8.2.3.2.2 - Minitab Express: 1 Sample Mean t Test, Summarized Data

MinitabExpress  – One Sample Mean t Test Using Summarized Data

Here we are testing \(H_{a}:\mu\neq72\) 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.

  1. On a PC: Select STATISTICS > One Sample > t
    On a Mac: Select Statistics > 1-Sample Inference > t
  2. Change Sample data in column to Summarized data
  3. The Sample size is 35
  4. The Sample mean is 76.8
  5. The Sample standard deviation is 11.62
  6. Check the box Perform a hypothesis test
  7. For the Hypothesized mean enter 72
  8. Click the Options tab
  9.  Use the default Alternative hypothesis of Mean ≠ hypothesized value 
  10. Use the default Confidence level of 95
  11. Click OK

This should result in the following output:

1-Sample t
Descriptive Statistics
N Mean StDev SE Mean 95% CI for \(\mu\)
35 76.800 11.620 1.964 (72.808, 80.792)

\(\mu\) : mean of Sample

Test
Null hypothesis H0: \(\mu\) = 72
Alternative hypothesis H1: \(\mu\) ≠ 72
T-Value P-Value
2.44 0.0199
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 shape of the population distribution is unknown, however with \(n \ge 30\) we can perform a one sample mean t test. 

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

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.