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Statistics By Jim

Making statistics intuitive

Z Test: Uses, Formula & Examples

By Jim Frost Leave a Comment

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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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).

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: Definition & Two Proportion Z-Test

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:

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 (left-tail) .

Two Proportion Z-Test

A Two Proportion Z-Test (or Z-interval) 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 Z-Interval 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 z-score falls into the “ rejection region .”

Step 5: Compare the calculated z-score from Step 3 with the table z-score from Step 4. If the calculated z-score 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 Z-Interval:

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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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.  

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Z-Test: Formula, Examples, Uses, Z-Test vs T-Test

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Z-test Definition

z-test 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 t-tests, z tests are also based on normal probability distribution.
• Z-test 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 z-test, the variance is usually known.
• Z-test is more convenient than t-test as the critical value at each significance level in the confidence interval is the sample for all sample sizes.
• A z-score is a number indicating how many standard deviations above or below the mean of the population is.

Z-test 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.

z-test 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 z-test (one-tailed z-test)

• One sample z-test 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 z-value of the sample being tested falls into the criteria for the one-sided tets, the alternative hypothesis will be accepted instead of the null hypothesis.
• A one-tailed 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 one-sample z-test 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 one-sided z-test can be either a left-sided z-test or a right-sided z-test. 
• For instance, if our H 0 : µ 0 = µ and H a : µ < µ 0 , such a test would be a one-sided test or more precisely, a left-tailed 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 one-tailed test (right tail), and the rejection region is present on the right tail of the curve.

Two sample z-test (two-tailed z-test)

• In the case of two sample z-test, two normally distributed independent samples are required.
• A two-tailed z-test is performed to determine the relationship between the population parameters of the two samples.
• In the case of the two-tailed z-test, the alternative hypothesis is accepted as long as the population parameter is not equal to the assumed value.
• The two-tailed 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 two-tailed test, there are two rejection regions, one on each tail of the curve.

Z-test 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:

z-test applications

• Z-test 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 z-test 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.

Z-test vs T-test (8 major differences)

References and Sources

• C.R. Kothari (1990) Research Methodology. Vishwa Prakasan. India.
• https://ncss-wpengine.netdna-ssl.com/wp-content/themes/ncss/pdf/Procedures/PASS/One-Sample_Z-Tests.pdf
• https://www.wallstreetmojo.com/z-test-vs-t-test/
• https://sites.google.com/site/fundamentalstatistics/chapter-13
• 3% – https://www.investopedia.com/terms/z/z-test.asp
• 2% – https://www.coursehero.com/file/61052903/Questions-statisticswpdf/
• 2% – https://towardsdatascience.com/everything-you-need-to-know-about-hypothesis-testing-part-i-4de9abebbc8a
• 2% – https://ncss-wpengine.netdna-ssl.com/wp-content/themes/ncss/pdf/Procedures/PASS/One-Sample_Z-Tests.pdf
• 1% – https://www.slideshare.net/MuhammadAnas96/ztest-with-examples
• 1% – https://www.mathandstatistics.com/learn-stats/hypothesis-testing/two-tailed-z-test-hypothesis-test-by-hand
• 1% – https://www.infrrr.com/proportions/difference-in-proportions-hypothesis-test-calculator
• 1% – https://keydifferences.com/difference-between-t-test-and-z-test.html
• 1% – https://en.wikipedia.org/wiki/Z-test
• 1% – http://www.sci.utah.edu/~arpaiva/classes/UT_ece3530/hypothesis_testing.pdf
• <1% – https://www.researchgate.net/post/Can-a-null-hypothesis-be-stated-as-a-difference
• <1% – https://www.isixsigma.com/tools-templates/hypothesis-testing/making-sense-two-sample-t-test/
• <1% – https://www.investopedia.com/terms/t/two-tailed-test.asp
• <1% – https://www.academia.edu/24313503/BIOSTATISTICS_AND_RESEARCH_METHODS_IN_PHARMACY_Pharmacy_C479_4_quarter_credits_A_Course_for_Distance_Learning_Prepared

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2 thoughts on “Z-Test: Formula, Examples, Uses, Z-Test vs T-Test”

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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A Z-test is a type of statistical hypothesis test used to test the mean of a normally distributed test statistic. It tests whether there is a significant difference between an observed population mean and the population mean under the null hypothesis, H 0 .

A Z-test can only be used when the population variance is known (or can be estimated with a high degree of accuracy), or if the sample size of the experiment is large (typically n>30). Also, the test statistic must exhibit a normal distribution; if it exhibits a distribution that is clearly not normal, the Z-test is not applicable. In many cases, population parameters may not be known, or it may not be possible to estimate them accurately. In such cases, or in cases where the sample size is small, a Student's t-test is more appropriate.

How to conduct a Z-test

The procedure for conducting a Z-test is similar to that of other statistical hypothesis tests, and is generally as follows:

• State the null (H 0 ) and alternative hypotheses (H a ).
• Select a significance level, α.
• Calculate the Z-score.
• Determine the critical value(s) of Z or the p-value.
• Compare the Z-score of the observed value to the critical value of Z (or compare the p-value to α) to determine if the null hypothesis should be rejected in favor of the alternative hypothesis, or if the null hypothesis should not be rejected.

