observations, the average temperature (𝑋̅ = 76.6 degrees) is statistically significantly higher than it is supposed to be, z = 5.77, p < .05.
d = (76.6074.00)/ 1= 2.60
The effect size you calculate is definitely large, meaning someone has some explaining to do!
Example: Different Significance Level
First, let’s take a look at an example phrased in generic terms, rather than in the context of a specific research question, to see the individual pieces one more time. This time, however, we will use a stricter significance level, α = 0.01, to test the hypothesis.
We will use 60 as an arbitrary null hypothesis value: H 0 : The average score does not differ from the population H 0 : μ = 50
We will assume a twotailed test: H A : The average score does differ H A : μ ≠ 50
We have seen the critical values for ztests at α = 0.05 levels of significance several times. To find the values for α = 0.01, we will go to the standard normal table and find the zscore cutting of 0.005 (0.01 divided by 2 for a twotailed test) of the area in the tail, which is z crit * = ±2.575. Notice that this cutoff is much higher than it was for α = 0.05. This is because we need much less of the area in the tail, so we need to go very far out to find the cutoff. As a result, this will require a much larger effect or much larger sample size in order to reject the null hypothesis.
We can now calculate our test statistic. The average of 10 scores is M = 60.40 with a µ = 60. We will use σ = 10 as our known population standard deviation. From this information, we calculate our zstatistic as:
Our obtained zstatistic, z = 0.13, is very small. It is much less than our critical value of 2.575. Thus, this time, we fail to reject the null hypothesis. Our conclusion would look something like:
Notice two things about the end of the conclusion. First, we wrote that p is greater than instead of p is less than, like we did in the previous two examples. This is because we failed to reject the null hypothesis. We don’t know exactly what the p value is, but we know it must be larger than the α level we used to test our hypothesis. Second, we used 0.01 instead of the usual 0.05, because this time we tested at a different level. The number you compare to the pvalue should always be the significance level you test at. Because we did not detect a statistically significant effect, we do not need to calculate an effect size. Note: some statisticians will suggest to always calculate effects size as a possibility of Type II error. Although insignificant, calculating d = (60.460)/10 = .04 which suggests no effect (and not a possibility of Type II error).
Review Considerations in Hypothesis Testing
Errors in hypothesis testing.
Keep in mind that rejecting the null hypothesis is not an allornothing decision. The Type I error rate is affected by the α level: the lower the α level the lower the Type I error rate. It might seem that α is the probability of a Type I error. However, this is not correct. Instead, α is the probability of a Type I error given that the null hypothesis is true. If the null hypothesis is false, then it is impossible to make a Type I error. The second type of error that can be made in significance testing is failing to reject a false null hypothesis. This kind of error is called a Type II error.
Statistical Power
The statistical power of a research design is the probability of rejecting the null hypothesis given the sample size and expected relationship strength. Statistical power is the complement of the probability of committing a Type II error. Clearly, researchers should be interested in the power of their research designs if they want to avoid making Type II errors. In particular, they should make sure their research design has adequate power before collecting data. A common guideline is that a power of .80 is adequate. This means that there is an 80% chance of rejecting the null hypothesis for the expected relationship strength.
Given that statistical power depends primarily on relationship strength and sample size, there are essentially two steps you can take to increase statistical power: increase the strength of the relationship or increase the sample size. Increasing the strength of the relationship can sometimes be accomplished by using a stronger manipulation or by more carefully controlling extraneous variables to reduce the amount of noise in the data (e.g., by using a withinsubjects design rather than a betweensubjects design). The usual strategy, however, is to increase the sample size. For any expected relationship strength, there will always be some sample large enough to achieve adequate power.
Inferential statistics uses data from a sample of individuals to reach conclusions about the whole population. The degree to which our inferences are valid depends upon how we selected the sample (sampling technique) and the characteristics (parameters) of population data. Statistical analyses assume that sample(s) and population(s) meet certain conditions called statistical assumptions.
It is easy to check assumptions when using statistical software and it is important as a researcher to check for violations; if violations of statistical assumptions are not appropriately addressed then results may be interpreted incorrectly.
Learning Objectives
Having read the chapter, students should be able to:
 Conduct a hypothesis test using a zscore statistics, locating critical region, and make a statistical decision including.
 Explain the purpose of measuring effect size and power, and be able to compute Cohen’s d.
Exercises – Ch. 10
 List the main steps for hypothesis testing with the zstatistic. When and why do you calculate an effect size?
 z = 1.99, twotailed test at α = 0.05
 z = 1.99, twotailed test at α = 0.01
 z = 1.99, onetailed test at α = 0.05
 You are part of a trivia team and have tracked your team’s performance since you started playing, so you know that your scores are normally distributed with μ = 78 and σ = 12. Recently, a new person joined the team, and you think the scores have gotten better. Use hypothesis testing to see if the average score has improved based on the following 8 weeks’ worth of score data: 82, 74, 62, 68, 79, 94, 90, 81, 80.
 A study examines selfesteem and depression in teenagers. A sample of 25 teens with a low selfesteem are given the Beck Depression Inventory. The average score for the group is 20.9. For the general population, the average score is 18.3 with σ = 12. Use a twotail test with α = 0.05 to examine whether teenagers with low selfesteem show significant differences in depression.
 You get hired as a server at a local restaurant, and the manager tells you that servers’ tips are $42 on average but vary about $12 (μ = 42, σ = 12). You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don’t know if you will make more or less in tips. After working 16 shifts, you find that your average nightly amount is $44.50 from tips. Test for a difference between this value and the population mean at the α = 0.05 level of significance.
Answers to Odd Numbered Exercises – Ch. 10
1. List hypotheses. Determine critical region. Calculate z. Compare z to critical region. Draw Conclusion. We calculate an effect size when we find a statistically significant result to see if our result is practically meaningful or important
5. Step 1: H 0 : μ = 42 “My average tips does not differ from other servers”, H A : μ ≠ 42 “My average tips do differ from others”
Introduction to Statistics for Psychology Copyright © 2021 by Alisa Beyer is licensed under a Creative Commons AttributionNonCommercialShareAlike 4.0 International License , except where otherwise noted.
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What Is a ZTest?
Understanding ztests, the bottom line.
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ZTest: Definition, Uses in Statistics, and Example
James Chen, CMT is an expert trader, investment adviser, and global market strategist.
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A ztest is a statistical test used to determine whether two population means are different when the variances are known and the sample size is large. It can also be used to compare one mean to a hypothesized value.
The data must approximately fit a normal distribution , otherwise the test doesn't work. Parameters such as variance and standard deviation should be calculated for a ztest to be performed.
Key Takeaways
 A ztest is a statistical test to determine whether two population means are different or to compare one mean to a hypothesized value when the variances are known and the sample size is large.
 A ztest is a hypothesis test for data that follows a normal distribution.
 A zstatistic, or zscore, is a number representing the result from the ztest.
 Ztests are closely related to ttests, but ttests are best performed when an experiment has a small sample size.
 Ztests assume the standard deviation is known, while ttests assume it is unknown.
The ztest is also a hypothesis test in which the zstatistic follows a normal distribution. The ztest is best used for greaterthan30 samples because, under the central limit theorem , as the number of samples gets larger, the samples are considered to be approximately normally distributed.
When conducting a ztest, the null and alternative hypotheses, and alpha level should be stated. The zscore , also called a test statistic, should be calculated, and the results and conclusion stated. A zstatistic, or zscore, is a number representing how many standard deviations above or below the mean population a score derived from a ztest is.
Examples of tests that can be conducted as ztests include a onesample location test, a twosample location test, a paired difference test, and a maximum likelihood estimate. Ztests are closely related to ttests, but ttests are best performed when an experiment has a small sample size. Also, ttests assume the standard deviation is unknown, while ztests assume it is known. If the standard deviation of the population is unknown, the assumption of the sample variance equaling the population variance is made.
Formula for ZScore
The Zscore is calculated with the formula:
z = ( x  μ ) / σ
 z = Zscore
 x = the value being evaluated
 μ = the mean
 σ = the standard deviation
OneSample ZTest Example
Assume an investor wishes to test whether the average daily return of a stock is greater than 3%. A simple random sample of 50 returns is calculated and has an average of 2%. Assume the standard deviation of the returns is 2.5%. Therefore, the null hypothesis is when the average, or mean, is equal to 3%.
Conversely, the alternative hypothesis is whether the mean return is greater or less than 3%. Assume an alpha of 0.05% is selected with a twotailed test . Consequently, there is 0.025% of the samples in each tail, and the alpha has a critical value of 1.96 or 1.96. If the value of z is greater than 1.96 or less than 1.96, the null hypothesis is rejected.
The value for z is calculated by subtracting the value of the average daily return selected for the test, or 3% in this case, from the observed average of the samples. Next, divide the resulting value by the standard deviation divided by the square root of the number of observed values.
Therefore, the test statistic is:
(0.02  0.03) ÷ (0.025 ÷ √ 50) = 2.83
The investor rejects the null hypothesis since z is less than 1.96 and concludes that the average daily return is less than 3%.
What's the Difference Between a TTest and ZTest?
Ztests are closely related to ttests, but ttests are best performed when the data consists of a small sample size, i.e., less than 30. Also, ttests assume the standard deviation is unknown, while ztests assume it is known.
When Should You Use a ZTest?
If the standard deviation of the population is known and the sample size is greater than or equal to 30, the ztest can be used. Regardless of the sample size, if the population standard deviation is unknown, a ttest should be used instead.
What Is a ZScore?
A zscore, or zstatistic, is a number representing how many standard deviations above or below the mean population the score derived from a ztest is. Essentially, it is a numerical measurement that describes a value's relationship to the mean of a group of values. If a zscore is 0, it indicates that the data point's score is identical to the mean score. A zscore of 1.0 would indicate a value that is one standard deviation from the mean. Zscores may be positive or negative, with a positive value indicating the score is above the mean and a negative score indicating it is below the mean.
What Is Central Limit Theorem (CLT)?
In the study of probability theory, the central limit theorem (CLT) states that the distribution of sample approximates a normal distribution (also known as a “bell curve”) as the sample size becomes larger, assuming that all samples are identical in size, and regardless of the population distribution shape. Sample sizes equal to or greater than 30 are considered sufficient for the CLT to predict the characteristics of a population accurately. The ztest's fidelity relies on the CLT holding.
What Are the Assumptions of the ZTest?
For a ztest to be effective, the population must be normally distributed, and the samples must have the same variance. In addition, all data points should be independent of one another.
