- Minitab: 1 Sample Mean t Test, Summarized Data

Minitab 18

Minitab®  – One Sample Mean t Test Using Summarized Data

Here we are testing \(H_{a}\colon\mu\neq72\) and are given \(n=35\), \(\bar{x}=76.8\), and \(s=11.62\).

We do not know the shape of the population, however the sample size is large (\(n \ge 30\)) therefore we can conduct a one sample mean \(t\) test.

To perform a one sample mean \(t\) test in Minitab using raw data:

  1. In Minitab, select Stat > Basic Statistics > 1-sample t
  2. Select Summarized data from the dropdown
  3. Enter 35 for the sample size, 76.8 for the sample mean and 11.62 for the standard deviation.
  4. Check the box Perform a hypothesis test
  5. For the Hypothesized mean enter 72
  6. Select Options
  7. Use the default Alternative hypothesis of Mean ≠ hypothesized value 
  8. Use the default Confidence level of 95
  9. Click OK and OK

This should result in the following output:

Descriptive Statistics

N Mean StDev SE Mean 95% CI for \(\mu\)
35 76.80 11.62 1.96 (72.81, 80.79)
\(\mu\): population mean of Sample
Null hypothesis H0: \(\mu\) = 72
Alternative hypothesis H1: \(\mu\) ≠ 72
T-Value P-Value
2.44 0.0199

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

1. Check assumptions and write hypotheses

The shape of the population distribution is unknown, however with \(n \ge 30\) we can perform a one sample mean t test. 

\(H_0\colon \mu = 72\)
\(H_a\colon \mu \ne 72\)

2. Calculate the test statistic

\(t (34) = 2.44\)

3. Determine the p-value

\(p = 0.0199\)

4. Make a decision

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

5. State a "real world" conclusion

There is evidence that the population mean is different from 72.