H 0 and H a

The null hypothesis is typically a statement of no difference. For example, assume that the average score received on the SAT by high schoolers in a given state was a 1200 with a known standard deviation. If the average score of students in a given high school is a 1230, we may use a Z-test to determine whether this result is better, statistically, than the state average. The null hypothesis in this case would be that the average score of students in the high school is not better than the state average, or H 0 : μ ≤ μ 0 , or μ ≤ 1200.

The alternative hypothesis is a statement of difference from the null hypothesis. It can take one of three forms:

• Given H 0 : μ ≤ μ 0 , H a : μ > μ 0
• Given H 0 : μ ≥ μ 0 , H a : μ 0
• Given H 0 : μ = μ 0 , H a : μ ≠ μ 0

In this example, it is believed that a score of 1230 is statistically significant, and that students in this high school performed better than the state average. Therefore, the alternative hypothesis takes on the first form in the list, H a : μ > μ 0 , or μ > 1200.

Significance level

The significance level, α, is the probability of a study rejecting the null hypothesis when the null hypothesis is true. Commonly used significance levels include 0.01, 0.05, and 0.10. A significance level of 0.05, or 5%, means that there is a 5% chance of concluding that a difference exists (thus rejecting H 0 ) when there is no actual difference. The lower the significance level, the more evidence required before the null hypothesis can be rejected. The significance level is compared to the p-value: if a p-value is less than the significance level, the null hypothesis is rejected in favor of the alternative hypothesis.

Calculating a Z-score is a necessary part of conducting a Z-test. A Z-score indicates the number of standard deviations that an observed value is from the mean in a standard normal distribution. For example, an observed value with a Z-score of 1.2 indicates that the observed value is 1.2 standard deviations from the mean. If the population mean and standard deviation are known, the Z-score is calculated using the following formula:

where μ is the mean of the population, σ is the standard deviation of the population, and x is the observed value. In many cases the population mean and standard deviation are not known. In such cases, these population parameters can be estimated using a sample mean and sample standard deviation, and the Z-score can be computed as follows:

where x is the sample mean, s is the sample standard deviation, and x is the observed value.

Critical value and p-value

Once a Z-score has been calculated, there are two methods for drawing conclusions about the test statistic: using the critical value(s), or using a p-value. To form a conclusion for a hypothesis test using a critical value, the Z-score of the observed value is compared to the critical value(s) of the selected significance level; to use a p-value, the p-value of the observed value is compared to the significance level.

Critical value

A critical value is a value that indicates the critical region(s) (or rejection region) of the standard normal distribution, where a critical region is the area of the distribution in which a value must lie in order to reject the null hypothesis.

The critical value is dependent on the significance level as well as whether a one-tailed or two-tailed test is being conducted. A one-tailed test is used when we want to know if a value is significantly larger or smaller than the Z-score. There is only one critical region in a one-tailed Z-test. It is either a left-tailed test (or lower-tailed) or right-tailed test (or upper-tailed) based on the position of the critical region, as shown in the figure below.

The critical regions are shown in pink. If a test statistic lies within the pink region, the null hypothesis is rejected in favor of the alternative hypothesis. Otherwise, the null hypothesis is not rejected.

If a test value lies in either of the critical regions shown in pink, the null hypothesis is rejected in favor of the alternative hypothesis; if it lies within the green region, the null hypothesis is not rejected.

After selecting the significance level and type of test, the critical Z value can be determined using a Z table by finding the Z value that corresponds to the selected significance level. For example, for a one-tailed test and a significance level of 0.05, find the probability closest to 0.05 and read the Z value that results in this probability; the Z value for α = 0.05 for a one-tailed Z-test is -1.96 for a left-tailed Z-test and 1.96 for a right-tailed Z-test. For a two-tailed Z-test, divide α by 2, then determine the corresponding Z-value. For α = 0.05, each tail will comprise an area of 0.025 in the standard normal distribution, which corresponds to Z-values of -1.645 and 1.645. Thus, the critical regions are Z 1.645. The critical values for common significance levels are shown in the table below:

The p-value indicates the probability of obtaining test results that are at least as extreme as the observed results, assuming that the null hypothesis is true. It tells us how likely it is for an outcome to occur solely based on chance. For example, a p-value of 0.05 means that there is a 5% chance that an outcome occurred solely by chance. The smaller the p-value, the less likely it is for an outcome to occur solely by chance, and the more evidence there is to reject the null hypothesis.

Like critical values, a p-value can be determined using a Z table. For a left-tailed Z-test, the p-value is the area under the standard normal distribution to the left of the Z-score of the observed value; for a right-tailed Z-test, it is the area to the right of the Z-score; for a two-tailed Z-test, it is the sum of the area to the left and right of the Z-score. If the p-value is less than or equal to the significance level, the null hypothesis is rejected in favor of the alternative hypothesis. Otherwise, the null hypothesis is not rejected.

It is important to note that the p-value is not the probability that the null hypothesis is true. It is the probability that the data could deviate from the null hypothesis as much, or more than it did. The calculation of the p-value assumes that the null hypothesis is true, so it is not a measure of whether or not the null hypothesis is correct. Rather, it is a measure of how well the data fits the null hypothesis. Also, the p-value (or critical value) may provide evidence that the null hypothesis should be rejected in favor of the alternative hypothesis at the chosen level of significance . This does not mean that the alternative hypothesis is being accepted, because it is possible that the null hypothesis would not be rejected at a different significance level. Similarly, if the p-value is greater than the significance level, this does not mean that the null hypothesis is being accepted, just that the null hypothesis is not rejected.

