8.2.3.3  One Sample Mean z Test (Optional)
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 collegeprep 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 z test in Minitab Express using summarized data:
 Open Minitab Express without data
 On a PC: Select STATISTICS > One Sample > z
On a Mac: Select Statistics > 1Sample Inference > z  Change Sample data in column to Summarized data
 The Sample size is 25
 The Sample mean is 110
 The Known standard deviation is 15
 Check the box Perform a hypothesis test
 For the Hypothesized mean enter 100
 Click the Options tab
 Change Alternative hypothesis to Mean > hypothesized value
 Use the default Confidence level of 95
 Click OK
This should result in the following output:
N  Mean  SE Mean  95% Lower Bound for \(\mu\) 

25  110.000  3.000  105.065 
\(\mu\) : mean of Sample
Known standard deviation = 15
Null hypothesis  H_{0}: \(\mu_d\) = 100 

Alternative hypothesis  H_{1}: \(\mu_d\) > 100 
ZValue  PValue 

3.33  0.0004 
Select your operating system below to see a stepbystep guide for this example.
We could summarize these results using the five step hypothesis testing procedure:
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
\(H_0: \mu = 100\)
\(H_a: \mu > 100\)
From the Minitab Express output, \(z = 3.33\)
From the Minitab Express output, \(p = 0.0004\)
\(p \le \alpha\), reject the null hypothesis
There is evidence that the mean IQ score of all students at this school is greater than 100.