## Introduction

The Z-test is a simple tool for hypothesis testing that can be used to identify whether a mean result, when compared to a larger set is statistically significant when the larger set is a normal distribution.

Many datasets (for instance population height, test scores, etc.) have normal distributions. If you’re unsure whether your dataset has a normal distribution you can approximate and assume that it does if you have at least 30 items to draw on (e.g. 30 students heights, 30 test scores.)

You will need to know the population mean and the standard deviation in order to perform the one-sample z-test. If you don’t know the standard deviation you should use a t-test instead.

## Hypothesis Testing

The following six steps are for a Z-test:

- Identify our population, comparison distribution, hypothesis and assumptions. Choose an appropriate test.
- State the null and research hypotheses.
- Determine the characteristics of the comparison distribution.
- Determine the cutoffs that indicate the points beyond which we reject the null hypothesis
- Calculate the test statistic.
- Decide whether to reject or accept (fail to reject) the null hypothesis.

## One-Sample Z-Test Formula

The following is the formula for the z-test:

Where x̄ (x-bar) is the sample mean, ∆ (delta) is the value you are comparing it with (the population mean), σ (sigma) is the population standard deviation and *n* is the number of values in the larger set.

## Z-Test Example

The example we will work with for our one-sample z-test is a set of students who received 4 hours of study strategies tutoring before beginning a statistics course and another set of students who did not. You can compare the grades of these students to find out if the tutoring has impacted their grades.

There are 150 students in the class. The mean score in the class (x-bar) is 72, with a SD of 6. The mean score of the 20 students who received tutoring (delta) is 75 with a SD of 5. These values seem too close for us to estimate purely by hand so we will use our formula. Plugging this into the formula, we get:

z = 75 – 73 / (5 / sqrt(150))

z = 75 – 73 / (5 / 12.25)

z = 75 – 73 / 0.41

z = 2 / 0.82

z = 2.44

Looking up 2.44 in our Z-Table gives us 49.27.

We subtract 49.27 from 50 (the mean) gives us 0.0073 (the % in tail value in the z-chart.) Because we are performing a two-tailed test (we want to know whether our value is significantly above or significantly below the mean), we multiply 0.0073 by to get a p value of 0.0146.

In order to reject the null hypothesis, our p value must be under 0.05. Because our p value is below 0.05, we reject the null hypothesis. This means that the students who received 4 hours of tutoring **did **have better grades than those who didn’t.