Finally, p-values and critical values only indicate statistical significance, and may not necessarily indicate that the study's findings are significant within their context. For example, if a new medicine and a placebo are tested on different populations, and the medicine is found to have a statistically significant effect, it may not necessarily mean that there is clinical significance. It is possible for a finding to be both statistically and clinically significant, or only one or the other. For large sample sizes, it is possible for results to indicate statistical significance even when the effect is actually small and unimportant. Conversely, a small sample may not exhibit statistical significance even when the effect is large and potentially important. Thus, it is important to fully understand the scope of a study, as well as the statistical methods used, in order to effectively interpret the results and draw accurate, unbiased conclusions.

The average score on a national mathematics exam taken by high school seniors is an 82 with a standard deviation of 8. A sample of 1000 seniors achieved an average score of 68. Perform a Z-test to determine whether there is a statistically significant difference between the national average and that of the sample of seniors at a significance level of 0.05.

We want to determine whether there is any difference, so the null hypothesis is that there is no difference, or

H 0 : μ = 82

and the alternative hypothesis is:

H a : μ ≠ 82

Thus, a two-tailed Z-test should be conducted since differences on either side of the distribution must be accounted for.

The selected significance level is:

α = 0.05

This value must be greater than the p-value in order to conclude that the difference in scores is statistically significant.

Since the population standard deviation and mean are known, the Z-score can be computed as:

Based on the selected significance level and the use of a two-tailed Z-test, the critical values are Z = ± 1.96. Since the Z-score of the observed value lies between both tails (rather than within one of them), we fail to reject the null hypothesis, as depicted in the figure below.

Thus, we conclude that the difference between the observed mean and the population mean is not statistically significant for a significance level of 0.05.

However, had we selected a significance level of 0.10, the critical values would be Z = ±1.645, and Z = -1.75 would lie within the left tail of the distribution. In this case, we would reject the null hypothesis in favor of the alternative hypothesis, and conclude that the observed value is statistically significant for a significance level of 0.10.

The above discussion involved hypothesis testing for one sample, where an observed value was compared to the expected population parameter. In certain cases, scientists may want to compare the means of two samples. In such cases, a two-sample Z-test is used instead.

Two-sample Z-test

A two-sample Z-test is conducted using the same procedures described above for a one-sample Z-test, with the exception that the Z-score is computed using the following formula:

where μ 1 and μ 2 are the means of the two respective populations, x 1 and x 2 are the sample means, and n 1 and n 2 are the sample sizes.

Researchers want to test whether a certain drug has any effect on the scores received by patients who are administered the drug prior to performing a physical stress test. The researchers place patients into 2 groups: 500 are placed into the experimental group and are administered the drug; 300 are placed into the control group and are administered a placebo. Both groups then perform the physical stress test, the results of which are as follows:

Determine whether or not there is a statistically significant difference between the two groups at a significance level of 0.05.

The null hypothesis is that there is no difference, so:

H 0 : μ 1 = μ 2

Also, since it is assumed that the null hypothesis is true, μ 1 - μ 2 = 0.

The alternative hypothesis is that there is a difference, so:

H a : μ 1 ≠ μ 2

The selected significance level is 0.05, and we conduct a two-tailed test since we are looking for any observable difference.

The Z-score is then calculated as follows:

Using a Z table (or a p-value calculator), the p-value for a two-tailed Z-test for a Z-score of 2.604 is 0.009214. Since the p-value is less than the selected significance level, we reject the null hypothesis in favor of the alternative hypothesis, and conclude that the drug has a statistically significant effect on the performance of the patients. Since the Z-score lies in the right tail, we may conclude that patients who received the drug scored significantly better than those who received the placebo. If the Z-score were to lie in left tail, we would conclude the opposite: that patients who received the drug performed significantly worse.

We could also have used the critical values Z = ±1.96 for a significance level of 0.05 to reach the same conclusion, since 2.604 lies within the critical region denoted by the right tail of the distribution, as shown in the figure below.

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 non-directional 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 non-directional hypothesis . To test the significance of a non-directional hypothesis, we have to consider the possibility that the sample could be extreme at either tail of the comparison distribution. We call this a two-tailed test .

Figure 1. showing a 2-tail test for non-directional 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 one-tail of the comparison distribution. We call this a one-tailed test .

Figure 2. showing a 1-tail test for a directional hypothesis (predicting an increase) for z for area C is the critical rejection region.

Determining Cutoff Scores with Two-Tailed 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 two-tailed 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 one-tailed test does have such an extreme value but with a one-tailed test only one side of the distribution is considered.

Figure 3. Critical value differences in one and two-tail tests. Photo Credit

Let’s re view th e set critical values for Z.

We discussed z-scores and probability in chapter 8.  If we revisit the z-score for 5% and 1%, we can identify the critical regions for the critical rejection areas from the unit standard normal table.

• A two-tailed test at the 5% level has a critical boundary Z score of +1.96 and -1.96
• A one-tailed test at the 5% level has a critical boundary Z score of +1.64 or -1.64
• A two-tailed test at the 1% level has a critical boundary Z score of +2.58 and -2.58
• A one-tailed test at the 1% level has a critical boundary Z score of +2.33 or -2.33.

Review: Critical values, p-values, 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 one-tailed 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 z-value, 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 (“z-crit”) or z* (hence the other name “critical region”). Finding the critical value works exactly the same as finding the z-score 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 z-score 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 non-directional 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 z-table, corresponds to critical values of z* = ±1.96. This is shown in Figure 5.

