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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 1-sample analysis to determine whether a population mean is different from a hypothesized value. Or use the 2-sample 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 T-Test Overview: How to Use & Examples and How T-Tests 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.

One-Sample 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 p-value 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

Two-Sample 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 p-value 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 two-sided analyses. You can use one-sided, directional hypotheses instead. Learn more in my post, One-Tailed and Two-Tailed 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:

• One-Sample : 20 or more observations.
• Two-Sample : At least 15 in each group.

Related posts : Central Limit Theorem and Skewed Distributions

Independent samples

For the two-sample 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 p-values.

The correct formula depends on whether you’re performing a one- or two-sample 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 Z-Scores 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 Z-scores in the context of the standard normal distribution. Unlike the t-distribution , 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 two-tailed.

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 two-tailed 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 Z-score 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 p-value. 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 p-value that corresponds to a Z-score from a two-tailed analysis, we need to find the negative value of our Z-score (even when it’s positive) and double it.

In the truncated Z-table below, I highlight the cell corresponding to a Z-score of -2.33.

The cell value of 0.00990 represents the area or probability to the left of the Z-score -2.33. We need to double it to include the area > +2.33 to obtain the p-value for a two-tailed analysis.

P-value = 0.00990 * 2 = 0.0198

That p-value is an approximation because it uses a Z-score of 2.33 rather than 2.333. Using an online calculator, the p-value for our Z test is a more precise 0.0196. This p-value is less than our significance level of 0.05, which reconfirms the statistically significant results.

See my full Z-table , which explains how to use it to solve other types of problems.

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One Sample Z-Test: Definition, Formula, and Example

A  one sample z-test 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 z-test.
• The assumptions of a one sample z-test.
• An example of how to perform a one sample z-test.

Let’s jump in!

One Sample Z-Test: Formula

A one sample z-test will always use one of the following null and alternative hypotheses:

1. Two-Tailed Z-Test

• 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. Left-Tailed Z-Test

• 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. Right-Tailed Z-Test

• 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 p-value 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 Z-Test: Assumptions

For the results of a one sample z-test 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 Z-Test : 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 z-test 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 z-test 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 p-value of the z test statistic.

According to the Z Score to P Value Calculator , the two-tailed p-value associated with z = 0.90933 is 0.36318 .

Step 5: Draw a conclusion.

Since the p-value (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 z-test by using the One Sample Z-Test Calculator .

Additional Resources

The following tutorials explain how to perform a one sample z-test using different statistical software:

How to Perform Z-Tests in Excel How to Perform Z-Tests in R How to Perform Z-Tests in Python

Pandas: How to Create Pivot Table with Sum of Values

How to use str_pad in r (with examples), related posts, three-way anova: definition & example, two sample z-test: 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 breusch-pagan test: definition & example, population vs. sample: what’s the difference, introduction to multiple linear regression, dunn’s test for multiple comparisons, qualitative vs. quantitative variables: what’s the difference.

One Sample Z Test: How to Run One

Hypothesis Testing > One Sample Z Test

Before reading this, you may find it helpful to review:

• What is a Normal distribution?
• What is a Z-score?

How to Run a One Sample Z Test

A one sample z test is one of the most basic types of hypothesis test . In order to run a one sample z test, you work through several steps:

Step 1: State the Null Hypothesis . This is one of the common stumbling blocks–in order to make sense of your sample and have the one sample z test give you the right information you must make sure you’ve written the null hypothesis and alternate hypothesis correctly. For example, you might be asked to test the hypothesis that the mean weight gain of pregnant women was more than 30 pounds. Your null hypothesis would be: H 0 : μ = 30 and your alternate hypothesis would be H,sub>1: μ > 30.

Example: 1,500 women followed the Atkin’s diet for a month. A random sample of 29 women gained an average of 6.7 pounds. Test the hypothesis that the average weight gain per woman for the month was over 5 pounds. The standard deviation for all women in the group was 7.1. Z = 6.7 – 5 / (7.1/√29) = 1.289.

Step 3: Look up your z score in the z-table . The sample score above gives you an area of 0.8997. This area is your probability up to that point (i.e. the area to the left of your z-score). For this one sample z test, you want the area in the right tail, so subtract from 1: 1 – 0.8997 = 0.1003.

If you have difficulty figuring out if you have a left- or-right tailed test, see: Left Tailed Test or Right Tailed Test? How to Decide in Easy Steps.

Next : several specific examples of one sample z tests (with answers): Hypothesis Testing Examples

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 left-tailed, right-tailed, and two-tailed 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.

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 two-sample z test are given below.

One-Sample Z Test

A one-sample 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 two-sample z test can be set up in the same way as the one-sample 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 = \(\frac{p-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}\). 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 =\(\frac{p_{1}-p_{2}-0}{\sqrt{p(1-p)\left ( \frac{1}{n_{1}} +\frac{1}{n_{2}}\right )}}\)

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 right-tailed two-sample 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 T-Test

Both z test and t-test are univariate tests used on the means of two datasets. The differences between both tests are outlined in the table given below:

Related Articles:

• Probability and Statistics
• Data Handling
• Summary Statistics

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 one-sample 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 right-tailed one-sample 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 left-tailed one-sample 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 two-tailed 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 two-tailed 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 =\(\frac{p_{1}-p_{2}-0}{\sqrt{p(1-p)\left ( \frac{1}{n_{1}} +\frac{1}{n_{2}}\right )}}\) = 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 One-Sample Z Test?

A one-sample 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 Two-Sample 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 \(\frac{p-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}\).

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 \(\frac{p_{1}-p_{2}-0}{\sqrt{p(1-p)\left ( \frac{1}{n_{1}} +\frac{1}{n_{2}}\right )}}\).

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 T-Test?

A z test is used on large samples n ≥ 30 and normally distributed data while a t-test 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.

Z-Test for Statistical Hypothesis Testing Explained

The Z-test 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 Z-test 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 Z-tests, 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 Z-Test?

A Z-test is a type of statistical hypothesis test where the test-statistic follows a normal distribution.  

The name Z-test comes from the Z-score 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.

Z-tests 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 Z-Test

In order to conduct a Z-test, 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?

Z-Test Steps

There are four steps to complete a Z-test. Let’s examine each one.

4 Steps to a Z-Test

• State the null hypothesis.
• State the alternate hypothesis.
• Choose your critical value.
• Calculate your Z-test statistics. 

1. State the Null Hypothesis

The first step in a Z-test 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 Z-test 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 Z-Test Statistic

Compute the Z-test 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

Z-Test Example

Let’s go through an example to fully understand the one-sample mean Z-test.

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 right-sided, one-tailed test as our sample mean is greater than the population’s mean. So, choosing a critical value of 5 percent, which equals a Z-score of 1.96 , we can only reject the null hypothesis if our Z-test statistic is greater than 1.96.

If the school claimed its students’ IQs were an average of 90, then we would use a left-tailed test, as shown in the figure above. We would then only reject the null hypothesis if our Z-test statistic is less than -1.96.

Computing our Z-test 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 Z-tests. In this post, we only addressed the most simple case, the one-sample 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

One-sample Z-test: Hypothesis Testing, Effect Size, and Power

Ke (kay) fang ( [email protected] ).

