# 8.3.3 - Minitab: Paired Means Test

8.3.3 - Minitab: Paired Means TestThe steps for constructing a confidence interval or conducting a paired means \(t\) in Minitab are identical. The output that the procedure provides includes both the confidence interval and the \(p\)-value for determining statistical significance.

##
Minitab^{®}
– Conducting a Paired Means Test

Let's compare students' SAT-Math scores to their SAT-Verbal scores.

- Open the Minitab file: class_survey.mpx
- Select
*Stat > Basic Statistics > Paired t* - Select
*Each sample is in a column*since we have the data in the worksheet - Double click the variable
*SATM*in the box on the left to insert the variable into the*Sample 1*box - Double click the variable
*SATV*in the box on the left to insert the variable into the*Sample 2*box - Click OK

This should result in the following output:

### Paired t: SATM, SATV

Sample | N | Mean | StDev | SE Mean |
---|---|---|---|---|

SATM | 215 | 599.81 | 84.70 | 5.78 |

SATV | 215 | 580.33 | 82.44 | 5.62 |

Mean | StDev | SE Mean | 95% CI for \(\mu_d\) |
---|---|---|---|

19.49 | 89.81 | 6.12 | (7.42, 31.56) |

\(\mu\)_difference*: population mean of (SATM - SATV)*

Null hypothesis | H_{0}: \(\mu\)_difference = 0 |
---|---|

Alternative hypothesis | H_{1}: \(\mu\)_difference ≠ 0 |

T-Value | P-Value |
---|---|

3.18 | 0.002 |

On the next page, the five-step hypothesis testing procedure is used to interpret this output.

# 8.3.3.1 - Example: SAT Scores

8.3.3.1 - Example: SAT Scores## Example: SAT Scores

This example uses the dataset from Lesson 8.3.3 to walk through the five-step hypothesis testing procedure using the Minitab output.

**Research question: **Do students score differently on the SAT-Math and SAT-Verbal tests?

Because the sample size is large (\(n \ge 30\)), the *t* distribution may be used to approximate the sampling distribution.

\(H_{0}:\mu_d=0\)

\(H_{a}:\mu_d \ne 0\)

Null hypothesis | H_{0}: \(\mu_d\) = 0 |
---|---|

Alternative hypothesis | H_{1}: \(\mu_d\) ≠ 0 |

T-Value | P-Value |
---|---|

3.18 | 0.002 |

The t test statistic is 3.18.

From the output, the p value is 0.002

\(p\leq .05\), therefore our decision is to reject the null hypothesis

There is evidence that in the population, on average, students' SAT-Math and their SAT-Verbal scores are different.

# 8.3.3.2 - Example: Marriage Age (Summarized Data)

8.3.3.2 - Example: Marriage Age (Summarized Data)In a sample of 105 married heterosexual couples, the average age difference (husband's age - wife's age) was 2.829 years with a standard deviation of 4.995 years. These summary statistics were taken from a data set from the Lock^{5} textbook. Is there evidence that, on average, in the population, husbands tend to be older than their wives?

First we need to check our assumptions. In this case the sample size is greater than 30 so we can use the t-distribution.

We know n = 105, \(\mu_{\text{husband's age}}-\mu_{\text{wife's age}}=2.829\), and \(s=4.995\). Since we want to know if the husbands are older than their wives then our difference in ages would be positive. So our alternative hypothesis is \(H_a\colon \gt 0\).

To complete this using Minitab...

- Select
*Stat > Basic Statistics > Paired t* - Select
*Summarized data (differences)* - For sample size enter 105, enter 2.829 for sample mean and 4.995 for standard deviation.
- Select
*Options* - The
*Hypothesized difference should be 0. (or 0.0)* - Select
*Difference > hypothesized difference*for the*Alternative hypothesis* - Click OK and OK

##### Estimation for Paired Difference

Mean | StDev | SE Mean | 95% CI for \(\mu_d\) |
---|---|---|---|

105 | 2.829 | 4.995 | (1.862, 3.796 |

\(\mu\)_difference*: population mean of (Sample 1 - Sample 2)*

##### Test

Null hypothesis | H_{0}: \(\mu\)_difference = 0 |
---|---|

Alternative hypothesis | H_{1}: \(\mu\)_difference ≠ 0 |

T-Value | P-Value |
---|---|

5.80 | 0.000 |

### Interpret the results

Because the sample size is large (\(n \ge 30\)), the *t* distribution may be used to approximate the sampling distribution.

\(H_{0}:\mu_d=0\)

\(H_{a}:\mu_d \gt 0\)

Null hypothesis | H_{0}: \(\mu_d\) = 0 |
---|---|

Alternative hypothesis | H_{1}: \(\mu_d\) > 0 |

T-Value | P-Value |
---|---|

5.80 | 0.000 |

The t test statistic is 5.80.

From the output, the p value is 0.000

\(p\leq .05\), therefore our decision is to reject the null hypothesis

There is evidence that in the population, on average, the husband's age in heterosexual couples is greater than the wife's age.