Figure 5: Two-tailed rejection region

Thus, any z-score falling outside ±1.96 (greater than 1.96 in absolute value) falls in the rejection region. When we use z-scores in this way, the obtained value of z (sometimes called z-obtained) 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 z-score 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 z-statistic to our critical z-value. 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 p-value, 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 z-statistic 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 four-step 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 two-tailed 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 two-tailed 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 two-tailed z-test 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 z-statistic 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 z-statistic and p-value. We don’t know the exact p-value, 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

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 one-tailed 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:

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 z-statistic

You compare your obtained z-statistic, 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 = (76.60-74.00)/ 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 two-tailed test: H A : The average score does differ H A : μ ≠ 50

We have seen the critical values for z-tests 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 z-score cutting of 0.005 (0.01 divided by 2 for a two-tailed 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 z-statistic as:

Our obtained z-statistic, 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 p-value 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 = (60.4-60)/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 all-or-nothing 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 within-subjects design rather than a between-subjects 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 z-score 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 z-statistic. When and why do you calculate an effect size?
• z = 1.99, two-tailed test at α = 0.05
• z = 1.99, two-tailed test at α = 0.01
• z = 1.99, one-tailed 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 self-esteem and depression in teenagers.  A sample of 25 teens with a low self-esteem 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 two-tail test with α = 0.05 to examine whether teenagers with low self-esteem 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 Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Regression in Machine Learning

Z-test 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 Z-Test?

Z-test formula, when to use z-test, hypothesis testing, steps to perform z-test, type of z-test.

Z-test 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).

Z-test 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 Z-test compares the difference between the sample mean and the population means by considering the standard deviation of the sampling distribution. The resulting Z-score represents the number of standard deviations that the sample mean deviates from the population mean. This Z-Score is also known as Z-Statistics, and can be formulated as:

[Tex]\text{Z-Score} = \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.

z-test assumes that the test statistic (z-score) 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 Z-Score for Delhi will be.

[Tex]\begin{aligned} \text{Z-Score}&=\frac{\bar{x}-\mu}{\sigma} \\&=\frac{300-200}{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 t-test.
• 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 z-test using
• n: sample size.
• Now compare with the hypothesis and decide whether to reject or not reject the null hypothesis

Left-tailed 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.

Right-tailed 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.

One-Tailed 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 Z-Score: X = 110 Mean = 100 Standard Deviation = 15 Number of samples = 50 [Tex]\begin{aligned} \text{Z-Score}&=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \\&=\frac{110-100}{15/\sqrt{50}} \\&=\frac{10}{2.12} \\&=4.71 \end{aligned} [/Tex]
• Now, we look up to the z-table. For the value of ∝=0.05, the z-score for the right-tailed test is 1.645.
• Here 4.71 >1.645, so we reject the null hypothesis. 
• If the z-test statistics are less than the z-score, then we will not reject the null hypothesis.

Code Implementations of One-Tailed Z-Test

# 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 z-score z_score = ( sample_mean - population_mean ) / ( population_std / np . sqrt ( 50 )) print ( 'Z-Score :' , z_score ) # Approach 1: Using Critical Z-Score # Critical Z-Score z_critical = stats . norm . ppf ( 1 - alpha ) print ( 'Critical Z-Score :' , z_critical ) # Hypothesis if z_score > z_critical : print ( "Reject Null Hypothesis" ) else : print ( "Fail to Reject Null Hypothesis" ) # Approach 2: Using P-value # P-Value : Probability of getting less than a Z-score p_value = 1 - stats . norm . cdf ( z_score ) print ( 'p-value :' , p_value ) # Hypothesis if p_value < alpha : print ( "Reject Null Hypothesis" ) else : print ( "Fail to Reject Null Hypothesis" )

Z-Score : 4.714045207910317Critical Z-Score : 1.6448536269514722Reject Null Hypothesisp-value : 1.2142337364462463e-06Reject Null Hypothesis

Two-tailed 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 z-test:

Two-sampled z-test

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 z-test 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 two-sample z-test 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: Z-Score

[Tex]\begin{aligned} \text{Z-score} &= \frac{(x_1-x_2)-(\mu_1 -\mu_2)} {\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_1}}} \\ &= \frac{(75-80)-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 Z-Score value in the Z-Table for apha/2 = 0.025

•  Critical Z-Score = 1.96

Step 5: Compare with the absolute Z-Score value

• absolute(Z-Score) > Critical Z-Score
• Reject the null hypothesis. There is a significant difference between the online and offline classes.

Code Implementations on Two-sampled Z-test

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_1-mu_2 = 0 # Hypothesized difference (under the null hypothesis) D = 0 # Set the significance level alpha = 0.05 # Calculate the test statistic (z-score) z_score = (( x1 - x2 ) - D ) / np . sqrt (( s1 ** 2 / n1 ) + ( s2 ** 2 / n2 )) print ( 'Z-Score:' , np . abs ( z_score )) # Calculate the critical value z_critical = stats . norm . ppf ( 1 - alpha / 2 ) print ( 'Critical Z-Score:' , 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 P-value # P-Value : Probability of getting less than a Z-score p_value = 2 * ( 1 - stats . norm . cdf ( np . abs ( z_score ))) print ( 'P-Value :' , p_value ) # Compare the p-value 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.""" )

Z-Score: 2.3836564731139807 Critical Z-Score: 1.959963984540054 Reject the null hypothesis. There is a significant difference between the online and offline classes. P-Value : 0.01714159544079563 Reject the null hypothesis. There is a significant difference between the online and offline classes.