Hey, I’m Kay! This guide provides an introduction to the fundamental concepts of and relationships between hypothesis testing, effect size, and power analysis, using the one-sample z-test as a prime example. While the primary goal is to elucidate the idea behind hypothesis testing, this guide does try to carefully derive the math details behind the test in the hope that it helps clarification. DISCLAIMER: It’s important to mention that the one-sample z-test is rarely used due to its restrictive assumptions. As such, there are limited resources on the subject, compelling me to derive most of the formulas, particularly those related to power, on my own. This self-reliance might increase the likelihood of errors. If you detect any inaccuracies or inconsistencies, please don’t hesitate to let me know, and I’ll make the necessary updates. Happy learning! ;)

Single sample Z-test

I. the data generating process.

In a single sample z-test, our data generating process (DGP) assumes that our observations of a random variable \(X\) are independently drawn from one identical distribution (i.i.d.) with mean \(\mu\) and variance \(\sigma^2\) .

Important Notation:

Here we use the capital \(\bar{X}\) to denote the sample mean to refer it as a random variable. And the \(X_i\) refer to each element in a sample also as a random variable.

Later, when we have an actual observed sample, we would use the lower case letter \(x_i\) to denote each observation/realization of the random variable \(X_i\) and calculate the observed sample mean \(\bar{x}\) and treat it as an realization of our sample mean \(\bar{X}\) .

The sample mean is defined as below. As indicated in previous guide, the sample mean is an unbiased estimator of population expectation under i.i.d. assumption.

\[\bar{X} = \frac{\sum^n_i X_i}{n}\]

The expectation of the sample mean should be:

\[ \begin{align*} E(\bar{X}) =& E(\frac{1}{n} \cdot \sum^n_i(X_i)) \\ =& \frac{1}{n} \cdot \sum^n_iE(X_i)\\ =&\frac{1}{n}\cdot n \cdot \mu\\ =& \mu \end{align*} \]

and the variance of the sample mean would be:

\[ \begin{align*} Var(\bar{X}) =& Var(\frac{1}{n} \cdot \sum^n_i(X_i))\\ =& \frac{1}{n^2} \cdot \sum^n_i Var(X_i)\\ =&\frac{1}{n^2} \cdot n \cdot \sigma^2\\ =& \frac{\sigma^2}{n}\\[2ex] *\text{Note: } & Var(X_1 +X_2) = Var(X_1) + Var(X_2) + Cov(X_1, X_2)\\ &\text{As the samples are drawn individually, } Cov(X_1, X_2) =0, \\ &Var(X_1 +X_2) = Var(X_1) + Var(X_2)\\ \end{align*} \]

More importantly, according to The Central Limit Theorem (CLT), even we did not specify the original distribution of \(x\) , if the original distributions of \(x\) have finite variances, as n become sufficiently large (rule of thumb: n >30), the distribution of \(\bar{x}\) become a normal distribution:

\[\bar{X} \sim N(\mu, \frac{\sigma^2}{n})\]

Given the nature of the normal distribution, we know the probability density function of \(\bar{X}\) would be

\[f_{pdf}(\bar{X}|\mu, \sigma, n) = \frac{1}{\left(\frac{\sigma}{\sqrt{n}}\right)\sqrt{2\pi}} \cdot \exp\left[-\frac{(\bar{X}-\mu)^2}{2 \cdot \left(\frac{\sigma^2}{n}\right)}\right]\]

This can be tedious to calculate so we could standardize the normal distribution to a standard normal distribution ( \(N(0, 1)\) ).

\[ Z = (\frac{\bar{X} - \mu}{\sigma/\sqrt{n}}) = (\frac{\sqrt{n} \cdot (\bar{X} - \mu)}{\sigma})\sim N(0, 1)\\ \]

Important Notation: Similar to \(\bar{X}\) and \(\bar{x}\) , we use \(Z\) to refer to the random variable and \(z\) to refer to the observation from a fixed sample.

Also we could get the theoretical probability of getting Z between an interval from the distribution by

\[ Pr(z_{min} < Z < z_{max}) = \Phi(z_{max}) - \Phi(z_{min})\\[2ex] \text{where } \Phi(k) = \int^k_{-\infty} f_{pdf}(Z|\mu, \sigma,n)\ dZ\\[2ex] f_{pdf}(Z|\mu, \sigma,n) = \frac{1}{\sqrt{2\pi}} \cdot exp(-\frac{1}{2}Z^2)\\[2ex] Z|\mu, \sigma,n = \frac{\sqrt{n} \cdot (\bar{X} - \mu)}{\sigma} \]

II. The Hypothesis Testing

1. logic of hypothesis testing: the null hypothesis.

For a one-sample Z-test, we assume we know the variance parameter \(\sigma^2\) of our data generating distribution (a very unrealistic assumption, but let’s stick with it for now)

Given a sample, we could also know the sample size n, the observed sample mean \(\bar{x}\) (remember we use lower case so it don’t get confused as we view the sample mean \(\bar{X}\) as a random variable in our DGP).

The aim of our hypothesis testing is then, given our knowledge about the \(\sigma\) , n and the \(\bar{x}\) , we can test hypothesis about our sample mean \(\mu\) . Specifically, the null hypothesis ( \(H_0\) ) stating that,

\(\mu = \mu_{H_0}\) (a two-tailed test)

\(\mu \geq \mu_{H_0}\) (a right-tailed test)

\(\mu \leq \mu_{H_0}\) (a left-tailed test)

We make this decision follow the logic that: if, given the null hypothesis is true, the probability of getting a sample mean \(\bar{X}\) (or its corresponding test statistics \(Z\) ) that is as extreme or more extreme as the observed sample mean \(\bar{x}\) (or its corresponding test statistics \(z\) ) is smaller than some threshold ( \(\alpha\) ), we would rather believe the null hypothesis is not true.

The p-value represents the probability of observing a test statistic \(Z = \frac{\sqrt{n} \cdot (\bar{X} - \mu_0)}{\sigma}\) as extreme as, or more extreme than, the one computed from the sample \(z = \frac{\sqrt{n} \cdot (\bar{x} - \mu_0)}{\sigma}\) , given that the null hypothesis is true.

The threshold we set is called significance level, denoted as \(\alpha\) . As we reject the null if the p-value is below \(\alpha\) , this also means that we have the probability of \(\alpha\) to falsely reject the null given our null is true and our observed case is indeed extreme (known as Type I error).

Moreover, given the distribution under the null, the \(\alpha\) correspond to a specific value(s) of z called the critical value(s), which we can denote as \(z_c\) .

There are two practical ways we could conduct this hypothesis testing (they are actually the same), we could either calculate the p-value and compare them to the \(\alpha\) , or compare the test statistics \(z\) with the critical value \(z_c\) .

2. Calculation of p-value

Two-tail test: p-value.

If we are concerned with the probability that our actual \(\mu\) is different (either larger or smaller) than \(\mu_{H_0}\) , we are doing a two-tail test .

For a two-tailed test, when we refer to values that are “as extreme or more extreme” than the observed test statistic, we’re considering deviations in both positive and negative directions from zero.

• Specifically, if \(z\) is positive, the p-value encompasses the probability of getting a \(Z\) that is greater than or equal to \(z\) and the probability of observing a z-value less than or equal to \(-z\) .

