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

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:

In Minitab, select Stat > Basic Statistics > 1-sample t
Select Summarized data from the dropdown
Enter 35 for the sample size, 76.8 for the sample mean and 11.62 for the standard deviation.
Check the box Perform a hypothesis test
For the Hypothesized mean enter 72
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:

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

Test
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 convincing evidence that the population mean is different from 72.