8.3.3 - Minitab Express: Paired Means Test

8.3.3 - Minitab Express: Paired Means Test

The steps for constructing a confidence interval or conducting a paired means $t$ in Minitab Express are identical. The output that procedure provides includes both the confidence interval and the $p$-value for determining statistical significance.

MinitabExpress – Conducting a Paired Means Test

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

1. Open the Minitab Express file:
2. On a PC: In the menu bar select STATISTICS > Two Samples > Paired t
3. On a Mac: In the menu bar select Statistics > 2-Sample Inference > Paired t
4. Double click the variable SATM in the box on the left to insert the variable into the Sample 1 box
5. Double click the variable SATV in the box on the left to insert the variable into the Sample 2 box
6. Click OK

This should result in the following output:

Paired t: SATM, SATV
Descriptive Statistics
Sample N Mean StDev SE Mean
SATM 215 599.814 84.700 5.776
SATV 215 580.326 82.436 5.622
Estimation for Paired Difference
Mean StDev SE Mean 95% CI for $\mu_d$
19.488 89.808 6.125 (7.416, 31.561)

$\mu_d$ : mean of (SATM - SATV)

Null hypothesis H0: $\mu_d$ = 0 H1: $\mu_d$ ≠ 0
T-Value P-Value
3.18 0.0017
Video Walkthrough

Select your operating system below to see a step-by-step guide for this example.

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 Express output.

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

1. Check assumptions and write hypotheses

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

$H_{0}:\mu_d=0$
$H_{a}:\mu_d \ne 0$

2. Calculate the test statistic
Null hypothesis H0: $\mu_d$ = 0 H1: $\mu_d$ ≠ 0
T-Value P-Value
3.18 0.0017

The t test statistic is 3.18.

3. Determine the p value associated with the test statistic

From the output, the p value is 0.0017

4. Make a decision

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

5. State a "real world" conclusion

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

8.3.3.2 - Video Example: Marriage Age (Summarized Data)

8.3.3.2 - Video 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 Lock5 textbook. Is there evidence that, on average, in the population, husbands tend to be older than their wives?

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