Therefore, the two-tailed p-value is:

\[ \begin{align*} \text{If}\ z > 0\ \text{and } & \text{alternative hypo: }\ \mu\neq \mu_{H_0}, \\[2ex] p\text{-value} =& P(Z > z) + P(Z < -z)\\ =& (1 - \Phi(z)) + \Phi(-z) =\int^{\infty}_{z} f_{pdf}(Z)\ dZ + \int^{-z}_{-\infty} f_{pdf}(Z)\ dZ,\\[2ex] & \text{As the distribution is symmetrical to 0}\\[2ex] =& 2 \cdot P(Z > z) = 2 \cdot (1-\Phi(z)) = 2 \cdot \int^{\infty}_{z}f(Z)dZ\\[2ex] =& 2 \cdot P(Z < -z) = 2 \cdot \Phi(-z)= 2 \cdot \int^{-z}_{-\infty}f(Z)dZ\\[2ex] & \text{In abosolute sense: }\\[2ex] =& 2 \cdot P(Z > |z|) = 2 \cdot (1-\Phi(|z|)) = 2 \cdot \int^{\infty}_{|z|}f(Z)dZ\\[2ex] z = &\frac{\sqrt{n} \cdot (\bar{x} - \mu_0)}{\sigma}\ \text{is calculated from the observed sample} \end{align*} \]

• Conversely, if \(z\) is negative, we consider values less than or equal to \(z\) and those greater than or equal to \(-z\) .

\[ \begin{align*} \text{If}\ z < 0\ \text{and } & \text{alternative hypo: }\ \mu\neq \mu_{H_0}, \\[2ex] p\text{-value} =& P(Z < z) + P(Z > -z) = \Phi(z) + (1-\Phi(-z))=\int^{z}_{-\infty} f_{pdf}(Z)\ dZ + \int^{\infty}_{-z} f_{pdf}(Z)\ dZ,\\[2ex] & \text{As the distribution is symmetrical to 0}\\[2ex] =& 2 \cdot P(Z < z) = 2 \cdot \Phi(z) = 2 \cdot \int^{z}_{-\infty}f(Z)dZ\\[2ex] =& 2 \cdot P(Z > -z) = 2 \cdot (1-\Phi(-z)) =2 \cdot \int^{\infty}_{-z}f(Z)dZ\\[2ex] & \text{In abosolute sense: }\\[2ex] =& 2 \cdot P(Z > |z|) = 2 \cdot (1-\Phi(|z|)) =2 \cdot \int^{\infty}_{|z|}f(Z)dZ\\[2ex] z = &\frac{\sqrt{n} \cdot (\bar{x} - \mu_0)}{\sigma}\ \text{is calculated from the observed sample} \end{align*} \]

• Overall, we can combine these two scenarios by using the absolute value of \(z\) .

\[ \text{Overall, for two-tailed test, alternative hypo: } \mu\neq \mu_{H_0}\\[2ex] p\text{-value} = 2 \cdot P(Z > |z|) = 2 \cdot (1-\Phi(|z|)) = 2 \cdot \int^{\infty}_{|z|}f_{pdf}(Z)dZ,\\[2ex] z = \frac{\sqrt{n} \cdot (\bar{x} - \mu_0)}{\sigma}\ \text{is calculated from the observed sample} \]

One-tail test: p-value

And if we are only concerned with the probability that our actual \(\mu\) is larger (or smaller) than \(\mu_{H_0}\) , we are doing a one-tail test .

For a one-tailed test, when we refer to values that are “as extreme or more extreme” than the observed test statistic, we’re considering deviations only in one direction from zero.

Therefore, the one-tailed p-value is:

\[ p-value= \begin{cases} P(Z > z) = 1 - \Phi(z)=\int^{\infty}_{z} f_{pdf}(Z)\ dZ,\quad \text{alternative hypo: } \mu> \mu_{H_0}\\[2ex] P(Z < z) = \Phi(z)= \int^{z}_{-\infty} f_{pdf}(Z)\ dZ, \quad \text{alternative hypo: } \mu < \mu_{H_0}\\[2ex] \end{cases} \\[2ex] z = \frac{\sqrt{n} \cdot (\bar{x} - \mu_0)}{\sigma}\ \text{is calculated from the observed sample} \]

If the p-value is smaller than our significance level \(\alpha\) , we can reject the null.

\[p-value(z) < \alpha \Rightarrow \text{reject } H_0: \mu = \mu_{H_0}\]

3. Critical value and rejection area

Alternatively, we could choose to not to calculate p-value for our observed \(z\) , but compare our \(z\) to the z value(s) corresponding to our \(\alpha\) .

Two-tailed test

Under a two-tailed test, we use:

\[ Pr(Z > |z|) < \frac{1}{2}\alpha \]

The critical value \(z_{\alpha/2}\) is defined as:

\[ z_{\alpha/2}= arg_{z_i} \Big[Pr(Z > z_{i}) = \frac{ \alpha}{2} \Big] = \Phi^{-1} \Big(1 -\frac{ \alpha}{2} \Big) \]

Due to the symmetry of the standard normal distribution:

\[ -z_{\alpha/2} = arg_{z_i} \Big[Pr(Z < -z_{i}) = \frac{ \alpha}{2} \Big] =\Phi^{-1} \Big(\frac{ \alpha}{2} \Big) \]

Our decision rule then implies:

\[ |z| > z_{\alpha/2},\ \text{if alternative hypo: } \mu \neq \mu_{H_0} \]

One-tailed test

Similarly for one-tailed test, the critical value \(z_{c}\) is:

\[ z_{\alpha} = \begin{cases} arg_{z_i}[Pr(Z > z_{i}) = \alpha] = \Phi^{-1}(1-\alpha), & \text{if alternative hypo: } \mu> \mu_{H_0}\\[2ex] arg_{z_i}[Pr(Z < z_{i}) = \alpha] = \Phi^{-1}(\alpha), & \text{if alternative hypo: } \mu < \mu_{H_0}\\[2ex] \end{cases} \]

Then, our conditions to reject the null hypothesis are equivalent to:

\[ \begin{cases} z > z_{\alpha}, & \text{if alternative hypo: } \mu> \mu_{H_0}\\[2ex] z < z_{\alpha}, & \text{if alternative hypo: } \mu < \mu_{H_0}\\[2ex] \end{cases}\\[2ex] \]

III. The Effect Size

The idea behind effect size is to calculate a statistic that measure how large the difference actually is and make this statistic comparable across different situations.

Our intuitive effect size in the single sample Z-test might be \(\bar{x} - \mu_0 = \bar{x} - \mu_{H_0}\) , given our hypothesized \(\mu_0 = \mu_{H_0}\) .

But this statistic is not comparable across situations, as the same difference should be more important for us to consider when the population standard deviation is very small.

So to adjust for this, we could use Cohen’s d, the magnitude of the difference between your sample mean and the hypothetical population mean, relative to the population standard deviation.

\[Cohen's\ d = \frac{\bar{x}-\mu_{H_0}}{\sigma}, \ \text{given } H_0:\mu=\mu_{H_0}\] \[ Cohen's\ d = \frac{z}{\sqrt{n}},\ \text{if}\ H_0:\mu=\mu_{H_0}\\ \text{given}\ z = \frac{\bar{x} - \mu_{H_0}}{\sigma/\sqrt{n}} =\frac{(\bar{x} - \mu_{H_0})\cdot \sqrt{n}}{\sigma} \]

IV. The Power

1. theoretical derivation of power.

The power indicate the probability that the Z-test correctly reject the null ( \(H_0: \mu = \mu_{H_0}\) ). In other word, if the \(\mu \neq \mu_{H_0}\) , what’s our chance of detecting this difference?.