A ztest is used in hypothesis testing to evaluate whether a finding or association is statistically significant or not. In particular, it tests whether two means are the same (the null hypothesis). A ztest can only be used if the population standard deviation is known and the sample size is 30 data points or larger. Otherwise, a ttest should be employed.
Newcastle University. " ZTest ."
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ZTest: Formula, Examples, Uses, ZTest vs TTest
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Ztest Definition
ztest is a statistical tool used for the comparison or determination of the significance of several statistical measures, particularly the mean in a sample from a normally distributed population or between two independent samples.
 Like ttests, z tests are also based on normal probability distribution.
 Ztest is the most commonly used statistical tool in research methodology, with it being used for studies where the sample size is large (n>30).
 In the case of the ztest, the variance is usually known.
 Ztest is more convenient than ttest as the critical value at each significance level in the confidence interval is the sample for all sample sizes.
 A zscore is a number indicating how many standard deviations above or below the mean of the population is.
Ztest formula
For the normal population with one sample:
where x̄ is the mean of the sample, and µ is the assumed mean, σ is the standard deviation, and n is the number of observations.
ztest for the difference in mean:
where x̄ 1 and x̄ 2 are the means of two samples, σ is the standard deviation of the samples, and n1 and n2 are the numbers of observations of two samples.
One sample ztest (onetailed ztest)
 One sample ztest is used to determine whether a particular population parameter, which is mostly mean, significantly different from an assumed value.
 It helps to estimate the relationship between the mean of the sample and the assumed mean.
 In this case, the standard normal distribution is used to calculate the critical value of the test.
 If the zvalue of the sample being tested falls into the criteria for the onesided tets, the alternative hypothesis will be accepted instead of the null hypothesis.
 A onetailed test would be used when the study has to test whether the population parameter being tested is either lower than or higher than some hypothesized value.
 A onesample ztest assumes that data are a random sample collected from a normally distributed population that all have the same mean and same variance.
 This hypothesis implies that the data is continuous, and the distribution is symmetric.
 Based on the alternative hypothesis set for a study, a onesided ztest can be either a leftsided ztest or a rightsided ztest.
 For instance, if our H 0 : µ 0 = µ and H a : µ < µ 0 , such a test would be a onesided test or more precisely, a lefttailed test and there is one rejection area only on the left tail of the distribution.
 However, if H 0 : µ = µ 0 and H a : µ > µ 0 , this is also a onetailed test (right tail), and the rejection region is present on the right tail of the curve.
Two sample ztest (twotailed ztest)
 In the case of two sample ztest, two normally distributed independent samples are required.
 A twotailed ztest is performed to determine the relationship between the population parameters of the two samples.
 In the case of the twotailed ztest, the alternative hypothesis is accepted as long as the population parameter is not equal to the assumed value.
 The twotailed test is appropriate when we have H 0 : µ = µ 0 and H a : µ ≠ µ 0 which may mean µ > µ 0 or µ < µ 0
 Thus, in a twotailed test, there are two rejection regions, one on each tail of the curve.
Ztest examples
If a sample of 400 male workers has a mean height of 67.47 inches, is it reasonable to regard the sample as a sample from a large population with a mean height of 67.39 inches and a standard deviation of 1.30 inches at a 5% level of significance?
Taking the null hypothesis that the mean height of the population is equal to 67.39 inches, we can write:
H 0 : µ = 67 . 39 “
H a : µ ≠ 67 . 39 “
x̄ = 67 . 47 “, σ = 1 . 30 “, n = 400
Assuming the population to be normal, we can work out the test statistic z as under:
ztest applications
 Ztest is performed in studies where the sample size is larger, and the variance is known.
 It is also used to determine if there is a significant difference between the mean of two independent samples.
 The ztest can also be used to compare the population proportion to an assumed proportion or to determine the difference between the population proportion of two samples.
Ztest vs Ttest (8 major differences)



The ttest is a test in statistics that is used for testing hypotheses regarding the mean of a small sample taken population when the standard deviation of the population is not known.  ztest is a statistical tool used for the comparison or determination of the significance of several statistical measures, particularly the mean in a sample from a normally distributed population or between two independent samples.  
The ttest is usually performed in samples of a smaller size (n≤30).  ztest is generally performed in samples of a larger size (n>30).  
ttest is performed on samples distributed on the basis of tdistribution.  ztets is performed on samples that are normally distributed.  
A ttest is not based on the assumption that all key points on the sample are independent.  ztest is based on the assumption that all key points on the sample are independent.  
Variance or standard deviation is not known in the ttest.  Variance or standard deviation is known in ztest.  
The sample values are to be recorded or calculated by the researcher.  In a normal distribution, the average is considered 0 and the variance as 1.  
In addition, to the mean, the ttest can also be used to compare partial or simple correlations among two samples.  In addition, to mean, ztest can also be used to compare the population proportion.  
ttests are less convenient as they have separate critical values for different sample sizes.  ztest is more convenient as it has the same critical value for different sample sizes. 
References and Sources
 C.R. Kothari (1990) Research Methodology. Vishwa Prakasan. India.
 https://ncsswpengine.netdnassl.com/wpcontent/themes/ncss/pdf/Procedures/PASS/OneSample_ZTests.pdf
 https://www.wallstreetmojo.com/ztestvsttest/
 https://sites.google.com/site/fundamentalstatistics/chapter13
 3% – https://www.investopedia.com/terms/z/ztest.asp
 2% – https://www.coursehero.com/file/61052903/Questionsstatisticswpdf/
 2% – https://towardsdatascience.com/everythingyouneedtoknowabouthypothesistestingparti4de9abebbc8a
 2% – https://ncsswpengine.netdnassl.com/wpcontent/themes/ncss/pdf/Procedures/PASS/OneSample_ZTests.pdf
 1% – https://www.slideshare.net/MuhammadAnas96/ztestwithexamples
 1% – https://www.mathandstatistics.com/learnstats/hypothesistesting/twotailedztesthypothesistestbyhand
 1% – https://www.infrrr.com/proportions/differenceinproportionshypothesistestcalculator
 1% – https://keydifferences.com/differencebetweenttestandztest.html
 1% – https://en.wikipedia.org/wiki/Ztest
 1% – http://www.sci.utah.edu/~arpaiva/classes/UT_ece3530/hypothesis_testing.pdf
 <1% – https://www.researchgate.net/post/Cananullhypothesisbestatedasadifference
 <1% – https://www.isixsigma.com/toolstemplates/hypothesistesting/makingsensetwosamplettest/
 <1% – https://www.investopedia.com/terms/t/twotailedtest.asp
 <1% – https://www.academia.edu/24313503/BIOSTATISTICS_AND_RESEARCH_METHODS_IN_PHARMACY_Pharmacy_C479_4_quarter_credits_A_Course_for_Distance_Learning_Prepared
About Author
Anupama Sapkota
2 thoughts on “ZTest: Formula, Examples, Uses, ZTest vs TTest”
The formula for Z test provided for testing the single mean is wrong. The correct formula is wrong. Please check and correct it. It should be Z = (𝑥̅−𝜇)/𝜎/√n
Hi Ramnath, Sorry for the mistake. Thank you so much for the correction. We have updated the page with correct formula.
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One Sample ZTest: Definition, Formula, and Example
A one sample ztest is used to test whether the mean of a population is less than, greater than, or equal to some specific value.
This test assumes that the population standard deviation is known.
This tutorial explains the following:
 The formula to perform a one sample ztest.
 The assumptions of a one sample ztest.
 An example of how to perform a one sample ztest.
Let’s jump in!
One Sample ZTest: Formula
A one sample ztest will always use one of the following null and alternative hypotheses:
1. TwoTailed ZTest
 H 0 : μ = μ 0 (population mean is equal to some hypothesized value μ 0 )
 H A : μ ≠ μ 0 (population mean is not equal to some hypothesized value μ 0 )
2. LeftTailed ZTest
 H 0 : μ ≥ μ 0 (population mean is greater than or equal to some hypothesized value μ 0 )
 H A : μ 0 (population mean is less than some hypothesized value μ 0 )
3. RightTailed ZTest
 H 0 : μ ≤ μ 0 (population mean is less than or equal to some hypothesized value μ 0 )
 H A : μ > μ 0 (population mean is greaer than some hypothesized value μ 0 )
We use the following formula to calculate the z test statistic:
z = ( x – μ 0 ) / (σ/√ n )
 x : sample mean
 μ 0 : hypothesized population mean
 σ: population standard deviation
 n: sample size
If the pvalue that corresponds to the z test statistic is less than your chosen significance level (common choices are 0.10, 0.05, and 0.01) then you can reject the null hypothesis .
One Sample ZTest: Assumptions
For the results of a one sample ztest to be valid, the following assumptions should be met:
 The data are continuous (not discrete).
 The data is a simple random sample from the population of interest.
 The data in the population is approximately normally distributed .
 The population standard deviation is known.
One Sample ZTest : Example
Suppose the IQ in a population is normally distributed with a mean of μ = 100 and standard deviation of σ = 15.
A scientist wants to know if a new medication affects IQ levels, so she recruits 20 patients to use it for one month and records their IQ levels at the end of the month:
To test this, she will perform a one sample ztest at significance level α = 0.05 using the following steps:
Step 1: Gather the sample data.
Suppose she collects a simple random sample with the following information:
 n (sample size) = 20
 x (sample mean IQ) = 103.05
Step 2: Define the hypotheses.
She will perform the one sample ztest with the following hypotheses:
 H 0 : μ = 100
 H A : μ ≠ 100
Step 3: Calculate the z test statistic.
The z test statistic is calculated as:
 z = (x – μ) / (σ√ n )
 z = (103.05 – 100) / (15/√ 20 )
 z = 0.90933
Step 4: Calculate the pvalue of the z test statistic.
According to the Z Score to P Value Calculator , the twotailed pvalue associated with z = 0.90933 is 0.36318 .
Step 5: Draw a conclusion.
Since the pvalue (0.36318) is not less than the significance level (.05), the scientist will fail to reject the null hypothesis.
There is not sufficient evidence to say that the new medication significantly affects IQ level.
Note: You can also perform this entire one sample ztest by using the One Sample ZTest Calculator .
