Question

In: Statistics and Probability

You wish to test the following claim (HaHa) at a significance level of α=0.05α=0.05. For the...

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.)

Solutions

Expert Solution

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


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