8.2.3.2 - Minitab: One Sample Mean t Tests

A hypothesis test for one group mean can be conducted in Minitab using raw data or summarized data.

If you have a data file with every individual's observation, then you have raw data.
If you do not have each individual's observation, but rather have the sample mean, sample standard deviation, and sample size, then you have summarized data.

The next two pages will show you how to use Minitab to conduct a one-sample mean t-test using either raw data or summarized data. There is also one example of using Minitab to conduct a one-sample mean z test which is only performed if the population is known to be normally distributed and the population standard deviation (\(\sigma\)) is available.

8.2.3.2.1 - Minitab: 1 Sample Mean t Test, Raw Data

Minitab^® – One Sample Mean t Test Using Raw Data

Research question: Is the mean GPA in the population different from 3.0?

Null hypothesis: \(\mu\) = 3.0
Alternative hypothesis: \(\mu\) ≠ 3.0

The GPAs of 226 students are available.

A one sample mean \(t\) test should be performed because the shape of the population is unknown, however the sample size is large (\(n\) ≥ 30).

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

Open the Minitab file: class_survey.mpx
Select Stat > Basic Statistics > 1-sample t
Select One or more samples, each in a column from the dropdown
Double-click on the variable GPA to insert it into the Sample box
Check the box Perform a hypothesis test
For the Hypothesized mean enter 3
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 \(\mu\)
226	3.2311	0.5104	0.0340	(3.1642, 3.2980)

\(\mu\): population mean of GPA

Test
Null hypothesis	H₀: \(\mu\) = 3
Alternative hypothesis	H₁: \(\mu\) ≠ 3

T-Value	P-Value
6.81	0.000

Summary of Results

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

1. Check assumptions and write hypotheses

We do not know if the population is normally distributed, however the sample size is large (\(n \ge 30\)) so we can perform a one sample mean t test.

\(H_0\colon \mu = 3.0\)
\(H_a\colon \mu \ne 3.0\)

2. Calculate the test statistic

\(t (225) = 6.81\)

3. Determine the p-value

\(p < 0.0001\)

4. Make a decision

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

5. State a "real world" conclusion

There is evidence that the mean GPA in the population is different from 3.0

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:

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

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.

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