8.3.3 - Minitab: Paired Means Test

The 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

Descriptive Statistics
Sample	N	Mean	StDev	SE Mean
SATM	215	599.81	84.70	5.78
SATV	215	580.33	82.44	5.62

Estimation for Paired Difference
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)

Test
Null hypothesis	H₀: \(\mu\)_difference = 0
Alternative hypothesis	H₁: \(\mu\)_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

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?

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

Test
Null hypothesis	H₀: \(\mu_d\) = 0
Alternative hypothesis	H₁: \(\mu_d\) ≠ 0

T-Value	P-Value
3.18	0.002

The t test statistic is 3.18.

3. Determine the p value associated with the test statistic

From the output, the p value is 0.002

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 - 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 convincing 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

You should get the following output:

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	H₀: \(\mu\)_difference = 0
Alternative hypothesis	H₁: \(\mu\)_difference ≠ 0

T-Value	P-Value
5.80	0.000

Interpret the results