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.