Suppose the true expectation is \(\mu_{H_1}\) , so the difference between the true expectation and our hypothetical expectation is:

\[ \Delta = \mu_{H_1} - \mu_{H_0} \\ \text{Thus } \mu_{H_0} = \mu_{H_1} - \Delta \] Our original statistics can be written as:

\[ \begin{align*} Z =& \frac{\sqrt{n} \cdot (\bar{X} - \mu_{H_0})}{\sigma}\\ =& \frac{\sqrt{n} \cdot [\bar{X} - (\mu_{H_1} - \Delta)]}{\sigma}\\ =& \frac{\sqrt{n} \cdot (\bar{X} - \mu_{H_1} + \Delta)}{\sigma}\\ =& \frac{\sqrt{n} \cdot (\bar{X} - \mu_{H_1})}{\sigma} + \frac{\sqrt{n} \cdot \Delta}{\sigma}\\ \end{align*} \]

The first term of \(Z\) can be seen as the z-statistics under the true expectation \(\mu_{H_1}\) , let’s denote it as \(Z'\) .

Let’s define \(\delta\) as below. \(\delta\) is referred to as the non-centrality parameter (NCP) because it measures how much the distribution of \(Z'\) diverge from the central distribution of \(z\)

\[ \delta = \frac{\Delta \sqrt{n}}{\sigma} \]

\[ Z = Z' + \delta \Rightarrow Z'=Z-\delta \]

Thus, the power would be the probability that the \(Z'\) is in the rejection area, or more simply, use \(Z'\) to replace the \(z\) in our decision rule above:

For two-tailed test:

\[ \begin{align*} Power =& Pr(|Z'| > z_{\alpha/2})\\ =& Pr(Z' > z_{\alpha/2}) + Pr(Z' < -z_{\alpha/2})\\ =& Pr(Z - \delta > z_{\alpha/2}) + Pr(Z - \delta < -z_{\alpha/2})\\ =& Pr(Z > \delta + z_{\alpha/2}) + Pr(Z < \delta-z_{\alpha/2})\\ =& 1 -\Phi(\delta + z_{\alpha/2}) + \Phi(\delta - z_{\alpha/2})\\ & \text{if alternative hypo: } \mu \neq \mu_{H_0}\\ & \delta = \frac{\sqrt{n} \cdot (\mu_{H_1} - \mu_{H_0})}{\sigma} \end{align*} \]

For one-tailed test:

\[ \begin{align*} Power =& \begin{cases} Pr(Z' > z_{\alpha}) =Pr(Z - \delta > z_{\alpha}) =Pr(Z > \delta + z_{\alpha}) = 1- \Phi(\delta + z_{\alpha}),\ \text{if alternative hypo: } \mu> \mu_{H_0}\\[2ex] Pr(Z' < z_{\alpha}) =Pr(Z - \delta < z_{\alpha}) =Pr(Z < \delta + z_{\alpha}) = \Phi(\delta + z_{\alpha}),\ \ \ \ \ \ \ \ \text{if alternative hypo: } \mu< \mu_{H_0}\\[2ex] \end{cases}\\[2ex] \text{with}\ \ \delta =& \frac{\sqrt{n} \cdot (\mu_{H_1} - \mu_{H_0})}{\sigma} \end{align*} \]

2. Post-hoc power analysis

The post-hoc power analysis indicates that, if the null hypothesis is false, the probability that the one-sample Z-test would correctly reject the null hypothesis based on the observed sample mean \(\bar{x}\) . Here the logic that we use the sample mean \(\bar{x}\) is that we do not know the ‘true’ distribution parameter and the sample mean is the best estimate we have.

\[ \text{When } \mu_{H_1} = \bar{x},\\ \delta = \frac{\sqrt{n} \cdot (\bar{x} - \mu_{H_0})}{\sigma} =z \]

Thus, for a one-sample Z-test, the NCP given observed sample mean \(\bar{x}\) actually is the same as the observed \(z\) .

\[ \begin{align*} Power =& Pr(Z > z + z_{\alpha/2}) + Pr(Z < z -z_{\alpha/2})\\ =& 1 -\Phi(z + z_{\alpha/2}) + \Phi(z - z_{\alpha/2})\\ \text{where }z &= \frac{\sqrt{n} \cdot (\bar{x} - \mu_{H_0})}{\sigma}\\ & \text{if alternative hypo: } \mu \neq \mu_{H_0}\\ \end{align*} \]

\[ \begin{align*} Power =& \begin{cases} Pr(Z > z + z_{\alpha}) = 1- \Phi(z + z_{\alpha}),\ \text{if alternative hypo: } \mu> \mu_{H_0}\\[2ex] Pr(Z < z + z_{\alpha}) = \Phi(z + z_{\alpha}),\ \ \ \ \ \ \ \ \text{if alternative hypo: } \mu< \mu_{H_0}\\[2ex] \end{cases}\\[2ex] \text{with}\ \ z &= \frac{\sqrt{n} \cdot (\bar{x} - \mu_{H_0})}{\sigma} \end{align*} \]

If the Z-test is already significant, a post-hoc power analysis may not be useful as we have already rejected the null. But if the Z-test is non-significant, a low power may indicate the possibility that the null is falselt accepted because low power of the test.

3. Priori power analysis

The priori power analysis is aimed to estimate the sample size n needed given a desired power and assumed \(\alpha\) and effect size d (let \(\mu = \bar{X}\) ).

\[ Cohen's\ d = \frac{\bar{X} - \mu_{H_0}}{\sigma}\\ \delta = \frac{\sqrt{n} \cdot (\bar{X} - \mu_{H_0})}{\sigma} = d \cdot \sqrt{n} \]

For two-tailed test, remind ourselve that its power is:

\[ \begin{align*} Power =& Pr(Z > \delta + z_{\alpha/2}) + Pr(Z < \delta -z_{\alpha/2})\\ =& 1 -\Phi(\delta + z_{\alpha/2}) + \Phi(\delta - z_{\alpha/2})\\ & \text{if alternative hypo: } \mu \neq \mu_{H_0}\\ \end{align*} \]

Thus, to determine the sample size, we have:

\[ \Rightarrow \Phi(d \cdot \sqrt{n} + z_{\alpha/2}) - \Phi(d \cdot \sqrt{n} - z_{\alpha/2}) = 1 -Power\\ \text{as the cdf of normal distribution is symmetrical of point (0, 0.5)}\\ \Rightarrow \Phi(z_{\alpha/2} + d \cdot \sqrt{n}) + \Phi(z_{\alpha/2} - d \cdot \sqrt{n}) = 2 -Power\\ \text{if alternative hypo: } \mu \neq \mu_{H_0}\\ \]

This equation is a a transcendental equation that cannot be solved analytically (using standard algebraic techniques or in terms of elementary functions) but can be solved numerically, so we could rely on computation to solve n.