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.

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7.4.1 - hypothesis testing, five step hypothesis testing procedure section  .

In the remaining lessons, we will use the following five step hypothesis testing procedure. This is slightly different from the five step procedure that we used when conducting randomization tests. 

• Check assumptions and write hypotheses.  The assumptions will vary depending on the test. In this lesson we'll be confirming that the sampling distribution is approximately normal by visually examining the randomization distribution. In later lessons you'll learn more objective assumptions. The null and alternative hypotheses will always be written in terms of population parameters; the null hypothesis will always contain the equality (i.e., \(=\)).
• Calculate the test statistic.  Here, we'll be using the formula below for the general form of the test statistic.
• Determine the p-value.  The p-value is the area under the standard normal distribution that is more extreme than the test statistic in the direction of the alternative hypothesis.
• Make a decision.  If \(p \leq \alpha\) reject the null hypothesis. If \(p>\alpha\) fail to reject the null hypothesis.
• State a "real world" conclusion.  Based on your decision in step 4, write a conclusion in terms of the original research question.

General Form of a Test Statistic Section  

When using a standard normal distribution (i.e., z distribution), the test statistic is the standardized value that is the boundary of the p-value. Recall the formula for a z score: \(z=\frac{x-\overline x}{s}\). The formula for a test statistic will be similar. When conducting a hypothesis test the sampling distribution will be centered on the null parameter and the standard deviation is known as the standard error.

This formula puts our observed sample statistic on a standard scale (e.g., z distribution). A z score tells us where a score lies on a normal distribution in standard deviation units. The test statistic tells us where our sample statistic falls on the sampling distribution in standard error units.

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.

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Two-Tailed z-test Hypothesis Test By Hand

Running a Two-Tailed z-test Hypothesis Test by Hand

HOW TO Video z-test Using Excel

Suppose it is up to you to determine if a certain state (Michigan) receives a significantly different amount of public school funding (per student) than the USA average. You know that the USA mean public school yearly funding is $6800 per student per year, with a standard deviation of $400.

Next, suppose you collect a sample (n = 100) from Michigan and determine that the sample mean for Michigan (per student per year) is $6873

Use the z-test and the correct Ho and Ha to run a hypothesis test to determine if Michigan receives a significantly different amount of funding for public school education (per student per year).

NOTE: This entire example works the same way if you have a dataset. Using the dataset, you would need to first calculate the sample mean. To run a z-test, it is generally expected that you have a larger sample size (30 or more) and that you have information about the population mean and standard deviation. If you do not have this information, it is sometimes best to use the t-test.

Step 1: Set up your hypothesis

Hypothesis: The mean per student per year funding in Michigan is significantly different than the average per student per year funding over the entire USA.

Step 2: Create Ho and Ha

NOTE: There are many ways to write out Ho.

Ho: mean per student per year funding for Michigan = mean per student per year funding for the USA

This can also be written as the following. Ho: Michigan mean – Population mean = 0

Ha: mean per student per year funding for Michigan ≠ mean per student per year funding for the USA

NOTICE1: The Ha in this example is TWO-TAILED because we are interested in seeing if Michigan is significantly different than the population mean. In a two-tailed test, the Ha contains a NOT EQUAL and the test will see if there is a significant difference (greater or smaller).

NOTICE2: The Ho is the null hypothesis and so always contains the equal sign as it is the case for which there is no significant difference between the two groups.

Step 3: Calculate the z-test statistic

Now, calculate the test statistic. In this example, we are using the z-test and are doing this by hand. However, there are many applications that run such tests. This Site has several examples under the Stats Apps link.

z = (sample mean – population mean) / [population standard deviation/sqrt(n)]

z = (6873 – 6800) / [400/sqrt(100)]

z = 73 / [400/10]

z = 73/ [40]

So, the z-test result, also called the test statistic is 1.825.

Step 4: Using the z-table, determine the rejection regions for you z-test. To do this, you must first select an alpha value . The alpha value is the percentage chance that you will reject the null (choose to go with your Ha research hypothesis as you conclusion) when in fact the Ho really true (and your research Ha should not be selected). This is also called a Type I error (choosing Ha when Ho is actually correct).

The smaller the alpha, the smaller the percentage of error, BUT the smaller the rejection regions and more difficult to reject Ho.

Most research uses alpha at .05, which creates only a 5% chance of Type I error. However, in cases such as medical research, the alpha is set much smaller.

In our case, we will use alpha = .05

This is TWO-TAILED test, therefore the rejection regions are denoted by + or – 1.96.

HOW TO Find Critical Values and Rejection Regions

NOTE: From the z-table, the critical values for a two-tailed z-test at alpha = .05 is +/- 1.96

Step 5: Create a conclusion

Our z-test result is 1.825

Because 1.825 < 1.96 it is NOT inside the rejection region.

Recall that the rejection regions for a two tailed test with alpha set to .05 is any value above 1.96 OR any value below – 1.96. Because 1.825 is not above 1.96 or below -1.96, it is NOT in the rejection region.

Therefore, this result is NOT significant. We CANNOT reject Ho. We CANNOT conclude that there is a significant difference between the funding for Michigan and the average funding for the USA.

http://www.ascd.org/publications/educational-leadership/may02/vol59/num08/Unequal-School-Funding-in-the-United-States.aspx

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What Is a Z-Test?

