# 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.