At the same time, the transcendental equation can be hard to interpret, but we could use some intuition, the two terms on the left is the sum of the y value of two points symmetrical to \(Z = z_{\alpha /2}\) (which is to the right of the x = 0), as \(\alpha\) is fixed, we could only decide the how spread these two points are from the center. As the cdf function increase slower and slower on the right side, the wider the spread, the sum tend to get smaller. If we fix the power, as desired effect size d decrease (we want to detect small effect), the sample size also need to increase quadratically (a \(k*d\) change in d lead to a \((1/k^2)*n\) change in n). Similarly, if we decide a specific effect size d to detect, we can see power increase (our test being more effective in rejecting the null), our sample size n need to increase roughly quadratically (not strictly as \(\Phi^{-1}\) is not linear).

\[ \begin{align*} Power =& \begin{cases} Pr(Z >\delta + z_{\alpha}) = 1- \Phi(\delta + z_{\alpha}),\ \text{if alternative hypo: } \mu> \mu_{H_0}\\[2ex] Pr(Z < \delta + z_{\alpha}) = \Phi(\delta + z_{\alpha}),\ \ \ \ \ \ \ \ \text{if alternative hypo: } \mu< \mu_{H_0}\\[2ex] \end{cases}\\[2ex] \end{align*} \]

Thus, for a right-tailed test, the sample size needed is:

\[ Power = 1- \Phi(d \cdot \sqrt{n} + z_{\alpha}) \\ \Rightarrow n= \bigg[\frac{\Phi^{-1}(1-Power)-z_{\alpha}}{d} \bigg]^2\\ \text{as } \Phi^{-1}(z) \text{ is symmetric to (0.5, 0)}, \Phi^{-1}(1-z)=-\Phi^{-1}(z),\\ \Rightarrow n= \bigg[\frac{-\Phi^{-1}(Power)-z_{\alpha}}{d} \bigg]^2 \\ \Rightarrow n = \frac{[\Phi^{-1}(Power)+z_{\alpha}]^2}{d^2}\\ \text{if alternative hypo: } \mu> \mu_{H_0}\\ \]

Similarly, for a left-tailed test, the sample size needed is:

\[ Power = \Phi(d \cdot \sqrt{n} + z_{\alpha}),\\ \Rightarrow n= \bigg[\frac{\Phi^{-1}(Power)-z_{\alpha}}{d} \bigg]^2\\ \Rightarrow n= \frac{[\Phi^{-1}(Power)-z_{\alpha}]^2}{d^2}\\ \text{if alternative hypo: } \mu< \mu_{H_0}\\ \]

These equations are more intuitive. As the effect size aimed to detect decrease, the sample size n need to increase quadratically (if \(d\) becomes half \(1/2*d\) i.e.  \(k *d, k=1/2\) , n becomes \(4 * n\) i.e., \((1/k^2) * n, k = 1/2\) ). As the power and significance level increase (for right-tailed \(z_{\alpha}\) become more positive and for left-tailed \(z_{\alpha}\) become more negative), the sample size n also roughly increase quadratically (not strictly as \(\Phi^{-1}\) is not linear and the numerator is a quadratic form of a sum).

Z-test Calculator

Table of contents

This Z-test calculator is a tool that helps you perform a one-sample Z-test on the population's mean . Two forms of this test - a two-tailed Z-test and a one-tailed Z-tests - exist, and can be used depending on your needs. You can also choose whether the calculator should determine the p-value from Z-test or you'd rather use the critical value approach!

Read on to learn more about Z-test in statistics, and, in particular, when to use Z-tests, what is the Z-test formula, and whether to use Z-test vs. t-test. As a bonus, we give some step-by-step examples of how to perform Z-tests!

Or you may also check our t-statistic calculator , where you can learn the concept of another essential statistic. If you are also interested in F-test, check our F-statistic calculator .

What is a Z-test?

A one sample Z-test is one of the most popular location tests. The null hypothesis is that the population mean value is equal to a given number, μ 0 \mu_0 μ 0 ​ :

We perform a two-tailed Z-test if we want to test whether the population mean is not μ 0 \mu_0 μ 0 ​ :

and a one-tailed Z-test if we want to test whether the population mean is less/greater than μ 0 \mu_0 μ 0 ​ :

Let us now discuss the assumptions of a one-sample Z-test.

When do I use Z-tests?

You may use a Z-test if your sample consists of independent data points and:

the data is normally distributed , and you know the population variance ;

the sample is large , and data follows a distribution which has a finite mean and variance. You don't need to know the population variance.

The reason these two possibilities exist is that we want the test statistics that follow the standard normal distribution N ( 0 , 1 ) \mathrm N(0, 1) N ( 0 , 1 ) . In the former case, it is an exact standard normal distribution, while in the latter, it is approximately so, thanks to the central limit theorem.

The question remains, "When is my sample considered large?" Well, there's no universal criterion. In general, the more data points you have, the better the approximation works. Statistics textbooks recommend having no fewer than 50 data points, while 30 is considered the bare minimum.

Z-test formula

Let x 1 , . . . , x n x_1, ..., x_n x 1 ​ , ... , x n ​ be an independent sample following the normal distribution N ( μ , σ 2 ) \mathrm N(\mu, \sigma^2) N ( μ , σ 2 ) , i.e., with a mean equal to μ \mu μ , and variance equal to σ 2 \sigma ^2 σ 2 .

We pose the null hypothesis, H 0  ⁣  ⁣ :  ⁣  ⁣   μ = μ 0 \mathrm H_0 \!\!:\!\! \mu = \mu_0 H 0 ​ :   μ = μ 0 ​ .

We define the test statistic, Z , as:

x ˉ \bar x x ˉ is the sample mean, i.e., x ˉ = ( x 1 + . . . + x n ) / n \bar x = (x_1 + ... + x_n) / n x ˉ = ( x 1 ​ + ... + x n ​ ) / n ;

μ 0 \mu_0 μ 0 ​ is the mean postulated in H 0 \mathrm H_0 H 0 ​ ;

n n n is sample size; and

σ \sigma σ is the population standard deviation.

In what follows, the uppercase Z Z Z stands for the test statistic (treated as a random variable), while the lowercase z z z will denote an actual value of Z Z Z , computed for a given sample drawn from N(μ,σ²).

If H 0 \mathrm H_0 H 0 ​ holds, then the sum S n = x 1 + . . . + x n S_n = x_1 + ... + x_n S n ​ = x 1 ​ + ... + x n ​ follows the normal distribution, with mean n μ 0 n \mu_0 n μ 0 ​ and variance n 2 σ n^2 \sigma n 2 σ . As Z Z Z is the standardization (z-score) of S n / n S_n/n S n ​ / n , we can conclude that the test statistic Z Z Z follows the standard normal distribution N ( 0 , 1 ) \mathrm N(0, 1) N ( 0 , 1 ) , provided that H 0 \mathrm H_0 H 0 ​ is true. By the way, we have the z-score calculator if you want to focus on this value alone.

If our data does not follow a normal distribution, or if the population standard deviation is unknown (and thus in the formula for Z Z Z we substitute the population standard deviation σ \sigma σ with sample standard deviation), then the test statistics Z Z Z is not necessarily normal. However, if the sample is sufficiently large, then the central limit theorem guarantees that Z Z Z is approximately N ( 0 , 1 ) \mathrm N(0, 1) N ( 0 , 1 ) .

In the sections below, we will explain to you how to use the value of the test statistic, z z z , to make a decision , whether or not you should reject the null hypothesis . Two approaches can be used in order to arrive at that decision: the p-value approach, and critical value approach - and we cover both of them! Which one should you use? In the past, the critical value approach was more popular because it was difficult to calculate p-value from Z-test. However, with help of modern computers, we can do it fairly easily, and with decent precision. In general, you are strongly advised to report the p-value of your tests!

p-value from Z-test

Formally, the p-value is the smallest level of significance at which the null hypothesis could be rejected. More intuitively, p-value answers the questions: provided that I live in a world where the null hypothesis holds, how probable is it that the value of the test statistic will be at least as extreme as the z z z - value I've got for my sample? Hence, a small p-value means that your result is very improbable under the null hypothesis, and so there is strong evidence against the null hypothesis - the smaller the p-value, the stronger the evidence.

