8.2.3.3 - One Sample Mean z Test (Optional)

A one sample mean \(z\) test is used when the population is known to be normally distributed and when the population standard deviation (\(\sigma\)) is known. This most frequently occurs in the social sciences when standardized measures are used such as IQ, SAT, ACT, or GRE scores, for which the population parameters are known. 

The formula for computing a \(z\) test statistic for one sample mean is identical to that of computing a \(t\) test statistic for one sample mean, except now the population standard deviation is known and can be used in computing the standard error.

z Test Statistic: One Group Mean
\(z=\dfrac{\overline{x}-\mu_0}{\frac{\sigma}{\sqrt{n}}}\)

\(\overline{x}\) = sample mean
\(\mu_{0}\) = hypothesized population mean
\(s\) = sample standard deviation
\(n\) = sample size

The other primary difference between the one sample mean \(t\) test and the one sample mean \(z\) test is the latter uses the standard normal distribution (i.e., \(z\) distribution) in determining the \(p\)-value. Below are the directions for conducting a one sample mean \(z\) test in Minitab Express. 

MinitabExpress  – Performing a One Sample Mean z Test

Research question: Are the IQ scores of students at one college-prep school above the national average?

Scores on one American IQ test are normed to have a mean of 100 and standard deviation of 15.  In a simple random sample of 25 students at this school the mean was 110. 

To perform a one sample mean test in Minitab Express using summarized data:

  1. Open Minitab Express without data
  2. On a PC: Select STATISTICS > One Sample > z
    On a Mac: Select Statistics > 1-Sample Inference > z
  3. Change Sample data in column to Summarized data
  4. The Sample size is 25
  5. The Sample mean is 110
  6. The Known standard deviation is 15
  7. Check the box Perform a hypothesis test
  8. For the Hypothesized mean enter 100
  9. Click the Options tab
  10.  Change Alternative hypothesis to Mean > hypothesized value 
  11. Use the default Confidence level of 95
  12. Click OK

This should result in the following output:

1-Sample Z
Descriptive Statistics
N Mean SE Mean 95% Lower Bound for \(\mu\)
25 110.000 3.000 105.065

\(\mu\) : mean of Sample
Known standard deviation = 15

Test
Null hypothesis H0: \(\mu_d\) = 100
Alternative hypothesis H1: \(\mu_d\) > 100
Z-Value P-Value
3.33 0.0004
Video Walkthrough

Select your operating system below to see a step-by-step guide for this example.

We could summarize these results using the five step hypothesis testing procedure:

1. Check assumptions and write hypotheses

The population is known to be normally distributed and the population standard deviation is known to be 15. With these two conditions met we can conduct a one sample mean z test

\(H_0: \mu = 100\)
\(H_a: \mu > 100\)

2. Calculate the test statistic

From the Minitab Express output, \(z = 3.33\)

3. Determine the p-value

From the Minitab Express output, \(p = 0.0004\)

4. Make a decision

\(p \le \alpha\), reject the null hypothesis

5. State a "real world" conclusion

There is evidence that the mean IQ score of all students at this school is greater than 100.