Additional Resources
The following tutorials explain how to perform a one sample ztest using different statistical software:
How to Perform ZTests in Excel How to Perform ZTests in R How to Perform ZTests in Python
Pandas: How to Create Pivot Table with Sum of Values
How to use str_pad in r (with examples), related posts, threeway anova: definition & example, two sample ztest: definition, formula, and example, how to find a confidence interval for a..., an introduction to the exponential distribution, an introduction to the uniform distribution, the breuschpagan test: definition & example, population vs. sample: what’s the difference, introduction to multiple linear regression, dunn’s test for multiple comparisons, how to use dunnett’s test for multiple comparisons.
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Ztests for Hypothesis testing: Formula & Examples
Ztests are statistical hypothesis testing techniques that are used to determine whether the null hypothesis relating to comparing sample means or proportions with that of population at a given significance level can be rejected or otherwise based on the zstatistics or zscore. As a data scientist , you must get a good understanding of the ztests and its applications to test the hypothesis for your statistical models. In this blog post, we will discuss an overview of different types of ztests and related concepts with the help of examples. You may want to check my post on hypothesis testing titled – Hypothesis testing explained with examples
Table of Contents
What are Ztests & Zstatistics?
Ztests can be defined as statistical hypothesis testing techniques that are used to quantify the hypothesis testing related to claim made about the population parameters such as mean and proportion. Ztest uses the sample data to test the hypothesis about the population parameters (mean or proportion). There are different types of Ztests which are used to estimate the population mean or proportion, or, perform hypotheses testing related to samples’ means or proportions.
Different types of Ztests
There are following different types of Ztests which are used to perform different types of hypothesis testing.
 Onesample Ztest for means
 Twosample Ztest for means
 One sample Ztest for proportion
 Two sample Ztest for proportions
Four variables are involved in the Ztest for performing hypothesis testing for different scenarios. They are as follows:
 An independent variable that is called the “sample” and assumed to be normally distributed;
 A dependent variable that is known as the test statistic (Z) and calculated based on sample data
 Different types of Ztest that can be used for performing hypothesis testing
 A significance level or “alpha” is usually set at 0.05 but can take the values such as 0.01, 0.05, 0.1
When to use Ztest – Explained with examples
The following are different scenarios when Ztest can be used:
 Compare the sample or a single group with that of the population with respect to the parameter, mean. This is called as onesample Ztest for means. For example, whether the student of a particular school has been scoring marks in Mathematics which is statistically significant than the other schools. This can also be thought of as a hypothesis test to check whether the sample belongs to the population or otherwise.
 Compare two groups with respect to the population parameter, mean. This is called as twosamples Ztest for means. For example, you want to compare class X students from different schools and determine if students of one school are better than others based on their score of Mathematics.
 Compare hypothesized proportion of the population to that of population theoritical proportion. For example, whether the unemployment rate of a given state is different than the wellestablished rate for the ccountry
 Compare the proportion of one population with the proportion of othe rproportion. For example, whether the efficacy rate of vaccination in two different population are statistically significant or otherwise.
Ztest Interview Questions
Here is a list of a few interview questions you may expect in your data scientists interview:
 What is Ztest?
 What is Zstatistics or Zscore?
 When to use Ztest vs other tests such as Ttest or Chisquare test?
 What is Zdistribution?
 What is the difference between Zdistribution and Tdistribution?
 What is sampling distribution?
 What are different types of Ztests?
 Explain different types of Ztests with the help of realworld examples?
 What’s the difference two samples Ztest for means and twosamples Ztest for proportions? Explain with one example each.
 As data scientists, give some scenarios when you would like to use Ztest when building machine learning models?
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Z Table. Z Score Table. Normal Distribution Table. Standard Normal Table.
What is ZTest?
ZTest is a statistical test which let’s us approximate the distribution of the test statistic under the null hypothesis using normal distribution .
ZTest is a test statistic commonly used in hypothesis test when the sample data is large.For carrying out the ZTest, population parameters such as mean, variance, and standard deviation should be known.
This test is widely used to determine whether the mean of the two samples are different when the variance is known. We make use of the Z score and the Z table for running the ZTest.
ZTest as Hypothesis Test
A test statistic is a random variable that we calculate from the sample data to determine whether to reject the null hypothesis. This random variable is used to calculate the Pvalue, which indicates how strong the evidence is against the null hypothesis. ZTest is such a test statistic where we make use of the mean value and z score to determine the Pvalue. ZTest compares the mean of two large samples taken from a population when the variance is known.
ZTest is usually used to conduct a hypothesis test when the sample size is greater than 30. This is because of the central limit theorem where when the sample gets larger, the distributed data graph starts resembling a bell curve and is considered to be distributed normally. Since the ZTest follows normal distribution under the null hypothesis, it is the most suitable test statistic for large sample data.
Why do we use a large sample for conducting a hypothesis test?
In a hypothesis test, we are trying to reject a null hypothesis with the evidence that we should collect from sample data which represents only a portion of the population. When the population has a large size, and the sample data is small, we will not be able to draw an accurate conclusion from the test to prove our null hypothesis is false. As sample data provide us a door to the entire population, it should be large enough for us to arrive at a significant inference. Hence a sufficiently large data should be considered for a hypothesis test especially if the population is huge.
How to Run a ZTest
ZTest can be considered as a test statistic for a hypothesis test to calculate the Pvalue. However, there are certain conditions that should be satisfied by the sample to run the ZTest.
The conditions are as follows:
 The sample size should be greater than 30.
This is already mentioned above. The size of the sample is an important factor in ZTesting as the ZTest follows a normal distribution and so should the data. If the same size is less than 30, it is recommended to use a ttest instead
 All the data point should be independent and doesn’t affect each other.
Each element in the sample, when considered single should be independent and shouldn’t have a relationship with another element.
 The data must be distributed normally.
This is ensured if the sample data is large.
 The sample should be selected randomly from a population.
Each data in the population should have an equal chance to be selected as one of the sample data.
 The sizes of the selected samples should be equal if at all possible.
When considering multiple sample data, ensuring that the size of each sample is the same for an accurate calculation of population parameters.
 The standard deviation of the population is known.
The population parameter, standard deviation must be given to run a ZTest as we cannot perform the calculation without knowing it. If it is not directly given, then it assumed that the variance of the sample data is equal to the variance of the entire population.
If the conditions are satisfied, the ZTest can be successfully implemented.
Following are steps to run the ZTest:
 State the null hypothesis
The null hypothesis is a statement of no effect and it supports the data which is already given. It is generally represented as :
 State the alternate hypothesis
The statement that we are trying to prove is the alternate hypothesis. It is represented as:
This is the representation of a bidirectional alternate hypothesis.
 H 1 :µ > k
This is the representation of a onedirectional alternate hypothesis that is represented in the right region of the graph.
 H 1 :µ < k
This is the representation of a onedirectional alternate hypothesis that is represented in the left region of the graph.
 Choose an alpha level for the test.
Alpha level or significant level is the probability of rejecting the null hypothesis when it is true. It is represented by ( α ). An alpha level must be chosen wisely so as to avoid the Type I and Type II errors.
If we choose a large alpha value such as 10%, it is likely to reject a null hypothesis when it is true. There is a probability of 10% for us to reject the null hypothesis. This is an error known as the Type I error.
On the other hand, if we choose an alpha level as low as 1%, there is a chance to accept the null hypothesis even if it is false. That is we reject the alternate hypothesis to favor the null hypothesis. This is the Type II error.
Hence the alpha level should be chosen in such a way that the chance of making Type I or Type II error is minimal. For this reason, the alpha level is commonly selected as 5% which is proven best to avoid errors.
 Determining the critical value of Z from the Z table.
The critical value is the point in the normal distribution graph that splits the graph into two regions: the acceptance region and the rejection regions. It can be also described as the extreme value for which a null hypothesis can be accepted. This critical value of Z can be found from the Z table .
 Calculate the test statistic.
The sample data that we choose to test is converted into a single value. This is known as the test statistic. This value is compared to the null value. If the test statistic significantly differs from the null value, the null value is rejected.
 Comparing the test statistic with the critical value.
Now, we have to determine whether the test statistic we have calculated comes under the acceptance region or the rejection region. For this, the test statistic is compared with the critical value to know whether we should accept or reject a null hypothesis.
Types of ZTest
ZTest can be used to run a hypothesis test for a single sample or to compare the mean of two samples. There are two common types of ZTest
OneSample ZTest
This is the most basic type of hypothesis test that is widely used. For running an onesample ZTest, all we need to know is the mean and standard deviation of the population. We consider only a single sample for a onesample ZTest. Onesample ZTest is used to test whether the population parameter is different from the hypothesized value i.e whether the mean of the population is equal to, less than or greater than the hypothesized value.
The equation for finding the value of Z is:
The following are the assumptions that are generally taken for a onesampled ZTest:
 The sample size is equal to or greater than 30.
 One normally distributed sample is considered with the standard deviation known.
 The null hypothesis is that the population mean that is calculated from the sample is equal to the hypothetically determined population mean.
TwoSample ZTest
A twosample ZTest is used whenever there is a comparison between two independent samples. It is used to check whether the difference between the means is equal to zero or not. Suppose if we want to know whether men or women prefer to drive more in a city, we use a twosample ZTest as it is the comparison of two independent samples of men and women.
 x 1 and x 2 represent the mean of the two samples.
 µ 1 and µ 2 are the hypothesized mean values.
 σ 1 and σ 2 are the standard deviations.
 n 1 and n 2 are the sizes of the samples.
The following are the assumptions that are generally taken for a twosample ZTest:
 Two independent, normally distributed samples are considered for the ZTest with the standard deviation known.
 Each sample is equal to or greater than 30.
 The null hypothesis is stated that the population mean of the two samples taken does not differ.
Critical value
A critical value is a line that splits a normally distributed graph into two different sections. Namely the ‘Rejection region’ and ‘Acceptance region’. If your test value falls in the ‘Rejection region’, then the null hypothesis is rejected and if your test value falls in the ‘Accepted region’, then the null hypothesis is accepted.
Critical Value Vs Significant Value
Significant level, alpha is the probability of rejecting a null hypothesis when it is actually true. While the critical value is the extreme value up to which a null hypothesis is true. There migh come a confusion regarding both of these parameters.
Critical value is a value that lies in critical region. It is in fact the boundary value of the rejection region. Also, it is the value up to which the null hypothesis is true. Hence the critical value is considered to be the point at which the null hypothesis is true or is rejected.
Critical value gives a point of extremity whose probability is indicated by the significant level. Significant level is preselected for a hypothesis test and critical value is calculated from this Alpha value. Critical value is a point represented as Z score and Significant level is a probability.