Understanding z-tests, the bottom line.

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Z-Test: Definition, Uses in Statistics, and Example

James Chen, CMT is an expert trader, investment adviser, and global market strategist.

Investopedia / Julie Bang

A z-test 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 z-test to be performed.

Key Takeaways

• A z-test 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 z-test is a hypothesis test for data that follows a normal distribution. 
• A z-statistic, or z-score, is a number representing the result from the z-test.
• Z-tests are closely related to t-tests, but t-tests are best performed when an experiment has a small sample size.
• Z-tests assume the standard deviation is known, while t-tests assume it is unknown.

The z-test is also a hypothesis test in which the z-statistic follows a normal distribution. The z-test is best used for greater-than-30 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 z-test, the null and alternative hypotheses, and alpha level should be stated. The z-score , also called a test statistic, should be calculated, and the results and conclusion stated. A z-statistic, or z-score, is a number representing how many standard deviations above or below the mean population a score derived from a z-test is.

Examples of tests that can be conducted as z-tests include a one-sample location test, a two-sample location test, a paired difference test, and a maximum likelihood estimate. Z-tests are closely related to t-tests, but t-tests are best performed when an experiment has a small sample size. Also, t-tests assume the standard deviation is unknown, while z-tests 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 Z-Score

The Z-score is calculated with the formula:

z = ( x - μ ) / σ

• z = Z-score
• x = the value being evaluated
• μ = the mean
• σ = the standard deviation

One-Sample Z-Test 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 two-tailed 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 T-Test and Z-Test?

Z-tests are closely related to t-tests, but t-tests are best performed when the data consists of a small sample size, i.e., less than 30. Also, t-tests assume the standard deviation is unknown, while z-tests assume it is known.

When Should You Use a Z-Test?

If the standard deviation of the population is known and the sample size is greater than or equal to 30, the z-test can be used. Regardless of the sample size, if the population standard deviation is unknown, a t-test should be used instead.

What Is a Z-Score?

A z-score, or z-statistic, is a number representing how many standard deviations above or below the mean population the score derived from a z-test is. Essentially, it is a numerical measurement that describes a value's relationship to the mean of a group of values. If a z-score is 0, it indicates that the data point's score is identical to the mean score. A z-score of 1.0 would indicate a value that is one standard deviation from the mean. Z-scores 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 z-test's fidelity relies on the CLT holding.

What Are the Assumptions of the Z-Test?

For a z-test 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 z-test 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 z-test can only be used if the population standard deviation is known and the sample size is 30 data points or larger. Otherwise, a t-test should be employed.

Newcastle University. " Z-Test ."

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

A  one proportion z-test  is used to compare an observed proportion to a theoretical one.

This tutorial explains the following:

• The motivation for performing a one proportion z-test.
• The formula to perform a one proportion z-test.
• An example of how to perform a one proportion z-test.

One Proportion Z-Test: Motivation

Suppose we want to know if the proportion of people in a certain county that are in favor of a certain law is equal to 60%. Since there are thousands of residents in the county, it would be too costly and time-consuming to go around and ask each resident about their stance on the law.

Instead, we might select a simple random sample of residents and ask each one whether or not they support the law:

However, it’s virtually guaranteed that the proportion of residents in the sample who support the law will be at least a little different from the proportion of residents in the entire population who support the law. The question is whether or not this difference is statistically significant . Fortunately, a one proportion z-test allows us to answer this question.

One Proportion Z-Test: Formula

A one proportion z-test always uses the following null hypothesis:

• H 0 :  p = p 0 (population proportion is equal to some hypothesized population proportion p 0 )

The alternative hypothesis can be either two-tailed, left-tailed, or right-tailed:

• H 1 (two-tailed):  p ≠ p 0 (population proportion is not equal to some hypothesized value p 0 )
• H 1 (left-tailed):  p < p 0 (population proportion is less than some hypothesized value p 0 )
• H 1 (right-tailed):  p > p 0 (population proportion is greater than some hypothesized value p 0 )

We use the following formula to calculate the test statistic z:

z = (p-p 0 ) / √ p 0 (1-p 0 )/n

• p:  observed sample proportion
• p 0 : hypothesized population proportion
• n:  sample size

If the p-value that corresponds to the test statistic z 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 Proportion Z-Test : Example

Suppose we want to know whether or not the proportion of residents in a certain county who support a certain law is equal to 60%. To test this, will perform a one proportion z-test at significance level α = 0.05 using the following steps:

Step 1: Gather the sample data.

Suppose  we survey a random sample of residents and end up with the following information:

• p:  observed sample proportion = 0.64
• p 0 : hypothesized population proportion = 0.60
• n:  sample size = 100

Step 2: Define the hypotheses.

We will perform the one sample t-test with the following hypotheses:

• H 0 :  p = 0.60 (population proportion is equal to 0.60)
• H 1 :  p ≠ 0.60 (population proportion is not equal to 0.60)

Step 3: Calculate the test statistic  z .

z = (p-p 0 ) / √ p 0 (1-p 0 )/n  = (.64-.6) / √ .6(1-.6)/100  = 0.816

Step 4: Calculate the p-value of the test statistic  z .

According to the Z Score to P Value Calculator , the two-tailed p-value associated with z = 0.816 is  0.4145 .

