In: Statistics and Probability
You wish to test the following claim (HaHa) at a significance
level of α=0.05α=0.05. For the context of this problem,
μd=μ2−μ1μd=μ2-μ1 where the first data set represents a pre-test and
the second data set represents a post-test.
Ho:μd=0Ho:μd=0
Ha:μd>0Ha:μd>0
You believe the population of difference scores is normally
distributed, but you do not know the standard deviation. You obtain
the following sample of data:
| pre-test | post-test |
|---|---|
| 52 | 69 |
| 51.6 | 34.7 |
| 48.3 | 55.4 |
| 31.6 | 46.8 |
| 46.4 | 42.6 |
| 31.6 | 27.1 |
| 49.3 | 49.3 |
| 33.1 | 57.3 |
| 22.3 | 44.4 |
| 56.6 | 43.6 |
| 62.5 | 63.7 |
| 66.4 | 118.1 |
| 62.5 | 64.3 |
| 31.6 | 49.2 |
| 41.8 | 81.2 |
| 52 | 53.8 |
| 65.7 | 106.5 |
| 55.4 | 94.8 |
| 67.3 | 66.7 |
| 45 | 43.8 |
| 35 | 30.5 |
| 27.6 | 37 |
| 56.6 | 33.4 |
| 39 | 63.9 |
What is the test statistic for this sample?
test statistic = (Report answer accurate to 4 decimal
places.)
What is the p-value for this sample?
p-value = (Report answer accurate to 4 decimal
places.)
Ho:μd=0
Ha:μd>0
| Number | Post Test | Pre Test | Difference |
![]() |
| 69 | 52 | 17 | 45.6187674 | |
| 34.7 | 51.6 | -16.9 | 736.896267 | |
| 55.4 | 48.3 | 7.1 | 9.89626736 | |
| 46.8 | 31.6 | 15.2 | 24.5437674 | |
| 42.6 | 46.4 | -3.8 | 197.285434 | |
| 27.1 | 31.6 | -4.5 | 217.439601 | |
| 49.3 | 49.3 | 0 | 104.977101 | |
| 57.3 | 33.1 | 24.2 | 194.718767 | |
| 44.4 | 22.3 | 22.1 | 140.521267 | |
| 43.6 | 56.6 | -13 | 540.368767 | |
| 63.7 | 62.5 | 1.2 | 81.8271007 | |
| 118.1 | 66.4 | 51.7 | 1718.44793 | |
| 64.3 | 62.5 | 1.8 | 71.3321007 | |
| 49.2 | 31.6 | 17.6 | 54.0837674 | |
| 81.2 | 41.8 | 39.4 | 849.965434 | |
| 53.8 | 52 | 1.8 | 71.3321007 | |
| 106.5 | 65.7 | 40.8 | 933.557101 | |
| 94.8 | 55.4 | 39.4 | 849.965434 | |
| 66.7 | 67.3 | -0.6 | 117.632101 | |
| 43.8 | 45 | -1.2 | 131.007101 | |
| 30.5 | 35 | -4.5 | 217.439601 | |
| 37 | 27.6 | 9.4 | 0.71543403 | |
| 33.4 | 56.6 | -23.2 | 1118.62377 | |
| 63.9 | 39 | 24.9 | 214.744601 | |
| Total | 1377.1 | 1131.2 | 245.9 | 8642.93958 |
Test Statistic :-
t = 2.5893
Test Criteria :-
Reject null hypothesis if
Result :- Reject null hypothesis
Decision based on P value
P - value = P ( t > 2.5893 ) = 0.0082
Reject null hypothesis if P value <
level of significance
P - value = 0.0082 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis