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
\(z=\dfrac{\overline{x}-\mu_0}{\dfrac{\sigma}{\sqrt{n}}}\)
\(\overline{x}\) = 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.
Minitab® – 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 z test in Minitab using summarized data:
- In Minitab, select Stat > Basic Statistics > 1-sample Z
- Select Summarized data from the dropdown
- Enter 25 for the sample size, 110 for the sample mean and 15 for the known standard deviation.
- Check the box Perform a hypothesis test
- For the Hypothesized mean enter 100
- Select Options
- Use the default Alternative hypothesis of Mean > hypothesized value
- Use the default Confidence level of 95
- Click OK and OK
This should result in the following output:
Descriptive Statistics
N | Mean | SE Mean | 95% Lower Bound for \(\mu\) |
|---|---|---|---|
25 | 110.00 | 3.00 | 105.07 |
\(\mu\): population mean of Sample
Known standard deviation = 15
Test
Null hypothesis | H0: \(\mu\) = 100 |
|---|---|
Alternative hypothesis | H1: \(\mu\) > 100 |
Z-Value | P-Value |
|---|---|
3.33 | 0.000 |
Summary of Results Section
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
\(H_0\colon \mu = 100\)
\(H_a\colon \mu > 100\)
From the Minitab output, \(z = 3.33\)
From the Minitab output, \(p = 0.000\)
\(p \le \alpha\), reject the null hypothesis
There is convincing evidence that the mean IQ score of all students at this school is greater than 100.