- Minitab: Independent Means t Test

Here we will use Minitab to conduct an independent means t-test. Note that Minitab uses a more complicated formula for computing the degrees of freedom for this test.

Within Minitab, the procedure for obtaining the test statistic and confidence interval for independent means is identical.

Minitab 18

Minitab®  – Conducting an Independent Means t Test

Let's compare the mean SAT-Math scores of students who have and have not ever cheated. Both sample sizes are at least 30 so the sampling distribution can be approximated using the \(t\) distribution. 

  1. Open the Minitab file: class_survey.mpx
  2. Select Stat > Basic Statistics > 2 Sample t...
  3. Enter the variable SATM into the Samples box
  4. Enter variable Ever_Cheat into the Sample IDs box
  5. Click OK

This should result in the following output:

2-Sample t: SATM by Ever Cheat

\(\mu_1\): mean of SATM when Ever_Cheat = No
\(\mu_2\): mean of SATM when Ever_Cheat = Yes
Difference: \(\mu_1-\mu_2\)

Equal variances are not assumed for this analysis.

Descriptive Statistics: SATM
Ever_Cheat N Mean StDev SE Mean
No 163 604.0 86.9 6.8
Yes 53 583.7 79.2 11
Estimation of Difference
Difference 95% CI for Difference
20.3 (-5.2, 45.8)
Null hypothesis \(H_0\): \(\mu_1-\mu_2=0\)
Alternative hypothesis \(H_1\): \(\mu_1-\mu_2\neq0\)
T-Value DF P-Value
1.58 95 0.117

The result of our two independent means t test is \(t(95) = 1.58, p = 0.117\). Our p-value is greater than the standard alpha level of 0.05 so we fail to reject the null hypothesis. There is not enough evidence to state that the mean SAT-Math scores of students who have and have not ever cheated are different. 

Note that we could also interpret the confidence interval in this output. We are 95% confident that the mean difference in the population is between -5.16 and 45.78.

The example above uses a dataset. The following examples show how you can conduct this type of test using summarized data.