To find the p-value, you have to calculate the probability that the test statistic, Z Z Z , is at least as extreme as the value we've actually observed, z z z , provided that the null hypothesis is true. (The probability of an event calculated under the assumption that H 0 \mathrm H_0 H 0 ​ is true will be denoted as P r ( event ∣ H 0 ) \small \mathrm{Pr}(\text{event} | \mathrm{H_0}) Pr ( event ∣ H 0 ​ ) .) It is the alternative hypothesis which determines what more extreme means :

• Two-tailed Z-test: extreme values are those whose absolute value exceeds ∣ z ∣ |z| ∣ z ∣ , so those smaller than − ∣ z ∣ -|z| − ∣ z ∣ or greater than ∣ z ∣ |z| ∣ z ∣ . Therefore, we have:

The symmetry of the normal distribution gives:

• Left-tailed Z-test: extreme values are those smaller than z z z , so
• Right-tailed Z-test: extreme values are those greater than z z z , so

To compute these probabilities, we can use the cumulative distribution function, (cdf) of N ( 0 , 1 ) \mathrm N(0, 1) N ( 0 , 1 ) , which for a real number, x x x , is defined as:

Also, p-values can be nicely depicted as the area under the probability density function (pdf) of N ( 0 , 1 ) \mathrm N(0, 1) N ( 0 , 1 ) , due to:

Two-tailed Z-test and one-tailed Z-test

With all the knowledge you've got from the previous section, you're ready to learn about Z-tests.

• Two-tailed Z-test:

From the fact that Φ ( − z ) = 1 − Φ ( z ) \Phi(-z) = 1 - \Phi(z) Φ ( − z ) = 1 − Φ ( z ) , we deduce that

The p-value is the area under the probability distribution function (pdf) both to the left of − ∣ z ∣ -|z| − ∣ z ∣ , and to the right of ∣ z ∣ |z| ∣ z ∣ :

• Left-tailed Z-test:

The p-value is the area under the pdf to the left of our z z z :

• Right-tailed Z-test:

The p-value is the area under the pdf to the right of z z z :

The decision as to whether or not you should reject the null hypothesis can be now made at any significance level, α \alpha α , you desire!

if the p-value is less than, or equal to, α \alpha α , the null hypothesis is rejected at this significance level; and

if the p-value is greater than α \alpha α , then there is not enough evidence to reject the null hypothesis at this significance level.

Z-test critical values & critical regions

The critical value approach involves comparing the value of the test statistic obtained for our sample, z z z , to the so-called critical values . These values constitute the boundaries of regions where the test statistic is highly improbable to lie . Those regions are often referred to as the critical regions , or rejection regions . The decision of whether or not you should reject the null hypothesis is then based on whether or not our z z z belongs to the critical region.

The critical regions depend on a significance level, α \alpha α , of the test, and on the alternative hypothesis. The choice of α \alpha α is arbitrary; in practice, the values of 0.1, 0.05, or 0.01 are most commonly used as α \alpha α .

Once we agree on the value of α \alpha α , we can easily determine the critical regions of the Z-test:

To decide the fate of H 0 \mathrm H_0 H 0 ​ , check whether or not your z z z falls in the critical region:

If yes, then reject H 0 \mathrm H_0 H 0 ​ and accept H 1 \mathrm H_1 H 1 ​ ; and

If no, then there is not enough evidence to reject H 0 \mathrm H_0 H 0 ​ .

As you see, the formulae for the critical values of Z-tests involve the inverse, Φ − 1 \Phi^{-1} Φ − 1 , of the cumulative distribution function (cdf) of N ( 0 , 1 ) \mathrm N(0, 1) N ( 0 , 1 ) .

How to use the one-sample Z-test calculator?

Our calculator reduces all the complicated steps:

Choose the alternative hypothesis: two-tailed or left/right-tailed.

In our Z-test calculator, you can decide whether to use the p-value or critical regions approach. In the latter case, set the significance level, α \alpha α .

Enter the value of the test statistic, z z z . If you don't know it, then you can enter some data that will allow us to calculate your z z z for you:

• sample mean x ˉ \bar x x ˉ (If you have raw data, go to the average calculator to determine the mean);
• tested mean μ 0 \mu_0 μ 0 ​ ;
• sample size n n n ; and
• population standard deviation σ \sigma σ (or sample standard deviation if your sample is large).

Results appear immediately below the calculator.

If you want to find z z z based on p-value , please remember that in the case of two-tailed tests there are two possible values of z z z : one positive and one negative, and they are opposite numbers. This Z-test calculator returns the positive value in such a case. In order to find the other possible value of z z z for a given p-value, just take the number opposite to the value of z z z displayed by the calculator.

Z-test examples

To make sure that you've fully understood the essence of Z-test, let's go through some examples:

• A bottle filling machine follows a normal distribution. Its standard deviation, as declared by the manufacturer, is equal to 30 ml. A juice seller claims that the volume poured in each bottle is, on average, one liter, i.e., 1000 ml, but we suspect that in fact the average volume is smaller than that...

Formally, the hypotheses that we set are the following:

H 0  ⁣ :   μ = 1000  ml \mathrm H_0 \! : \mu = 1000 \text{ ml} H 0 ​ :   μ = 1000  ml

H 1  ⁣ :   μ < 1000  ml \mathrm H_1 \! : \mu \lt 1000 \text{ ml} H 1 ​ :   μ < 1000  ml

We went to a shop and bought a sample of 9 bottles. After carefully measuring the volume of juice in each bottle, we've obtained the following sample (in milliliters):

1020 , 970 , 1000 , 980 , 1010 , 930 , 950 , 980 , 980 \small 1020, 970, 1000, 980, 1010, 930, 950, 980, 980 1020 , 970 , 1000 , 980 , 1010 , 930 , 950 , 980 , 980 .

Sample size: n = 9 n = 9 n = 9 ;

Sample mean: x ˉ = 980   m l \bar x = 980 \ \mathrm{ml} x ˉ = 980   ml ;

Population standard deviation: σ = 30   m l \sigma = 30 \ \mathrm{ml} σ = 30   ml ;

And, therefore, p-value = Φ ( − 2 ) ≈ 0.0228 \text{p-value} = \Phi(-2) \approx 0.0228 p-value = Φ ( − 2 ) ≈ 0.0228 .

As 0.0228 < 0.05 0.0228 \lt 0.05 0.0228 < 0.05 , we conclude that our suspicions aren't groundless; at the most common significance level, 0.05, we would reject the producer's claim, H 0 \mathrm H_0 H 0 ​ , and accept the alternative hypothesis, H 1 \mathrm H_1 H 1 ​ .

We tossed a coin 50 times. We got 20 tails and 30 heads. Is there sufficient evidence to claim that the coin is biased?

Clearly, our data follows Bernoulli distribution, with some success probability p p p and variance σ 2 = p ( 1 − p ) \sigma^2 = p (1-p) σ 2 = p ( 1 − p ) . However, the sample is large, so we can safely perform a Z-test. We adopt the convention that getting tails is a success.