ZTest Vs TTest
ZTest are used when the sample size exceeds 30. As ZTest follows normal distribution, large sample size can be taken for the ZTest. ZTest indicates the distance of a data point from the mean of the data set in terms of standard deviation. Also. this test can only be used if the standard deviation of the data set is known.
TTest is based on T distribution in which the mean value is known and the variance could be calculated from the sample. TTest is most preferred to know the difference between the statistical parameters of two samples as the standard deviation of the samples are not usually given in a twosample test for running the ZTest. Also, if the sample size is less than 30, TTest is preferred.
Approximate Hypothesis Tests: the z Test and the t Test
This chapter presents two common tests of the hypothesis that a population mean equals a particular value and of the hypothesis that two population means are equal: the z test and the t test. These tests are approximate : They are based on approximations to the probability distribution of the test statistic when the null hypothesis is true, so their significance levels are not exactly what they claim to be. If the sample size is reasonably large and the population from which the sample is drawn has a nearly normal distribution —a notion defined in this chapter—the nominal significance levels of the tests are close to their actual significance levels. If these conditions are not met, the significance levels of the approximate tests can differ substantially from their nominal values. The z test is based on the normal approximation ; the t test is based on Student's t curve, which approximates some probability histograms better than the normal curve does. The chapter also presents the deep connection between hypothesis tests and confidence intervals, and shows how to compute approximate confidence intervals for the population mean of nearly normal populations using Student's t curve.
where \(\phi\) is the pooled sample percentage of the two samples. The estimate of \(SE(\phi^{tc})\) under the null hypothesis is
\[ se = s^*\times(1/n_t + 1/n_c)^{1/2}, \]
where \(n_t\) and \(n_c\) are the sizes of the two samples. If the null hypothesis is true, the Z statistic,
\[ Z=\phi^{tc}/se, \]
is the original test statistic \(\phi^{tc}\) in approximately standard units , and Z has a probability histogram that is approximated well by the normal curve , which allowed us to select the rejection region for the approximate test.
This strategy—transforming a test statistic approximately to standard units under the assumption that the null hypothesisis true, and then using the normal approximation to determine the rejection region for the test—works to construct approximate hypothesis tests in many other situations, too. The resulting hypothesis test is called a z test. Suppose that we are testing a null hypothesis using a test statistic \(X\) , and the following conditions hold:
 We have a probability model for how the observations arise, assuming the null hypothesis is true. Typically, the model is that under the null hypothesis, the data are like random draws with or without replacement from a box of numbered tickets.
 Under the null hypothesis, the test statistic \(X\) , converted to standard units, has a probability histogram that can be approximated well by the normal curve.
 Under the null hypothesis, we can find the expected value of the test statistic, \(E(X)\) .
 Under the null hypothesis, either we can find the SE of the test statistic, \(SE(X)\) , or we can estimate \(SE(X)\) accurately enough to ignore the error of the estimate of the SE. Let se denote either the exact SE of \(X\) under the null hypothesis, or the estimated value of \(SE(X)\) under the null hypothesis.
Then, under the null hypothesis, the probability histogram of the Z statistic
\[ Z = (XE(X))/se \]
is approximated well by the normal curve, and we can use the normal approximation to select the rejection region for the test using \(Z\) as the test statistic. If the null hypothesis is true,
\[ P(Z < z_a) \approx a \]
\[ P(Z > z_{1a} ) \approx a, \]
\[ P(Z > z_{1a/2} ) \approx a. \]
These three approximations yield three different z tests of the hypothesis that \(\mu = \mu_0\) at approximate significance level \(a\) :
 Reject the null hypothesis whenever \(Z (lefttail z test)
 Reject the null hypothesis whenever \(Z > z_{1a}\) (righttail z test)
 Reject the null hypothesis whenever \(Z> z_{1a/2}\) (twotail z test)
The word "tail" refers to the tails of the normal curve: In a lefttail test, the probability of a Type I error is approximately the area of the left tail of the normal curve, from minus infinity to \(z_a\) . In a righttail test, the probability of a Type I error is approximately the area of the right tail of the normal curve, from \(z_{1a}\) to infinity. In a twotail test, the probability of a Type I error is approximately the sum of the areas of both tails of the normal curve, the left tail from minus infinity to \(z_{a/2}\) and the right tail from \(z_{1a/2}\) to infinity. All three of these tests are called z tests. The observed value of Z is called the z score .
Which of these three tests, if any, should one use? The answer depends on the probability distribution of Z when the alternative hypothesis is true. As a rule of thumb, if, under the alternative hypothesis, \(E(Z) , use the lefttail test. If, under the alternative hypothesis, \(E(Z) > 0\) , use the righttail test. If, under the alternative hypothesis, it is possible that \(E(Z) and it is possible that \(E(Z) > 0\) , use the twotail test. If, under the alternative hypothesis, \(E(Z) = 0\) , consult a statistician. Generally (but not always), this rule of thumb selects the test with the most power for a given significance level.
P values for z tests
Each of the three z tests gives us a family of procedures for testing the null hypothesis at any (approximate) significance level \(a\) between 0 and 100%—we just use the appropriate quantile of the normal curve. This makes it particularly easy to find the P value for a z test. Recall that the P value is the smallest significance level for which we would reject the null hypothesis, among a family of tests of the null hypothesis at different significance levels.
Suppose the z score (the observed value of \(Z\) ) is \(x\) . In a lefttail test, the P value is the area under the normal curve to the left of \(x\) : Had we chosen the significance level \(a\) so that \(z_a=x\) , we would have rejected the null hypothesis, but we would not have rejected it for any smaller value of \(a\) , because for all smaller values of \(a\) , \(z_a . Similarly, for a righttail z test, the P value is the area under the normal curve to the right of \(x\) : If \(x=z_{1a}\) we would reject the null hypothesis at approximate significance level \(a\) , but not at smaller significance levels. For a twotail z test, the P value is the sum of the area under the normal curve to the left of \(x\) and the area under the normal curve to the right of \(x\) .
Finding P values and specifying the rejection region for the z test involves the probability distribution of \(Z\) under the assumption that the null hypothesis is true. Rarely is the alternative hypothesis sufficiently detailed to specify the probability distribution of \(Z\) completely, but often the alternative does help us choose intelligently among lefttail, righttail, and twotail z tests. This is perhaps the most important issue in deciding which hypothesis to take as the null hypothesis and which as the alternative: We calculate the significance level under the null hypothesis, and that calculation must be tractable.
However, to construct a z test, we need to know the expected value and SE of the test statistic under the null hypothesis. Usually it is easy to determine the expected value, but often the SE must be estimated from the data. Later in this chapter we shall see what to do if the SE cannot be estimated accurately, but the shape of the distribution of the numbers in the population is known. The next section develops z tests for the population percentage and mean, and for the difference between two population means.
Examples of z tests
The central limit theorem assures us that the probability histogram of the sample mean of random draws with replacement from a box of tickets—transformed to standard units—can be approximated increasingly well by a normal curve as the number of draws increases. In the previous section, we learned that the probability histogram of a sum or difference of independent sample means of draws with replacement also can be approximated increasingly well by a normal curve as the two sample sizes increase. We shall use these facts to derive z tests for population means and percentages and differences of population means and percentages.
z Test for a Population Percentage
Suppose we have a population of \(N\) units of which \(G\) are labeled "1" and the rest are labeled "0." Let \(p = G/N\) be the population percentage. Consider testing the null hypothesis that \(p = p_0\) against the alternative hypothesis that \(p \ne p_0\) , using a random sample of \(n\) units drawn with replacement. (We could assume instead that \(N >> n\) and allow the draws to be without replacement.)
Under the null hypothesis, the sample percentage
\[ \phi = \frac{\mbox{# tickets labeled "1" in the sample}}{n} \]
has expected value \(E(\phi) = p_0\) and standard error
\[ SE(\phi) = \sqrt{\frac{p_0 \times (1  p_0)}{n}}. \]
Let \(Z\) be \(\phi\) transformed to standard units :
\[ Z = (\phi  p_0)/SE(\phi). \]
Provided \(n\) is large and \(p_0\) is not too close to zero or 100% (say \(n \times p > 30\) and \(n \times (1p) > 30)\) , the probability histogram of \(Z\) will be approximated reasonably well by the normal curve, and we can use it as the Z statistic in a z test. For example, if we reject the null hypothesis when \(Z > 1.96\) , the significance level of the test will be about 95%.
z Test for a Population Mean
The approach in the previous subsection applies, mutatis mutandis , to testing the hypothesis that the population mean equals a given value, even when the population contains numbers other than just 0 and 1. However, in contrast to the hypothesis that the population percentage equals a given value, the null hypothesis that a more general population mean equals a given value does not specify the SD of the population, which poses difficulties that are surmountable (by approximation and estimation) if the sample size is large enough. (There are also nonparametric methods that can be used.)
Consider testing the null hypothesis that the population mean \(\mu\) is equal to a specific null value \(\mu_0\) , against the alternative hypothesis that \(\mu , on the basis of a random sample with replacement of size \(n\) . Recall that the sample mean \(M\) of \(n\) random draws with or without replacement from a box of numbered tickets is an unbiased estimator of the population mean \(\mu\) : If
\[ M = \frac{\mbox{sum of sample values}}{n}, \]
\[ E(M) = \mu = \frac{\mbox{sum of population values}}{N}, \]
where \(N\) is the size of the population. The population mean determines the expected value of the sample mean. The SE of the sample mean of a random sample with replacement is
\[ \frac{SD(\mbox{box})}{\sqrt{n}}, \]
where SD(box) is the SD of the list of all the numbers in the box, and \(n\) is the sample size. As a special case, the sample percentage \phi of \(n\) independent random draws from a 01 box is an unbiased estimator of the population percentage p , with SE equal to
\[ \sqrt{\frac{p\times(1p)}{n}}. \]
In testing the null hypothesis that a population percentage \(p\) equals \(p_0\) , the null hypothesis specifies not only the expected value of the sample percentage \phi, it automatically specifies the SE of the sample percentage as well, because the SD of the values in a 01 box is determined by the population percentage \(p\) :
\[ SD(box) = \sqrt{p\times(1p)}. \]
The null hypothesis thus gives us all the information we need to standardize the sample percentage under the null hypothesis. In contrast, the SD of the values in a box of tickets labeled with arbitrary numbers bears no particular relation to the mean of the values, so the null hypothesis that the population mean \(\mu\) of a box of tickets labeled with arbitrary numbers equals a specific value \(\mu_0\) determines the expected value of the sample mean, but not the standard error of the sample mean. To standardize the sample mean to construct a z test for the value of a population mean, we need to estimate the SE of the sample mean under the null hypothesis. When the sample size is large, the sample standard deviation s> is likely to be close to the SD of the population, and
\[ se=\frac{s}{\sqrt{n}} \]
is likely to be an accurate estimate of \(SE(M)\) . The central limit theorem tells us that when the sample size \(n\) is large, the probability histogram of the sample mean, converted to standard units, is approximated well by the normal curve. Under the null hypothesis,
\[ E(M) = \mu_0, \]
and thus when \(n\) is large
\[ Z = \frac{M\mu_0}{s/\sqrt{n}} \]
has expected value zero, and its probability histogram is approximated well by the normal curve, so we can use \(Z\) as the Z statistic in a z test. If the alternative hypothesis is true, the expected value of \(Z\) could be either greater than zero or less than zero, so it is appropriate to use a twotail z test. If the alternative hypothesis is \(\mu > \mu_0\) , then under the alternative hypothesis, the expected value of \(Z\) is greater than zero, and it is appropriate to use a righttail z test. If the alternative hypothesis is \(\mu , then under the alternative hypothesis, the expected value of \(Z\) is less than zero, and it is appropriate to use a lefttail z test.
