8.3.3 - Minitab Express: Paired Means Test
8.3.3 - Minitab Express: Paired Means TestThe 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.
- Open the Minitab Express file:
- On a PC: In the menu bar select STATISTICS > Two Samples > Paired t
- On a Mac: In the menu bar select Statistics > 2-Sample Inference > Paired t
- 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:
Sample | N | Mean | StDev | SE Mean |
---|---|---|---|---|
SATM | 215 | 599.814 | 84.700 | 5.776 |
SATV | 215 | 580.326 | 82.436 | 5.622 |
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 | H_{0}: \(\mu_d\) = 0 |
---|---|
Alternative hypothesis | H_{1}: \(\mu_d\) ≠ 0 |
T-Value | P-Value |
---|---|
3.18 | 0.0017 |
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 ScoresExample: 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?
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 | 0.0017 |
The t test statistic is 3.18.
From the output, the p value is 0.0017
\(p\leq .05\), 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 - 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 Lock^{5} textbook. Is there evidence that, on average, in the population, husbands tend to be older than their wives?