Let us state the null and alternative hypotheses:

H 0  ⁣ :   p = 0.5 \mathrm H_0 \! : p = 0.5 H 0 ​ :   p = 0.5 (the coin is fair - the probability of tails is 0.5 0.5 0.5 )

H 1  ⁣ :   p ≠ 0.5 \mathrm H_1 \! : p \ne 0.5 H 1 ​ :   p  = 0.5 (the coin is biased - the probability of tails differs from 0.5 0.5 0.5 )

In our sample we have 20 successes (denoted by ones) and 30 failures (denoted by zeros), so:

Sample size n = 50 n = 50 n = 50 ;

Sample mean x ˉ = 20 / 50 = 0.4 \bar x = 20/50 = 0.4 x ˉ = 20/50 = 0.4 ;

Population standard deviation is given by σ = 0.5 × 0.5 \sigma = \sqrt{0.5 \times 0.5} σ = 0.5 × 0.5 ​ (because 0.5 0.5 0.5 is the proportion p p p hypothesized in H 0 \mathrm H_0 H 0 ​ ). Hence, σ = 0.5 \sigma = 0.5 σ = 0.5 ;

• And, therefore

Since 0.1573 > 0.1 0.1573 \gt 0.1 0.1573 > 0.1 we don't have enough evidence to reject the claim that the coin is fair , even at such a large significance level as 0.1 0.1 0.1 . In that case, you may safely toss it to your Witcher or use the coin flip probability calculator to find your chances of getting, e.g., 10 heads in a row (which are extremely low!).

What is the difference between Z-test vs t-test?

We use a t-test for testing the population mean of a normally distributed dataset which had an unknown population standard deviation . We get this by replacing the population standard deviation in the Z-test statistic formula by the sample standard deviation, which means that this new test statistic follows (provided that H₀ holds) the t-Student distribution with n-1 degrees of freedom instead of N(0,1) .

When should I use t-test over the Z-test?

For large samples, the t-Student distribution with n degrees of freedom approaches the N(0,1). Hence, as long as there are a sufficient number of data points (at least 30), it does not really matter whether you use the Z-test or the t-test, since the results will be almost identical. However, for small samples with unknown variance, remember to use the t-test instead of Z-test .

How do I calculate the Z test statistic?

To calculate the Z test statistic:

• Compute the arithmetic mean of your sample .
• From this mean subtract the mean postulated in null hypothesis .
• Multiply by the square root of size sample .
• Divide by the population standard deviation .
• That's it, you've just computed the Z test statistic!

Here, we perform a Z-test for population mean μ. Null hypothesis H₀: μ = μ₀.

Alternative hypothesis H₁

Significance level α

The probability that we reject the true hypothesis H₀ (type I error).

One Sample z-Test (Jump to: Lecture | Video )

Let's perform a one sample z-test: In the population, the average IQ is 100 with a standard deviation of 15. A team of scientists wants to test a new medication to see if it has either a positive or negative effect on intelligence, or no effect at all. A sample of 30 participants who have taken the medication has a mean of 140. Did the medication affect intelligence, using alpha = 0.05?

Let's begin.

1. Define Null and Alternative Hypotheses

2. State Alpha

Using an alpha of 0.05 with a two-tailed test, we would expect our distribution to look something like this:

Here we have 0.025 in each tail. Looking up 1 - 0.025 in our z-table , we find a critical value of 1.96. Thus, our decision rule for this two-tailed test is:

If Z is less than -1.96, or greater than 1.96, reject the null hypothesis.

4. Calculate Test Statistic

5. State Results

Result: Reject the null hypothesis.

6. State Conclusion

Medication significantly affected intelligence, z = 14.60, p < 0.05.

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Hypothesis testing.

Key Topics:

• Basic approach
• Null and alternative hypothesis
• Decision making and the p -value
• Z-test & Nonparametric alternative

Basic approach to hypothesis testing

• State a model describing the relationship between the explanatory variables and the outcome variable(s) in the population and the nature of the variability. State all of your assumptions .
• Specify the null and alternative hypotheses in terms of the parameters of the model.
• Invent a test statistic that will tend to be different under the null and alternative hypotheses.
• Using the assumptions of step 1, find the theoretical sampling distribution of the statistic under the null hypothesis of step 2. Ideally the form of the sampling distribution should be one of the “standard distributions”(e.g. normal, t , binomial..)
• Calculate a p -value , as the area under the sampling distribution more extreme than your statistic. Depends on the form of the alternative hypothesis.
• Choose your acceptable type 1 error rate (alpha) and apply the decision rule : reject the null hypothesis if the p-value is less than alpha, otherwise do not reject.

• \(\frac{\bar{X}-\mu_0}{\sigma / \sqrt{n}}\)
• general form is: (estimate - value we are testing)/(st.dev of the estimate)
• z-statistic follows N(0,1) distribution
• 2 × the area above |z|, area above z,or area below z, or
• compare the statistic to a critical value, |z| ≥ z α/2 , z ≥ z α , or z ≤ - z α
• Choose the acceptable level of Alpha = 0.05, we conclude …. ?

Making the Decision

It is either likely or unlikely that we would collect the evidence we did given the initial assumption. (Note: “likely” or “unlikely” is measured by calculating a probability!)

If it is likely , then we “ do not reject ” our initial assumption. There is not enough evidence to do otherwise.

If it is unlikely , then:

• either our initial assumption is correct and we experienced an unusual event or,
• our initial assumption is incorrect

In statistics, if it is unlikely, we decide to “ reject ” our initial assumption.

Example: Criminal Trial Analogy

First, state 2 hypotheses, the null hypothesis (“H 0 ”) and the alternative hypothesis (“H A ”)

• H 0 : Defendant is not guilty.
• H A : Defendant is guilty.

Usually the H 0 is a statement of “no effect”, or “no change”, or “chance only” about a population parameter.

While the H A , depending on the situation, is that there is a difference, trend, effect, or a relationship with respect to a population parameter.

• It can one-sided and two-sided.
• In two-sided we only care there is a difference, but not the direction of it. In one-sided we care about a particular direction of the relationship. We want to know if the value is strictly larger or smaller.

Then, collect evidence, such as finger prints, blood spots, hair samples, carpet fibers, shoe prints, ransom notes, handwriting samples, etc. (In statistics, the data are the evidence.)

Next, you make your initial assumption.

• Defendant is innocent until proven guilty.

In statistics, we always assume the null hypothesis is true .

Then, make a decision based on the available evidence.

• If there is sufficient evidence (“beyond a reasonable doubt”), reject the null hypothesis . (Behave as if defendant is guilty.)
• If there is not enough evidence, do not reject the null hypothesis . (Behave as if defendant is not guilty.)

If the observed outcome, e.g., a sample statistic, is surprising under the assumption that the null hypothesis is true, but more probable if the alternative is true, then this outcome is evidence against H 0 and in favor of H A .

An observed effect so large that it would rarely occur by chance is called statistically significant (i.e., not likely to happen by chance).

Using the p -value to make the decision

The p -value represents how likely we would be to observe such an extreme sample if the null hypothesis were true. The p -value is a probability computed assuming the null hypothesis is true, that the test statistic would take a value as extreme or more extreme than that actually observed. Since it's a probability, it is a number between 0 and 1. The closer the number is to 0 means the event is “unlikely.” So if p -value is “small,” (typically, less than 0.05), we can then reject the null hypothesis.

Significance level and p -value

Significance level, α, is a decisive value for p -value. In this context, significant does not mean “important”, but it means “not likely to happened just by chance”.

α is the maximum probability of rejecting the null hypothesis when the null hypothesis is true. If α = 1 we always reject the null, if α = 0 we never reject the null hypothesis. In articles, journals, etc… you may read: “The results were significant ( p <0.05).” So if p =0.03, it's significant at the level of α = 0.05 but not at the level of α = 0.01. If we reject the H 0 at the level of α = 0.05 (which corresponds to 95% CI), we are saying that if H 0 is true, the observed phenomenon would happen no more than 5% of the time (that is 1 in 20). If we choose to compare the p -value to α = 0.01, we are insisting on a stronger evidence!

