# 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 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*

Null hypothesis | H |
---|---|

Alternative hypothesis | H |

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 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*

Null hypothesis | H |
---|---|

Alternative hypothesis | H |

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.