z Test for a Difference of Population Means
Consider the problem of testing the hypothesis that two population means are equal, using random samples from the two populations. Different sampling designs lead to different hypothesis testing procedures. In this section, we consider two kinds of random samples from the two populations: paired samples and independent samples , and construct z tests appropriate for each.
Paired Samples
Consider a population of \(N\) individuals, each of whom is labeled with two numbers. For example, the \(N\) individuals might be a group of doctors, and the two numbers that label each doctor might be the annual payments to the doctor by an HMO under the terms of the current contract and under the terms of a proposed revision of the contract. Let the two numbers associated with individual \(i\) be \(c_i\) and \(t_i\) . (Think of \(c\) as control and \(t\) as treatment . In this example, control is the current contract, and treatment is the proposed contract.) Let \(\mu_c\) be the population mean of the \(N\) values
\[ \{c_1, c_2, \ldots, c_N \}, \]
and let \(\mu_t\) be the population mean of the \(N\) values
\[ \{t_1, t_2, \ldots, t_N\}. \]
Suppose we want to test the null hypothesis that
\[ \mu = \mu_t  \mu_c = \mu_0 \]
against the alternative hypothesis that \(\mu . With \(\mu_0=\$0\) , this null hypothesis is that the average annual payment to doctors under the proposed revision would be the same as the average payment under the current contract, and the alternative is that on average doctors would be paid less under the new contract than under the current contract. With \(\mu_0=\$5,000\) , this null hypothesis is that the proposed contract would save the HMO an average of $5,000 per doctor, compared with the current contract; the alternative is that under the proposed contract, the HMO would save even more than that. With \(\mu_0=\$1,000\) , this null hypothesis is that doctors would be paid an average of $1,000 more per year under the new contract than under the old one; the alternative hypothesis is that on average doctors would be paid less than an additional $1,000 per year under the new contract—perhaps even less than they are paid under the current contract. For the remainder of this example, we shall take \(\mu_0=\$1,000\) .
The data on which we shall base the test are observations of both \(c_i\) and \(t_i\) for a sample of \(n\) individuals chosen at random with replacement from the population of \(N\) individuals (or a simple random sample of size \(n ): We select \(n\) doctors at random from the \(N\) doctors under contract to the HMO, record the current annual payments to them, and calculate what the payments to them would be under the terms of the new contract. This is called a paired sample , because the samples from the population of control values and from the population of treatment values come in pairs: one value for control and one for treatment for each individual in the sample. Testing the hypothesis that the difference between two population means is equal to \(\mu_0\) using a paired sample is just the problem of testing the hypothesis that the population mean \(\mu\) of the set of differences
\[ d_i = t_i  c_i, \;\; i= 1, 2, \ldots, N, \]
is equal to \(\mu_0\) . Denote the \(n\) (random) observed values of \(c_i\) and \(t_i\) by \(\{C_1, C_2, \ldots, C_n\}\) and \(\{T_1, T_2, \ldots, T_n \}\) , respectively. The sample mean \(M\) of the differences between the observed values of \(t_i\) and \(c_i\) is the difference of the two sample means:
\[ M = \frac{(T_1C_1)+(T_2C_2) + \cdots + (T_nC_n)}{n} = \frac{T_1+T_2+ \cdots + T_n}{n}  \frac{C_1+C_2+ \cdots + C_n}{n} \]
\[ = (\mbox{sample mean of observed values of } t_i)  (\mbox{sample mean of observed values of } c_i). \]
\(M\) is an unbiased estimator of \(\mu\) , and if n is large, the normal approximation to its probability histogram will be accurate. The SE of \(M\) is the population standard deviation of the \(N\) values \(\{d_1, d_2, \ldots, d_N\}\) , which we shall denote \(SD_d\) , divided by the square root of the sample size, \(n^{1/2}\) . Let \(sd\) denote the sample standard deviation of the \(n\) observed differences \((T_i  C_i), \;\; i=1, 2, \ldots, n\) :
\[ sd = \sqrt{\frac{(T_1C_1M)^2 + (T_2C_2M)^2 + \cdots + (T_nC_nM)^2}{n1}} \]
(recall that \(M\) is the sample mean of the observed differences). If the sample size \(n\) is large, sd is very likely to be close to SD( d ), and so, under the null hypothesis,
\[ Z = \frac{M\mu_0}{sd/n^{1/2}} \]
has expected value zero, and when \(n\) is large the probability histogram of \(Z\) can be approximated well by the normal curve. Thus we can use \(Z\) as the Z statistic in a z test of the null hypothesis that \(\mu=\mu_0\) . Under the alternative hypothesis that \(\mu (doctors on the average are paid less than an additional $1,000 per year under the new contract), the expected value of \(Z\) is less than zero, so we should use a lefttail z test. Under the alternative hypothesis \(\mu\ne\mu_0\) (on average, the difference in average annual payments to doctors is not an increase of $1,000, but some other number instead), the expected value of \(Z\) could be positive or negative, so we would use a twotail z test. Under the alternative hypothesis that \(\mu>\mu_0\) (on average, under the new contract, doctors are paid more than an additional $1,000 per year), the expected value of \(Z\) would be greater than zero, so we should use a righttail z test.
Independent Samples
Consider two separate populations of numbers, with population means \(\mu_t\) and \(\mu_c\) , respectively. Let \(\mu=\mu_t\mu_c\) be the difference between the two population means. We would like to test the null hypothesis that \(\mu=\mu_0\) against the alternative hypothesis that \(\mu>0\) . For example, let \(\mu_t\) be the average annual payment by an HMO to doctors in the Los Angeles area, and let \(\mu_c\) be the average annual payment by the same HMO to doctors in the San Francisco area. Then the null hypothesis with \(\mu_0=0\) is that the HMO pays doctors in the two regions the same amount annually, on average; the alternative hypothesis is that the average annual payment by the HMO to doctors differs between the two areas. Suppose we draw a random sample of size \(n_t\) with replacement from the first population, and independently draw a random sample of size \(n_c\) with replacement from the second population. Let \(M_t\) and \(M_c\) be the sample means of the two samples, respectively, and let
\[ M = M_t  M_c \]
be the difference between the two sample means. Because the expected value of \(M_t\) is \(\mu_t\) and the expected value of \(M_c\) is \(\mu_c\) , the expected value of \(M\) is
\[ E(M) = E(M_t  M_c) = E(M_t)  E(M_c) = \mu_t  \mu_c = \mu. \]
Because the two random samples are independent , \(M_t\) and \(M_c\) are independent random variables, and the SE of their sum is
\[ SE(M) = (SE^2(M_t) + SE^2(M_c))^{1/2}. \]
Let \(s_t\) and \(s_c\) be the sample standard deviations of the two samples, respectively. If \(n_t\) and \(n_c\) are both very large, the two sample standard deviations are likely to be close to the standard deviations of the corresponding populations, and so \(s_t/n_t^{1/2}\) is likely to be close to \(SE(M_t)\) , and \(s_c/n_c^{1/2}\) is likely to be close to \(SE(M_c)\) . Therefore, the pooled estimate of the standard error
\[ se_\mbox{diff} = ( (s_t/n_t^{1/2})^2 + (s_c/n_c^{1/2})^2)^{1/2} = \sqrt{ s_t^2/n_t + s_c^2/n_c} \]
is likely to be close to \(SE(M)\) . Under the null hypothesis, the statistic
\[ Z = \frac{M  \mu_0}{se_\mbox{diff}} = \frac{M_1  M_2  \mu_0}{\sqrt{ s_t^2/n_t + s_c^2/n_c}} \]
has expected value zero and its probability histogram is approximated well by the normal curve, so we can use it as the Z statistic in a z test.
Under the alternative hypothesis
\[ \mu = \mu_t  \mu_c > \mu_0, \]
the expected value of \(Z\) is greater than zero, so it is appropriate to use a righttail z test.
If the alternative hypothesis were \(\mu \ne \mu_0\) , under the alternative the expected value of \(Z\) could be greater than zero or less than zero, so it would be appropriate to use a twotail z test. If the alternative hypothesis were \(\mu , under the alternative the expected value of \(Z\) would be less than zero, so it would be appropriate to use a lefttail z test.
The following exercises check that you can compute the z test for a population mean or a difference of population means. The exercises are dynamic: the data will tend to change when you reload the page.
For the nominal significance level of the z test for a population mean to be approximately correct, the sample size typically must be large. When the sample size is small, two factors limit the accuracy of the z test: the normal approximation to the probability distribution of the sample mean can be poor, and the sample standard deviation can be an inaccurate estimate of the population standard deviation, so se is not an accurate estimate of the SE of the test statistic Z . For nearly normal populations , defined in the next subsection, the probability distribution of the sample mean is nearly normal even when the sample size is small, and the uncertainty of the sample standard deviation as an estimate of the population standard deviation can be accounted for by using a curve that is broader than the normal curve to approximate the probability distribution of the (approximately) standardized test statistic. The broader curve is Student's t curve . Student's t curve depends on the sample size: The smaller the sample size, the more spread out the curve.