So, what kind of error could we make? No matter what decision we make, there is always a chance we made an error.

Errors in Criminal Trial:

Errors in Hypothesis Testing

Type I error (False positive): The null hypothesis is rejected when it is true.

• α is the maximum probability of making a Type I error.

Type II error (False negative): The null hypothesis is not rejected when it is false.

• β is the probability of making a Type II error

There is always a chance of making one of these errors. But, a good scientific study will minimize the chance of doing so!

The power of a statistical test is its probability of rejecting the null hypothesis if the null hypothesis is false. That is, power is the ability to correctly reject H 0 and detect a significant effect. In other words, power is one minus the type II error risk.

\(\text{Power }=1-\beta = P\left(\text{reject} H_0 | H_0 \text{is false } \right)\)

Which error is worse?

Type I = you are innocent, yet accused of cheating on the test. Type II = you cheated on the test, but you are found innocent.

This depends on the context of the problem too. But in most cases scientists are trying to be “conservative”; it's worse to make a spurious discovery than to fail to make a good one. Our goal it to increase the power of the test that is to minimize the length of the CI.

We need to keep in mind:

• the effect of the sample size,
• the correctness of the underlying assumptions about the population,
• statistical vs. practical significance, etc…

(see the handout). To study the tradeoffs between the sample size, α, and Type II error we can use power and operating characteristic curves.

What type of error might we have made?

Type I error is claiming that average student height is not 65 inches, when it really is. Type II error is failing to claim that the average student height is not 65in when it is.

We rejected the null hypothesis, i.e., claimed that the height is not 65, thus making potentially a Type I error. But sometimes the p -value is too low because of the large sample size, and we may have statistical significance but not really practical significance! That's why most statisticians are much more comfortable with using CI than tests.

There is a need for a further generalization. What if we can't assume that σ is known? In this case we would use s (the sample standard deviation) to estimate σ.

If the sample is very large, we can treat σ as known by assuming that σ = s . According to the law of large numbers, this is not too bad a thing to do. But if the sample is small, the fact that we have to estimate both the standard deviation and the mean adds extra uncertainty to our inference. In practice this means that we need a larger multiplier for the standard error.

We need one-sample t -test.

One sample t -test

• Assume data are independently sampled from a normal distribution with unknown mean μ and variance σ 2 . Make an initial assumption, μ 0 .

• t-statistic: \(\frac{\bar{X}-\mu_0}{s / \sqrt{n}}\) where s is a sample st.dev.
• t-statistic follows t -distribution with df = n - 1
• Alpha = 0.05, we conclude ….

Testing for the population proportion

Let's go back to our CNN poll. Assume we have a SRS of 1,017 adults.

We are interested in testing the following hypothesis: H 0 : p = 0.50 vs. p > 0.50

What is the test statistic?

If alpha = 0.05, what do we conclude?

We will see more details in the next lesson on proportions, then distributions, and possible tests.

Z Table. Z Score Table. Normal Distribution Table. Standard Normal Table.

What is Z-Test?

Z-Test is a statistical test which let’s us approximate the distribution of the test statistic under the null hypothesis using normal distribution .

Z-Test is a test statistic commonly used in hypothesis test when the sample data is large.For carrying out the Z-Test, 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 Z-Test.

Z-Test 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 P-value, which indicates how strong the evidence is against the null hypothesis. Z-Test is such a test statistic where we make use of the mean value and z score to determine the P-value. Z-Test compares the mean of two large samples taken from a population when the variance is known.

Z-Test 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 Z-Test 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 Z-Test

Z-Test can be considered as a test statistic for a hypothesis test to calculate the P-value. However, there are certain conditions that should be satisfied by the sample to run the Z-Test.

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 Z-Testing as the Z-Test follows a normal distribution and so should the data. If the same size is less than 30, it is recommended to use a t-test 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 Z-Test 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 Z-Test can be successfully implemented.

Following are steps to run the Z-Test:

• 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 one-directional alternate hypothesis that is represented in the right region of the graph.

• H 1 :µ < k

This is the representation of a one-directional 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 Z-Test

Z-Test 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 Z-Test

One-Sample Z-Test

This is the most basic type of hypothesis test that is widely used. For running an one-sample Z-Test, all we need to know is the mean and standard deviation of the population. We consider only a single sample for a one-sample Z-Test. One-sample Z-Test 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 one-sampled Z-Test:

• 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.

Two-Sample Z-Test

A two-sample Z-Test 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 two-sample Z-Test 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 two-sample Z-Test:

• Two independent, normally distributed samples are considered for the Z-Test 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 pre-selected 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.

Z-Test Vs T-Test

Z-Test are used when the sample size exceeds 30. As Z-Test follows normal distribution, large sample size can be taken for the Z-Test. Z-Test 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.

T-Test is based on T distribution in which the mean value is known and the variance could be calculated from the sample. T-Test 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 two-sample test for running the Z-Test. Also, if the sample size is less than 30, T-Test is preferred.

One-Sample Z Test

The one-sample Z test is used when we want to know whether our sample comes from a particular population . For instance, we are doing research on data collected from successive cohorts of students taking the Elementary Statistics class. We may want to know if this particular sample of college students is similar to or different from college students in general. The one-sample Z test is used only for tests of the sample mean . Thus, our hypothesis tests whether the average of our sample ( M ) suggests that our students come from a population with a know mean ( m ) or whether it comes from a different population.

Study Design

The name of the one-sample Z test tells us the general research design of studies in which this statistic is selected to test hypotheses. We use the one-sample Z test when we collect data on a single sample drawn from a defined population . In this design, we have one group of subjects , collect data on these subjects and compare our sample statistic ( M ) to the population parameter ( m ). The population parameter tells us what to expect if our sample came from that population. If our sample statistic is very different (beyond what we would expect from sampling error), then our statistical test allows us to conclude that our sample came from a different population. Again, in the one-sample Z test , we are comparing the mean ( M ) calculated on a single set of scores (one sample) to a known population mean ( m ) .

Available Information

The one-sample Z test compares a sample to a defined population. When we say "defined" population, we are saying that the parameters of the population are known . We typically define a population distribution in terms of central tendency and variability/dispersion. Thus, for the one-sample Z test , the population m and s must be known. The one-sample Z test cannot be done if we do not have m and s . Population information is available in the technical manuals of measurement instruments or in research publications. Population information for the attachment scales used in the class dataset is available in the articles on reserve.

Test Assumptions

All parametric statistics have a set of assumptions that must be met in order to properly use the statistics to test hypotheses. The assumptions of the one-sample Z test are listed below.

When reading the psychological literature, we can find many studies in which all of these assumptions are violated . Random sampling is required for all statistical inference because it is based on probability. Random samples are difficult to find , however, and psychologists and researchers in other fields will use inferential statistics but discuss the sampling limitations in the article. We learned in our scale of measurement tutorial that psychologists will apply parametric statistics like the Z test on approximately interval scales even though the tests require interval or ratio data . This is an accepted practice in psychology and one that we use when we analyze our class data. Finally, the assumption of normal distribution in the population is considered "robust" . This means that the the statistic has been shown to yield useful results even when the assumption is violated. The central limit theorem tells us that even if the population distribution is unknown, we know that the sampling distribution of the mean will be approximately normally distributed if the sample size is large. This helps to contribute to the Z test being robust for violations of normal distribution.

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