8.2.3.2 - Minitab Express: One Sample Mean t Tests
8.2.3.2 - Minitab Express: One Sample Mean t TestsA hypothesis test for one group mean can be conducted in Minitab Express 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 Express to conduct a one sample mean t test using either raw data or summarized data. There is also one example of using Minitab Express 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 Express: 1 Sample Mean t Test, Raw Data
8.2.3.2.1 - Minitab Express: 1 Sample Mean t Test, Raw DataMinitabExpress – 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 \(n) = 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 Express using raw data:
- Open Minitab data set:
- On a PC: Select STATISTICS > One Sample > t
On a Mac: Select Statistics > 1-Sample Inference > t - 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
- Click the Options tab
- Use the default Alternative hypothesis of Mean ≠ hypothesized value
- Use the default Confidence level of 95
- Click OK
This should result in the following output:
N | Mean | StDev | SE Mean | 95% CI for \(\mu\) |
---|---|---|---|---|
226 | 3.23106 | 0.51040 | 0.03395 | (3.16416, 3.29796) |
\(\mu\): mean of GPA
Null hypothesis | H_{0}: \(\mu\) = 3 |
---|---|
Alternative hypothesis | H_{1}: \(\mu\) ≠ 3 |
T-Value | P-Value |
---|---|
6.81 | <0.0001 |
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:
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.
\(H_0: \mu = 3.0\)
\(H_a: \mu \ne 3.0\)
\(t (225) = 6.81\)
\(p < 0.0001\)
\(p \le \alpha\), reject the null hypothesis
There is evidence that the mean GPA in the population is different from 3.0
8.2.3.2.2 - Minitab Express: 1 Sample Mean t Test, Summarized Data
8.2.3.2.2 - Minitab Express: 1 Sample Mean t Test, Summarized DataMinitabExpress – 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.
- On a PC: Select STATISTICS > One Sample > t
On a Mac: Select Statistics > 1-Sample Inference > t - Change Sample data in column to Summarized data
- The Sample size is 35
- The Sample mean is 76.8
- The Sample standard deviation is 11.62
- Check the box Perform a hypothesis test
- For the Hypothesized mean enter 72
- Click the Options tab
- Use the default Alternative hypothesis of Mean ≠ hypothesized value
- Use the default Confidence level of 95
- Click OK
This should result in the following output:
N | Mean | StDev | SE Mean | 95% CI for \(\mu\) |
---|---|---|---|---|
35 | 76.800 | 11.620 | 1.964 | (72.808, 80.792) |
\(\mu\) : mean of Sample
Null hypothesis | H_{0}: \(\mu\) = 72 |
---|---|
Alternative hypothesis | H_{1}: \(\mu\) ≠ 72 |
T-Value | P-Value |
---|---|
2.44 | 0.0199 |
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:
The shape of the population distribution is unknown, however with \(n \ge 30\) we can perform a one sample mean t test.
\(H_0: \mu = 72\)
\(H_a: \mu \ne 72\)
\(t (34) = 2.44\)
\(p = 0.0199\)
\(p \le \alpha\), reject the null hypothesis
There is evidence that the population mean is different from 72.