# 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 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:

1. In Minitab, select Stat > Basic Statistics > 1-sample t
2. Select Summarized data from the dropdown
3. Enter 35 for the sample size, 76.8 for the sample mean and 11.62 for the standard deviation.
4. Check the box Perform a hypothesis test
5. For the Hypothesized mean enter 72
6. Select Options
7. Use the default Alternative hypothesis of Mean ≠ hypothesized value
8. Use the default Confidence level of 95
9. 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 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 evidence that the population mean is different from 72.

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