Nearly Normally Distributed Populations
A list of numbers is nearly normally distributed if the fraction of values in any range is close to the area under the normal curve for the corresponding range of standard units—that is, if the list has mean \(\mu\) and standard deviation SD, and for every pair of values \(a < b\) ,
\[ \mbox{ the fraction of numbers in the list between } a \mbox{ and } b \approx \mbox{the area under the normal curve between } (a  \mu)/SD \mbox{ and } (b  \mu)/SD. \]
A list is nearly normally distributed if the normal curve is a good approximation to the histogram of the list transformed to standard units. The histogram of a list that is approximately normally distributed is (nearly) symmetric about some point, and is (nearly) bellshaped.
No finite population can be exactly normally distributed, because the area under the normal curve between every two distinct values is strictly positive—no matter how large or small the values nor how close together they are. No population that contains only a finite number of distinct values can be exactly normally distributed, for the same reason. In particular, populations that contain only zeros and ones are not approximately normally distributed, so results for the sample mean of samples drawn from nearly normally distributed populations need not apply to the sample percentage of samples drawn from 01 boxes. Such results will be more accurate for the sample percentage when the population percentage is close to 50% than when the population percentage is close to 0% or 100%, because then the histogram of population values is more nearly symmetric.
Suppose a population is nearly normally distributed. Then a histogram of the population is approximately symmetric about the mean of the population. The fraction of numbers in the population within ±1 SD of the mean of the population is about 68%, the fraction of numbers within ±2 SD of the mean of the population is about 95%, and the fraction of numbers in the population within ±3 SD of the mean of the population is about 99.7%.
The following exercises check that you understand what it means for a list to be nearly normally distributed. The exercises are dynamic: the data tend to change when you reload the page.
Student's t curve
Student's t curve is similar to the normal curve, but broader. It is positive, has a single maximum, and is symmetric about zero. The total area under Student's t curve is 100%. Student's t curve approximates some probability histograms more accurately than the normal curve does. There are actually infinitely many Student t curves, one for each positive integer value of the degrees of freedom. As the degrees of freedom increases, the difference between Student's t curve and the normal curve decreases.
Consider a population of \(N\) units labeled with numbers. Let \(\mu\) denote the population mean of the \(N\) numbers, and let SD denote the population standard deviation of the \(N\) numbers. Let \(M\) denote the sample mean of a random sample of size \(n\) drawn with replacement from a population, and let s> denote the sample standard deviation of the sample. The expected value of \(M\) is \(\mu\) , and the SE of \(M\) is \(SD/n^{1/2}\) . Let
\[ Z = (M  \mu)/(SD/n^{1/2}). \]
Then the expected value of \(Z\) is zero, the SE of \(Z\) is 1, and if \(n\) is large enough, the normal curve is a good approximation to the probability histogram of \(Z\) . The closer to normal the distribution of values in the population is, the smaller \(n\) needs to be for the normal curve to be a good approximation to the distribution of \(Z\) . Consider the statistic
\[ T = \frac{M  \mu}{s/n^{1/2}}, \]
which replaces SD by its estimated value (the sample standard deviation \(s\) ). If \(n\) is large enough, \(s\) is very likely to be close to SD, so \(T\) will be close to \(Z\) ; the normal curve will be a good approximation to the probability histogram of \(T\) ; and we can use \(T\) as the Z statistic in a z test of hypotheses about \(\mu\) .
For many populations, when the sample size is small—say less than 25, but the accuracy depends on the population—the normal curve is not a good approximation to the probability histogram of \(T\) . For nearly normally distributed populations, when the sample size is intermediate—say 25–100, but again this depends on the population—the normal curve is a good approximation to the probability histogram of \(Z\) , but not to the probability histogram of \(T\) , because of the variability of the sample standard deviation s> from sample to sample, which tends to broaden the probability distribution of \(T\) (i.e., to make \(SE(T)>1\) ).
When you first load this page, the degrees of freedom will be set to 25, and the region from 1.96 to 1.96 will be hilighted. The area under the normal curve between ±1.96 is 95%, but for Student's t curve with 25 degrees of freedom, the area is about 93.9%: Student's t curve with d.f.=25 is broader than the normal curve. Increase the degrees of freedom to 200; you will see that the Student t curve gets slightly narrower, and the area under the curve between ±1.96 is about 94.9%.
We define quantiles of Student t curves in the same way we defined quantiles of the normal curve: For any number a between 0 and 100%, the a quantile of Student's t curve with \(d.f.=d\) , \(t_{d,a}\) , is the unique value such that the area under the Student t curve with d degrees of freedom from minus infinity to \(t_{d,a}\) is equal to \(a\) . For example, \(t_{d,0.5} = 0\) for all values of \(d\) . Generally, the value of \(t_{d,a}\) depends on the degrees of freedom \(d\) . The probability calculator allows you to find quantiles of Student's t curve.
t test for the Mean of a Nearly Normally Distributed Population
We can use Student's t curve to construct approximate tests of hypotheses about the population mean \(\mu\) when the population standard deviation is unknown, for intermediate values of the sample size \(n\) . The approach is directly analogous to the z test, but instead of using a quantile of the normal curve, we use the corresponding quantile of Student's t curve (with the appropriate number of degrees of freedom). However, for the test to be accurate when \(n\) is small or intermediate, the distribution of values in the population must be nearly normal for the test to have approximately its nominal level. This is a somewhat bizarre restriction: It may require a very large sample to detect that the population is not nearly normal—but if the sample is very large, we can use the z test instead of the t test, so we don't need to rely as much on the assumption. It is my opinion that the t test is overtaught and overused—because its assumptions are not verifiable in the situations where it is potentially useful.
Consider testing the null hypothesis that \(\mu=\mu_0\) using the sample mean \(M\) and sample standard deviation s> of a random sample of size \(n\) drawn with replacement from a population that is known to have a nearly normal distribution. Define
\[ T = \frac{M  \mu_0}{s/n^{1/2}}. \]
Under the null hypothesis, if \(n\) is not too small, Student's t curve with \(n1\) degrees of freedom will be an accurate approximation to the probability histogram of \(T\) , so
\[ P(T < t_{n1,a}), \]
\[ P(T > t_{n1,1a}), \]
\[ P(T > t_{n1,1a/2}) \]
all are approximately equal to \(a\) . As we saw earlier in this chapter for the Z statistic, these three approximations give three tests of the null hypothesis \(\mu=\mu_0\) at approximate significance level \(a\) —a lefttail t test, a righttail t test, and a twotail t test:
 Reject the null hypothesis if \(T (lefttail)
 Reject the null hypothesis if \(T > t_{n1,1a}\) (righttail)
 Reject the null hypothesis if \(T > t_{n1,1a/2}\) (twotail)
To decide which t test to use, we can apply the same rule of thumb we used for the z test:
 Use a lefttail t test if, under the alternative hypothesis, the expected value of \(T\) is less than zero.
 Use a righttail t test if, under the alternative hypothesis, the expected value of \(T\) is greater than zero.
 Use a twotail t test if, under the alternative hypothesis, the expected value of \(T\) is not zero, but could be less than or greater than zero.
 Consult a statistician for a more appropriate test if, under the alternative hypothesis, the expected value of \(T\) is zero.
Pvalues for t tests are computed in much the same way as Pvalues for z tests. Let t be the observed value of \(T\) (the t score). In a lefttail t test, the Pvalue is the area under Student's t curve with \(n1\) degrees of freedom, from minus infinity to \(t\) . In a righttail t test, the Pvalue is the area under Student's t curve with \(n1\) degrees of freedom, from \(t\) to infinity. In a twotail t test, the Pvalue is the total area under Student's t curve with \(n1\) degrees of freedom between minus infinity and \(t\) and between \(t\) and infinity.
There are versions of the t test for comparing two means, as well. Just like for the z test, the method depends on how the samples from the two populations are drawn. For example, if the two samples are paired (if we are sampling individuals labeled with two numbers and for each individual in the sample, we observe both numbers), we may base the t test on the sample mean of the paired differences and the sample standard deviation of the paired differences. Let \(\mu_1\) and \(\mu_2\) be the means of the two populations, and let
\[ \mu = \mu_1  \mu_2. \]
The \(T\) statistic to test the null hypothesis that \(\mu=\mu_0\) is
\[ T = \frac{(\mbox{sample mean of differences})  \mu_0 }{(\mbox{sample standard deviation of differences})/n^{1/2}}, \]
and the appropriate curve to use to find the rejection region for the test is Student's t curve with \(n1\) degrees of freedom, where \(n\) is the number of individuals (differences) in the sample.
Twosample t tests for a difference of means using independent samples depend on additional assumptions, such as equality of the two population standard deviations; we shall not present such tests here. The following exercises check your ability to compute t tests. The exercises are dynamic: the data tend to change when you reload the page.
Hypothesis Tests and Confidence Intervals
There is a deep connection between hypothesis tests about parameters, and confidence intervals for parameters. If we have a procedure for constructing a level \(100\% \times (1a)\) confidence interval for a parameter \(\mu\) , then the following rule is a twosided significance level \(a\) test of the null hypothesis that \(\mu = \mu_0\) :
reject the null hypothesis if the confidence interval does not contain \(\mu_0\).
Similarly, suppose we have an hypothesistesting procedure that lets us test the null hypothesis that \(\mu=\mu_0\) for any value of \(\mu_0\) , at significance level \(a\) . Define
\(A\) = (all values of \(\mu_0\) for which we would not reject the null hypothesis that \(\mu = \mu_0\)).
Then \(A\) is a \(100\% \times (1a)\) confidence set for \(\mu\) :
\[ P( A \mbox{ contains the true value of } \mu ) = 100\% \times (1a). \]
(A confidence set is a generalization of the idea of a confidence interval: a \(1a\) confidence set for the parameter \(\mu\) is a random set that has probability \(1a\) of containing \(\mu\) . As is the case with confidence intervals, the probability makes sense only before collecting the data.) The set \(A\) might or might not be an interval, depending on the nature of the test. If one starts with a twotail z test or twotail t test, one ends up with a confidence interval rather than a more general confidence set.
