Final Project Instructions for Performing TwoSample t Tests and Paired t Tests Using Minitab
MTH 1210: Introduction to Statistics
These instructions are for performing a twosample t test or a paired t test. To decide which test is appropriate for your project, you need to decide whether your data consists of two independent samples (i.e. the values in one sample are not related to those in the other sample) or paired samples (each observation in one sample can be matched with an observation in the other, for example if the data consist of prices of the same 30 items at two different stores). See also below, where the assumptions required for each test are described.
Numerical examples of each type of test are given below, along with the Minitab output and a writeup of the conclusion based on the values produced in the output.
For more Minitab instructions for performing statistical inference (onesample confidence intervals and hypothesis tests), see the file minitab_inference.doc on the course website.
An engineer in a garment factory must compare two different work sequences for measuring the shear strength of polyester fibers to see if one sequence is, on average, faster than the other. Two groups of 6 employees each measure the shear strength of the fabric, the first group using Work Sequence 1 and the second using Work Sequence 2.
The following data represent the 6 completion times (in seconds) for each group:
Group 1
(Work Sequence 1) 
Group 2
(Work Sequence 2) 
220  247 
235  223 
214  215 
197  219 
206  207 
214  236 
A twosample t test was performed with this data in Minitab using the instructions below. The null hypothesis was that there is no difference between the mean times for the two work sequences, and since the engineer had no prior idea as to which work sequence he suspected would be quicker, the alternative hypothesis was twosided (i.e. “not equal”). The Minitab output is:
TwoSample TTest and CI: Work Sequence 1, Work Sequence 2
Twosample T for Work Sequence 1 vs Work Sequence 2
N Mean StDev SE Mean Work Sequence 1 6 214.3 12.9 5.3 Work Sequence 2 6 224.5 14.6 6.0
Difference = mu (Work Sequence 1) – mu (Work Sequence 2) Estimate for difference: 10.1667 95% CI for difference: (28.1544, 7.8211) TTest of difference = 0 (vs not =): TValue = 1.28 PValue = 0.233 DF = 9 
From the output, the summary statistics are:
The computed value of the test statistic is
,
and the Pvalue is 0.233. Thus, using a level of significance α = 0.05, since the Pvalue is greater than α, we fail to reject the null hypothesis. Apparently there is no significant difference between the true mean completion times for the two work sequences, and the observed difference between sample means (10.1667) can be explained by chance variation.
Suppose that the engineer the previous example instead conducts his study in the following manner: each of 8 employees measures the shear strength of the fabric using both Work Sequence 1 and Work Sequence 2, so that observations in the two samples can be matched according to which employee the observation represents.
The following data represent the 8 pairs of completion times (in seconds), along with the differences (Work Sequence 1 minus Work Sequence 2):
Employee  Work Sequence 1  Work Sequence 2  Difference 
1  225  230  5 
2  215  223  8 
3  210  215  5 
4  197  197  0 
5  202  207  5 
6  200  226  26 
7  236  230  6 
8  225  224  1 
Before performing the paired t test, it’s useful to look at a histogram of the 8 differences. See below for instructions for computing the differences in Minitab from the two data columns.
A paired t test was performed with this data in Minitab using the instructions below. The null hypothesis was that there is no difference between the mean times for the two work sequences, and since the engineer had no prior idea as to which work sequence he suspected would be quicker, the alternative hypothesis was twosided (i.e. “not equal”). The output is
Paired TTest and CI: Work Sequence 1, Work Sequence 2
Paired T for Work Sequence 1 – Work Sequence 2
N Mean StDev SE Mean Work Sequence 1 8 213.750 13.997 4.949 Work Sequence 2 8 219.000 11.808 4.175 Difference 8 5.25000 9.49812 3.35809
95% CI for mean difference: (13.19063, 2.69063) TTest of mean difference = 0 (vs not = 0): TValue = 1.56 PValue = 0.162

From the output, the summary statistics for the two samples are:
and the summary statistics for the differences are:
The computed value of the test statistic is
,
and the Pvalue is 0.162. Thus, using a level of significance α = 0.05, since the Pvalue is greater than α, we fail to reject the null hypothesis. Apparently there is no difference between the true mean completion times for the two work sequences and the observed sample mean of the differences (5.25) can be explained by chance variation.
Minitab Instructions
Assumptions:
Either
or,
Make sure the data are in two separate columns, each column representing a sample.
In the dialog box: Click the circle next to “Samples in different columns.”
FIRST: Select the variable (double click).
SECOND: Select the variable (double click).
< OPTIONS >
TEST DIFFERENCE: Enter the nullhypothesized value (if
different from 0.0).
ALTERNATIVE: Choose the form of the alternative hypothesis from the dropdown list using the following guideline (the subscripts 1 and 2 correspond to the variables previously chosen as FIRST and SECOND):
Hypothesis  Minitab Notation  
Null  H_{0}: µ_{1 – }µ_{2} = 0  
Alternative  H_{a}: µ_{1 – }µ_{2} < 0  less than 
H_{a}: µ_{1 – }µ_{2} ≠ 0  not equal  
H_{a}: µ_{1 – }µ_{2} > 0  greater than 
< OK >
< OK >
The desired calculation will appear in the Session window.
Assumptions: Data consists of two samples of size n that can be paired in a meaningful way (for example if there are two measurements on each of n individuals) so that a difference can be computed for each pair
and either
or,
Make sure the data are in two separate columns, each column representing a sample and each row representing a distinct pair.
In the dialog box: Click the circle next to “Samples in columns.”
FIRST SAMPLE: Select the variable (double click).
SECOND SAMPLE: Select the variable (double click).
< OPTIONS >
TEST MEAN: Enter the nullhypothesized value (if different from 0.0).
ALTERNATIVE: Choose the form of the alternative hypothesis from the dropdown list using the following guideline (the subscripts 1 and 2 correspond to the variables previously chosen as FIRST SAMPLE and SECOND SAMPLE):
Hypothesis  Equivalent Hypothesis  Minitab Notation  
Null  H_{0}: µ_{1 – }µ_{2} = 0  H_{0}: µ_{d} = 0  
Alternative  H_{a}: µ_{1 – }µ_{2} < 0  H_{a}: µ_{d} < 0  less than 
H_{a}: µ_{1 – }µ_{2} ≠ 0  H_{a}: µ_{d} ≠ 0  not equal  
H_{a}: µ_{1 – }µ_{2} > 0  H_{a}: µ_{d} > 0  greater than 
< OK >
< OK >
The desired calculation will appear in the Session window.
Computing the Difference between Values in Two Columns
When two columns in a Minitab worksheet represent samples of data that can be paired, it’s often useful to compute a column of differences, for example to examine a histogram of the differences, etc.
The following commands create a variable called “Difference” that’s the difference between two hypothetical variables called “Pretest Score” and “Posttest Score”:
Type the variable name “Difference” at the top of a vacant column.
CALC > CALCULATOR
In the dialogue box: STORE RESULT IN VARIABLE: Enter “Difference” (double click).
EXPRESSION: Enter ‘Pretest Score’ – ‘Posttest Score’ (by typing or
clicking).
< OK >
The differences should now appear in the appropriate column.
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