8.2.3.2 - Minitab: One Sample Mean t Tests
8.2.3.2 - Minitab: One Sample Mean t TestsA 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
8.2.3.2.1 - Minitab: 1 Sample Mean t Test, Raw DataMinitab® – 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
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:
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.
\(H_0\colon \mu = 3.0\)
\(H_a\colon \mu \ne 3.0\)
\(t (225) = 6.81\)
\(p < 0.0001\)
\(p \le \alpha\), reject the null hypothesis
There is convincing 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
8.2.3.2.2 - Minitab: 1 Sample Mean t Test, Summarized DataMinitab® – 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
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:
The shape of the population distribution is unknown, however with \(n \ge 30\) we can perform a one sample mean t test.
\(H_0\colon \mu = 72\)
\(H_a\colon \mu \ne 72\)
\(t (34) = 2.44\)
\(p = 0.0199\)
\(p \le \alpha\), reject the null hypothesis
There is convincing evidence that the population mean is different from 72.