Confidence Intervals Using Student's t curve
The t test lets us test the hypothesis that the population mean \(\mu\) is equal to \(\mu_0\) at approximate significance level a using a random sample with replacement of size n from a population with a nearly normal distribution. If the sample size n is small, the actual significance level is likely to differ considerably from the nominal significance level. Consider a twosided t test of the hypothesis \(\mu=\mu_0\) at significance level \(a\) . If the sample mean is \(M\) and the sample standard deviation is \(s\) , we would not reject the null hypothesis at significance level \(a\) if
\[ \frac{M\mu_0}{s/n^{1/2}} \le t_{n1,1a/2}. \]
We rearrange this inequality:
\[ t_{n1,1a/2} \le \frac{M\mu_0}{s/n^{1/2}} \le t_{n1,1a/2} \]
\[ t_{n1,1a/2} \times s/n^{1/2} \le M  \mu_0 \le t_{n1,1a/2} \times s/n^{1/2} \]
\[ M  t_{n1,1a/2} \times s/n^{1/2} \le  \mu_0 \le M + t_{n1,1a/2} \times s/n^{1/2} \]
\[ M + t_{n1,1a/2} \times s/n^{1/2} \le \mu_0 \le M  t_{n1,1a/2} \times s/n^{1/2} \]
That is, we would not reject the hypothesis \(\mu = \mu_0\) provided \(\mu_0\) is in the interval
\[ [M  t_{n1,1a/2} \times s/n^{1/2}, M + t_{n1,1a/2} \times s/n^{1/2}]. \]
Therefore, that interval is a \(100\%a\) confidence interval for \(\mu\) :
\[ P([M  t_{n1,1a/2} \times s/n^{1/2}, M + t_{n1,1a/2} \times s/n^{1/2}] \mbox{ contains } \mu) \approx 1a. \]
The following exercise checks that you can use Student's t curve to construct a confidence interval for a population mean. The exercise is dynamic: the data tend to change when you reload the page.
In hypothesis testing, a Z statistic is a random variable whose probability histogram is approximated well by the normal curve if the null hypothesis is correct: If the null hypothesis is true, the expected value of a Z statistic is zero, the SE of a Z statistic is approximately 1, and the probability that a Z statistic is between \(a\) and \(b\) is approximately the area under the normal curve between \(a\) and \(b\) . Suppose that the random variable \(Z\) is a Z statistic. If, under the alternative hypothesis, \(E(Z) , the appropriate z test to test the null hypothesis at approximate significance level \(a\) is the lefttailed z test: Reject the null hypothesis if \(Z , where \(z_a\) is the \(a\) quantile of the normal curve. If, under the alternative hypothesis, \(E(Z)>0\) , the appropriate z test to test the null hypothesis at approximate significance level \(a\) is the righttailed z test: Reject the null hypothesis if \(Z>z_{1a}\) . If, under the alternative hypothesis, \(E(Z)\ne 0 \) but could be greater than 0 or less than 0, the appropriate z test to test the null hypothesis at approximate significance level \(a\) is the twotailed z test: reject the null hypothesis if \(Z>z_{1a/2}\) . If, under the alternative hypothesis, \(E(Z)=0\) , a z test probably is not appropriate—consult a statistician. The exact significance levels of these tests differ from \(a\) by an amount that depends on how closely the normal curve approximates the probability histogram of \(Z\) .
Z statistics often are constructed from other statistics by transforming approximately to standard units, which requires knowing the expected value and SE of the original statistic on the assumption that the null hypothesis is true. Let \(X\) be a test statistic; let \(E(X)\) be the expected value of \(X\) if the null hypothesis is true, and let \(se\) be approximately equal to the SE of \(X\) if the null hypothesis is true. If \(X\) is a sample sum of a large random sample with replacement, a sample mean of a large random sample with replacement, or a sum or difference of independent sample means of large samples with replacement,
\[ Z = \frac{XE(X)}{se} \]
is a Z statistic.
Consider testing the null hypothesis that a population percentage \(p\) is equal to the value \(p_0\) on the basis of the sample percentage \phi of a random sample of size \(n\) with replacement. Under the null hypothesis, \(E(\phi)=p_0\) and
\[ SE(\phi) = \sqrt{\frac{p_0\times(1p_0)}{n}}, \]
and if \(n\) is sufficiently large (say \(n \times p > 30\) and \(n \times (1p)>30\) , but this depends on the desired accuracy), the normal approximation to
\[ Z = \frac{\phip_0}{\sqrt{(p_0 \times (1p_0))/n}} \]
will be reasonably accurate, so \(Z\) can be used as the Z statistic in a z test of the null hypothesis \(p=p_0\) .
Consider testing the null hypothesis that a population mean \(\mu\) is equal to the value \(\mu_0\) , on the basis of the sample mean \(M\) of a random sample of size \(n\) with replacement. Let \(s\) denote the sample standard deviation. Under the null hypothesis, \(E(M)=\mu_0\) , and if \(n\) is large,
\[ SE(M)=SD/n^{1/2} \approx s/n^{1/2}, \]
and the normal approximation to
\[ Z = \frac{M\mu_0}{s/n^{1/2}} \]
will be reasonably accurate, so \(Z\) can be used as the Z statistic in a z test of the null hypothesis \(\mu=\mu_0\) .
Consider a population of \(N\) individuals, each labeled with two numbers. The \(i\) th individual is labeled with the numbers \(c_i\) and \(t_i\) , \(i=1, 2, \ldots, N\) . Let \(\mu_c\) be the population mean of the \(N\) values \(\{c_1, \ldots, c_N\}\) and let \(\mu_t\) be the population mean of the \(N\) values \(\{t_1, \ldots, t_N \}\) . Let \(\mu=\mu_t\mu_c\) be the difference between the two population means. Consider testing the null hypothesis that \(\mu=\mu_0\) on the basis of a paired random sample of size \(n\) with replacement from the population: that is, a random sample of size \(n\) is drawn with replacement from the population, and for each individual \(i\) in the sample, \(c_i\) and \(t_i\) are observed. This is equivalent to testing the hypothesis that the population mean of the \(N\) values \(\{(t_1c_1), \ldots, (t_Nc_N)\}\) is equal to \(\mu_0\) , on the basis of the random sample of size \(n\) drawn with replacement from those \(N\) values. Let \(M_t\) be the sample mean of the \(n\) observed values of \(t_i\) and let \(M_c\) be the sample mean of the \(n\) observed values of \(c_i\) . Let \(sd\) denote the sample standard deviation of the \(n\) observed differences \(\{(t_ic_i)\}\) . Under the null hypothesis, the expected value of \(M_tM_c\) is \(\mu_0\) , and if \(n\) is large,
\[ SE(M_tM_c) \approx sd/n^{1/2}, \]
and the normal approximation to the probability histogram of
\[ Z = \frac{M_tM_c\mu_0}{sd/n^{1/2}} \]
will be reasonably accurate, so \(Z\) can be used as the Z statistic in a z test of the null hypothesis that \(\mu_t\mu_c=\mu_0\) .
Consider testing the hypothesis that the difference ( \(\mu_t\mu_c\) ) between two population means, \(\mu_c\) and \(\mu_t\) , is equal to \(\mu_0\) , on the basis of the difference ( \(M_tM_c\) ) between the sample mean \(M_c\) of a random sample of size \(n_c\) with replacement from the first population and the sample mean \(M_t\) of an independent random sample of size \(n_t\) with replacement from the second population. Let \(s_c\) denote the sample standard deviation of the sample of size \(n_c\) from the first population and let \(s_t\) denote the sample standard deviation of the sample of size \(n_t\) from the second population. If the null hypothesis is true,
\[ E(M_tM_c)=\mu_0, \]
and if \(n_c\) and \(n_t\) are both large,
\[ SE(M_tM_c) \approx \sqrt{s_t^2/n_t + s_c^2/n_c} \]
\[ Z = \frac{M_tM_c\mu_0}{\sqrt{s_t^2/n_t + s_c^2/n_c}} \]
A list of numbers is nearly normally distributed if the fraction of numbers between any pair of values, \(a , is approximately equal to the area under the normal curve between \((a\mu)/SD\) and \((b\mu)/SD\) , where \(\mu\) is the mean of the list and SD is the standard deviation of the list.
Student's t curve with \(d\) degrees of freedom is symmetric about 0, has a single bump centered at 0, and is broader and flatter than the normal curve. The total area under Student's t curve is 1, no matter what \(d\) is; as \(d\) increases, Student's t curve gets narrower, its peak gets higher, and it becomes closer and closer to the normal curve.
Let \(M\) be the sample mean of a random sample of size \(n\) with replacement from a population with mean \(\mu\) and a nearly normal distribution, and let \(s\) be the sample standard deviation of the random sample. For moderate values of \(n\) ( \(n or so), Student's t curve approximates the probability histogram of \((M\mu)/(s/n^{1/2})\) better than the normal curve does, which can lead to an approximate hypothesis test about \(\mu\) that is more accurate than the z test.
Consider testing the null hypothesis that the mean \(\mu\) of a population with a nearly normal distribution is equal to \(\mu_0\) from a random sample of size \(n\) with replacement. Let
\[ T=\frac{M\mu_0}{s/n^{1/2}}, \]
where \(M\) is the sample mean and \(s\) is the sample standard deviation. The tests that reject the null hypothesis if \(T (lefttail t test), if \(T>t_{n1,1a}\) (righttail t test), or if \(T>t_{n1,1a/2}\) (twotail t test) all have approximate significance level \(a\) . How close the nominal significance level \(a\) is to the true significance level depends on the distribution of the numbers in the population, the sample size \(n\) , and \(a\) . The same rule of thumb for selecting whether to use a left, right, or twotailed z test (or not to use a z test at all) works to select whether to use a left, right, or twotailed t test: If, under the alternative hypothesis, \(E(T) , use a lefttail test. If, under the alternative hypothesis, \(E(T) > 0 \) , use a righttail test. If, under the alternative hypothesis, \(E(T)\) could be less than zero or greater than zero, use a twotail test. If, under the alternative hypothesis, \(E(T) = 0 \) , consult an expert. Because the t test differs from the z test only when the sample size is small, and from a small sample it is not possible to tell whether the population has a nearly normal distribution, the t test should be used with caution.
A \(1a\) confidence set for a parameter \(\mu\) is like a \(1a\) confidence interval for a parameter \(\mu\) : It is a random set of values that has probability \(1a\) of containing the true value of \(\mu\) . The difference is that the set need not be an interval.