Step 5: Draw a conclusion.

Since this p-value is not less than our significance level α = 0.05, we fail to reject the null hypothesis. We do not have sufficient evidence to say that the proportion of residents who support the law is different from 0.60.

Note:  You can also perform this entire one proportion z-test by simply using the One Proportion Z-Test Calculator .

Additional Resources

How to Perform a One Proportion Z-Test in Excel One Proportion Z-Test Calculator

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Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike.  My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

2 Replies to “One Proportion Z-Test: Definition, Formula, and Example”

Can we get this for Python? Thank You!

Also, are there any assumptions here as well?

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Two samples Z-test for Means: Formula & Examples

Last updated: 21st Nov, 2023

Statistical hypothesis testing is an essential tool in inferential statistics that enables researchers to make informed decisions about the population parameters based on sample statistics. One common hypothesis test for comparing two sample means is the Two-Sample Z-Test .

In statistics , a two-sample z-test for means is used to determine if the means of two populations are equal. This test is used when the population standard deviations are known. As data scientists , it is of utmost importance to be able to understand and conduct this test accurately. In this blog, we will delve deeper into the Two-Sample Z-Test for means , exploring its formula, assumptions, and examples of how to apply it in real-world scenarios. Whether you are a student of statistics, a researcher in a scientific field, or a data analyst looking to build your statistical toolset, this blog will provide you with the foundation to understand and use the Two-Sample Z-Test for means.

Table of Contents

What is a Two-Sample Z-test for Means?

Two-sample Z-test for means is a statistical hypothesis testing technique that compares two independent samples to determine whether the means of the populations that generated them are different or not. It relies on the assumption that the populations have normal distributions, known population variances or equal variances, and that the samples are randomly and independently drawn from the respective populations. This test is used when the standard deviations (σ) of the two populations are known. This test can be used when we have a sample from each population and we know the variance for these populations.

The Two-Sample Z-Test is used when the sample sizes are large (typically, n ≥ 30), and the population standard deviations are known or can be estimated from the sample data. When these assumptions are met, the sampling distribution of the sample means follows a normal distribution, and the Z-test statistic can be used to test the null hypothesis that there is no significant difference between the means of the two populations.

However, when the population standard deviations are unknown, or the sample sizes are small, the Two-Sample T-Test for independence samples is more appropriate. The T-test is a robust statistical test that can handle the situation of unknown population standard deviations or small sample sizes by estimating the standard deviation from the sample data and using a t-distribution instead of standard normal distribution. The T-test has a higher degree of freedom and is more accurate in estimating the population parameters than the Z-test.

Here are some of the real-world examples where a two-sample z-test for means can be used:

• Comparing the performance of students in two different classes
• Comparing the average salaries of men and women in a company
• Comparing the KPIs of two different teams
• Comparing the performance of employees in two different departments
• Comparing the average IQ scores of two groups of people
• Determining if there is a significant difference in the amount of rainfall between two cities
• Investigating whether the mean daily energy intake of men and women are different

Null & Alternate Hypothesis for Two-Sample Z-test for Means

The null hypothesis for the Two-Sample Z-Test for means is that there is no significant difference between the means of two independent populations. Mathematically, it can be expressed as:

H0: μ1 – μ2 = 0

Where μ1 and μ2 represent the population means of the two independent populations.

The alternative hypothesis, on the other hand, states that there is a significant difference between the means of the two populations. The alternative hypothesis can be one-tailed or two-tailed, depending on the research question and the directionality of the expected difference.

For a two-tailed test, the alternative hypothesis can be expressed as:

Ha: μ1 – μ2 ≠ 0

This indicates that we are interested in detecting any significant difference between the two population means, regardless of its direction.

For a one-tailed test, the alternative hypothesis can be expressed as:

Ha: μ1 – μ2 > 0 or Ha: μ1 – μ2 < 0

This indicates that we are interested in detecting a significant difference between the two population means in a specific direction, either positive or negative.

In both cases, we use the Two-Sample Z-Test to determine whether there is sufficient evidence to reject the null hypothesis and accept the alternative hypothesis.

Z-statistics Formula: Two-sample Z-test for means?

The following is the formula for z-statistics for two-sample z-test for means given the population standard deviation is known .

x̄1 is the mean of the first sample

x̄2 is the mean of the second sample

μ1 is the mean of the first population

μ2 is the mean of the second population

(μ1 – μ2) is hypothesized difference between the population means

σ1 is the standard deviation of the first population

σ2 is the standard deviation of the second population

n1 is the number of the data points in the first sample

n2 is the number of the data points in the second sample

Once the value of z-statistics is calculated, we can use this value to calculate the p-value, which indicates the probability of obtaining a z-value as extreme as the one calculated under the null hypothesis. The value of p-value is then compared to the significance level (α) to determine whether to reject or fail to reject the null hypothesis. 

If the p-value is less than α, we reject the null hypothesis and conclude that there is a significant difference between the means of the two populations. If the p-value is greater than or equal to α, we fail to reject the null hypothesis and conclude that there is insufficient evidence to suggest that the means of the two populations are different.

Example: Two-sample Z-test for Means

A company wanted to compare the performance of its call center employees in two different centers located in two different parts of the country – Hyderabad, and Bengaluru, in terms of the number of tickets resolved in a day (hypothetically speaking). The company randomly selected 30 employees from the call center in Hyderabad and 30 employees from the call center in Bengaluru. The following data was collected:

Hyderabad: x̄1 = 750, σ1 = 20

Bengaluru: x̄2 = 780, σ2 = 25

The company wants to determine if the performance of the employees in Hyderabad is different from the performance of the employees in the Bengaluru center. To do this, we will use a two-sample z-test for means. We will perform two-tailed test .

