Download Paired-t test

Hypothesis Test for Comparing Two Population Means
Using Dependent Random Samples (Paired-t Test)
Example An assertiveness training course has just been added to the services offered by a counseling center. To measure its
effectiveness, ten students are given a test at the beginning of the course and again at the end. A high score on the test implies high
assertiveness. Do the data provide sufficient evidence to conclude that people are more assertive after taking the course? =.05
Before Course
50
62
51
41
63
56
49
67
42
57
After Course
65
68
52
43
60
70
48
69
53
61
Entering the data in Minitab:
Before
50
62
.
.
57
After
65
68
.
.
61
Difference
-15
-6
.
.
-4
Is Normality Assumption Reasonable?
MINITAB COMMANDS: STAT > BASIC STATISTICS > NORMALITY TESTS > VARIABLE Difference
Normal Probability Plot
.999
.99
Probability
.95
.80
.50
.20
.05
.01
.001
-15
-10
-5
0
DIFF
Average: -5.1
StDev: 6.26188
N: 10
Anderson-Darling Normality Test
A-Squared: 0.393
P-Value: 0.306
Since the plotted points fall approximately along a straight line, we can conclude that the normal distribution assumption for the
differences is reasonable. So, we can go ahead and conduct paired-t test.
Ho: d=before-after=0
Ha: d<0
Minitab Commands: STAT > BASIC STATISTICS > paired-t > Sample 1 Before
TEST MEAN 0 > OPTIONS > ALTERNATIVE > LESS THAN
Sample 2 After>
Minitab Output:
Paired T for Before - After
Before
After
Difference
N
10
10
10
Mean
53.80
58.90
-5.10
StDev
8.75
9.46
6.26
SE Mean
2.77
2.99
1.98
95% upper bound for mean difference: -1.47
T-Test of mean difference = 0 (vs < 0): T-Value = -2.58
P-Value = 0.015
Conclusion: p-value <  thus we reject Ho at =.05. We have sufficient evidence to
conclude that the population mean score for students before taking the course is less
than the mean score after taking the course.