There is a deep duality between hypothesis tests about a parameter \(\mu\) and confidence sets for \(\mu\) . Given a procedure for constructing a \(1a\) confidence set for \(\mu\) , the rule reject the null hypothesis that \(\mu=\mu_0\) if the confidence set does not contain \(\mu\) is a significance level \(a\) test of the null hypothesis that \(\mu=\mu_0\) . Conversely, given a family of significance level \(a\) hypothesis tests that allow one to test the hypothesis that \(\mu=\mu_0\) for any value of \(\mu_0\) , the set of all values \(\mu_0\) for which the test does not reject the null hypothesis that \(\mu=\mu_0\) is a \(1a\) confidence set for \(\mu\) .
 alternative hypothesis
 central limit theorem
 confidence interval
 confidence set
 expected value
 independent
 independent random variable
 mutatis mutandis
 nearly normal distribution
 normal approximation
 normal curve
 null hypothesis
 pooled bootstrap estimate of the population SD
 pooled bootstrap estimate of the SE
 population mean
 population percentage
 population standard deviation
 probability
 probability distribution
 probability histogram
 random sample
 random variable
 rejection region
 sample mean
 sample percentage
 sample size
 sample standard deviation
 significance level
 simple random sample
 standard deviation (SD)
 standard error (SE)
 standard unit
 Student's t curve
 test statistic
 twotailed test
 Type I error
 Z statistic
Six Sigma Study Guide
Study notes and guides for Six Sigma certification tests
One Sample Z Hypothesis Test
Posted by Ted Hessing
One sample Z hypothesis test is one of the basic tests of inferential statistics. One sample Z test is a parametric procedure for hypothesis testing. It tests whether the sample mean is significantly different (greater than, less than or not equal) than a population mean when the population’s standard deviation is known. Since we use standard normal distribution to compute the critical values, we often call it a one sample Z test.
When to use one sample Z test
One sample Z test is a robust hypothesis test for violations of standard normal distribution . Z test is similar to student ttest, z test is basically used for relatively large samples (say n>30) and the population standard deviation is known. Whereas, student ttest is for a small sample size, and also ttest assumes the population standard deviation is unknown.
Assumptions of one sample Z hypothesis test
 Population data is continuous
 Population follows a standard normal distribution
 The mean and standard deviation of the population is known
 Samples are independent of each other
 The sample should be randomly selected from the population
One sample Ztest Formula
 x̅ = sample mean
 µ= population mean
 σ= population standard deviation
 n= sample size
Steps to Calculate One Sample Z hypothesis test
 Select appropriate statistic onetailed or twotailed?
 Determine the null hypothesis and alternative hypothesis
 Determine the level of significance
 Find the critical value
 Calculate the test statistics
 x̅ is observed sample mean
 μ is population mean
 σ is population standard deviation
 n is sample size
 Then make a decision, reject the null hypothesis; If the test statistic falls in the critical region.
 Finally, interpret the decision in the context of the original claim.
Hypothesis Testing
A tailed hypothesis is an assumption about a population parameter. The assumption may or may not be true. Onetailed hypothesis is a test of hypothesis where the area of rejection is only in one direction. Whereas twotailed, the area of rejection is in two directions. The selection of one or twotailed tests depends upon the problem.
One Sample z test mostly performed in Analyze phase of DMAIC to check the sample mean is significantly different than a population mean when the population’s standard deviation is known.
Example of righttailed test
Example: A coaching institute claims that the students’ mean scores in their institute are greater than the 82 marks with a standard deviation of 20. A sample of 81 students is selected, and the mean score is 90 marks. At 95% confidence level, is there enough evidence to support the claim?
 Null hypothesis H 0 : µ ≤ 82
 Alternative hypothesis H 1 : µ>82
 Select appropriate statistic Since the claim is student marks are greater than 82 marks; it is a righttailed test.
 Level of significance: α= 0.05
 Find the critical value: 1α= 10.05=0.95
 Look at the 0.95 in z table= 1.645
Interpret the results : Compare Z calc to Z critical . In hypothesis testing, a critical value is a point on the test distribution compares to the test statistic to determine whether to reject the null hypothesis. Since zcal value is greater than z critical value and it is in the rejection region. Hence we can reject the null hypothesis.
There is enough evidence to support the students’ scores in their institute is greater than the 82 marks.
One Sample Z Excel Right Tail template
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Related posts: Null Hypothesis: Definition, Rejecting & Examples and Understanding Significance Levels. TwoSample Z Test Hypotheses. Null hypothesis (H 0): Two population means are equal (µ 1 = µ 2).; Alternative hypothesis (H A): Two population means are not equal (µ 1 ≠ µ 2).; Again, when the pvalue is less than or equal to your significance level, reject the null hypothesis.
Assuming a 5% significance level, perform a twosample ztest to determine if there is a significant difference between the online and offline classes. Solution: Step 1: Null & Alternate Hypothesis ... The z test is a commonly used hypothesis test in inferential statistics that allows us to compare two populations using the mean values of ...
A one sample ztest is used to test whether the mean of a population is less than, greater than, ... 0.05, and 0.01) then you can reject the null hypothesis. One Sample ZTest: Assumptions. For the results of a one sample ztest to be valid, the following assumptions should be met: The data are continuous (not discrete).
The Ztest January 9, 2021 Contents Example 1: (one tailed ztest) Example 2: (two tailed ztest) Questions Answers The ztest is a hypothesis test to determine if a single observed mean is signi cantly di erent (or greater or less than) the mean under the null hypothesis, hypwhen you know the standard deviation of the population.
The z test formula to set up the required hypothesis tests for a one sample and a twosample z test are given below. OneSample Z Test. A onesample z test is used to check if there is a difference between the sample mean and the population mean when the population standard deviation is known. The formula for the z test statistic is given as ...
Critical Values: Test statistic values beyond which we will reject the null hypothesis (cutoffs) p levels (α): Probabilities used to determine the critical value 5. Calculate test statistic (e.g., z statistic) 6. Make a decision Statistically Significant: Instructs us to reject the null hypothesis because the pattern in the data differs from
A Ztest is a type of statistical hypothesis test where the teststatistic follows a normal distribution. The name Ztest comes from the Zscore of the normal distribution. This is a measure of how many standard deviations away a raw score or sample statistics is from the populations' mean. Ztests are the most common statistical tests ...
The zscore associated with a 5% alpha level / 2 is 1.96.. Step 5: Compare the calculated zscore from Step 3 with the table zscore from Step 4. If the calculated zscore is larger, you can reject the null hypothesis. 8.99 > 1.96, so we can reject the null hypothesis.. Example 2: Suppose that in a survey of 700 women and 700 men, 35% of women and 30% of men indicated that they support a ...
In this chapter, we'll introduce hypothesis testing with examples from a 'ztest', when we're comparing a single mean to what we'd expect from a population with known mean and standard deviation. In this case, we can convert our observed mean into a zscore for the standard normal distribution. Hence the name ztest.
Chapter 10: Hypothesis Testing with Z. This chapter lays out the basic logic and process of hypothesis testing using a z. We will perform a test statistics using z, we use the z formula from chapter 8 and data from a sample mean to make an inference about a population.
The null hypothesis is retained when the sample mean is associated with a high probability of occurrence. Such probability of occurrence is better known as pvalue. ... Great, now that we know what hypothesis testing is when to apply the ztest, and the orientations of the hypotheses according to the alternative hypothesis, it's time to see a ...
A ztest is a hypothesis test for data that follows a normal distribution. A zstatistic, or zscore, is a number representing the result from the ztest. Ztests are closely related to ttests ...
Similarly, a large test statistic is an evidence that the population parameter is larger than assumed in the null hypothesis. For example, in case when a \(z\)score is used as the test statistic, we know that the average is zero, so negative test statistic means "small" and positive test statistic means "large".
Ztest is the most commonly used statistical tool in research methodology, with it being used for studies where the sample size is large (n>30). In the case of the ztest, the variance is usually known. Ztest is more convenient than ttest as the critical value at each significance level in the confidence interval is the sample for all sample ...
A one sample ztest is used to test whether the mean of a population is less than, greater than, ... 0.05, and 0.01) then you can reject the null hypothesis. One Sample ZTest: Assumptions. For the results of a one sample ztest to be valid, the following assumptions should be met: The data are continuous (not discrete).
When we do so, the zscore for our sample mean is. zX¯ = X¯−μ0 SE(X¯) z X ¯ = X ¯ − μ 0 S E (X ¯) or, equivalently. zX¯ = X¯−μ0 σ/ N√ z X ¯ = X ¯ − μ 0 σ / N. This zscore is our test statistic. The nice thing about using this as our test statistic is that like all zscores, it has a standard normal distribution:
Ztests are statistical hypothesis testing techniques that are used to determine whether the null hypothesis relating to comparing sample means or proportions with that of population at a given significance level can be rejected or otherwise based on the zstatistics or zscore. As a data scientist, you must get a good understanding of the z ...
If the pvalue that corresponds to the z test statistic is less than your chosen significance level (common choices are 0.10, 0.05, and 0.01) then you can reject the null hypothesis. Two Sample ZTest: Assumptions. For the results of a two sample ztest to be valid, the following assumptions should be met:
ZTest can be used to run a hypothesis test for a single sample or to compare the mean of two samples. There are two common types of ZTest. OneSample ZTest. This is the most basic type of hypothesis test that is widely used. For running an onesample ZTest, all we need to know is the mean and standard deviation of the population.
Approximate Hypothesis Tests: the z Test and the t Test . This chapter presents two common tests of the hypothesis that a population mean equals a particular value and of the hypothesis that two population means are equal: the z test and the t test. These tests are approximate: They are based on approximations to the probability distribution of the test statistic when the null hypothesis is ...
When we do so, the z score for our sample mean is. zX¯ = X¯−μ0 SE(X¯) z X ¯ = X ¯ − μ 0 S E (X ¯) or, equivalently. zX¯ = X¯−μ0 σ/ N√ z X ¯ = X ¯ − μ 0 σ / N. This z score is our test statistic. The nice thing about using this as our test statistic is that like all z scores, it has a standard normal distribution:
Part I. Start Here. This module will cover the basics of an inferential statistics technique called hypothesis testing. It is also known by its full name, null hypothesis significance testing, or NHST. This module introduces the onesample zTest to demonstrate hypothesis testing concepts. Along the way, we will show you how you can use Python ...
One sample Z test is a robust hypothesis test for violations of standard normal distribution. Z test is similar to student ttest, z test is basically used for relatively large samples (say n>30) and the population standard deviation is known. Whereas, student ttest is for a small sample size, and also ttest assumes the population standard ...
Any number of statistical tests may be used to calculate the value of the test statistic. For example, a onesample ttest may be used to evaluate the difference between the sample mean and the population mean (Chapter 8.5) or the independent sample ttest may be used to evaluate the difference between means of the control group and the ...