First, we will formulate the null and alternate hypotheses and set the level of significance for the test.

H0: There is no difference between the performance of employees at different call centers.

H1: There is a difference in the performance of the employees.

The level of significance is set as 0.05.

Next, the mean and standard deviation for each sample will need to be determined.

Next, we will calculate the hypothesized difference between the two population means. In this case, the company is hypothesizing that the mean performance in Hyderabad is the same as that of Bengaluru. So, (μ1 – μ2 ) = 0

Finally, we will use the formula for two-sample z-test for means to calculate the test statistic.

z= (x̄1 – x̄2 ) / √((σ1 )²/n1 + (σ2)²/n2)

z = (-30) / √((20)²/30 + (25)²/30))

At a significance level of 0.05, the p-value is less than 0.00001. You can calculate the same by using P-value from Z-score calculator . As the p-value is lot less than the critical value of 0.05, the result is statistically significant and hence you can reject the null hypothesis. Hence, the performance of Hyderabad’s team is considered to be not equal to the performance of Bengaluru’s team.

The following plot also demonstrates how the Z-statistics fall beyond the critical region given the two-tailed test performed. From the plot, you can understand that null hypothesis is rejected because Z-statistics is less than the critical region on the negative side of the plot.

Two Sample Z-test for Means: Python Example

The following Python code can be used for two samples Z-test for means.

The above method can be used to calculate z-statistics and p-value as like the following:

Two-sample Z-Test for Means & Data-Driven Decision Making

The Two-Sample Z-Test for means can be applied in various real-life scenarios related to decision making. For example, a company may want to test whether a new marketing strategy has a significant impact on sales by comparing the mean sales before and after the implementation of the strategy.

Suppose a clothing store introduced a new marketing strategy to increase sales by sending discount coupons to customers via email. To evaluate the effectiveness of the new strategy, the store can use the Two-Sample Z-Test for means to compare the average sales of a randomly selected group of customers who received the discount coupons to the sales of another randomly selected group of customers who did not receive any coupons.

The null hypothesis would be that there is no significant difference between the mean sales of the two groups, while the alternative hypothesis would be that there is a significant difference in the mean sales of the two groups. The significance level (α) would be set to 0.05.

After collecting the data and calculating the test statistic, if the p-value is less than 0.05, the company can reject the null hypothesis and conclude that the new marketing strategy has a significant impact on sales. This would enable the company to make an informed decision on whether to continue or modify the marketing strategy.

Overall, the Two-Sample Z-Test for means can be a useful tool for decision making in various real-life scenarios, enabling individuals and organizations to make data-driven decisions with confidence.

You may wonder whether we could also have used two-sample T-test for means? Well, the two-sample Z-Test for means is appropriate when the sample size is large (typically, n ≥ 30) and the population standard deviation is known or can be estimated from the sample data. In the example of the clothing store, it is possible that the population standard deviation of sales is known or can be estimated accurately from the available data. In such a case, the two-sample z-test would be more appropriate than the two-sample z-test because it has higher statistical power and can provide more accurate results.

However, if the population standard deviation is unknown or cannot be estimated accurately from the available data, the two-sample t-test would be more appropriate. Additionally, if the sample size is small (typically, n < 30), the t-test is preferred over the z-test because the normal distribution assumption of the z-test may not hold for small sample sizes.

Performing Two-Sample Z-Test for Means using Excel

In this section, we will learn step-by-step method of how to perform two-sample Z-test for means using an excel spreadsheet. Taking example discussed earlier in this blog, let’s say you have 50 data samples for both Hyderabad and Bengaluru call centers. The following steps would need to be followed:

• Open Excel and create two columns: one for Hyderabad and one for Bengaluru.
• Enter the generated data for each group in their respective columns.
• Mean: = AVERAGE(A2:A51)
• Standard Deviation: =STDEV.S(A2:A51)
• Mean: = AVERAGE(B2:B51)
• Standard Deviation: = STDEV.S(B2:B51)

• Use cell references for the means and standard deviations you calculated.
• Use Excel’s NORM.S.DIST function to find the two-tailed p-value 2∗(1− NORM . S . DIST ( ABS ( Z ), TRUE ))
• This will give you the probability of observing a Z-statistic as extreme as the one calculated, assuming the null hypothesis is true.
• Choose a significance level (commonly 0.05).
• If the p-value is less than the significance level, reject the null hypothesis, indicating a significant difference in performance between the two centers.
• If the p-value is greater than the significance level, you do not have sufficient evidence to reject the null hypothesis, suggesting no significant difference in performance.

When two samples are taken from two populations, the two-sample z-test for means is used to determine whether or not there is a significant difference between the two means. The null hypothesis states that there isn’t any statistical significance between the two population means (H0) and the alternate hypothesis says otherwise (H1). In order to perform the hypothesis testing to determine whether the difference exists between the two groups or samples and that the difference is statistically significant, the two-samples Z-test for means is used.

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Ajitesh Kumar

2 responses.

In this case why is standard error not considered as pooled one i.e sqrt((n1*σ^2 + n2*σ^2)/(n1 + n2 – 2))

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