Question

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=PostTest−PreTestμd=PostTest-PreTest where the first data set represents a pre-test and the second data set represents a post-test. (Each row represents the pre and post test scores for an individual. Be careful when you enter your data and specify what your μ1μ1 and μ2μ2 are so that the differences are computed correctly.)

      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
51	57.6
57.6	52.9
72	64.8
57.6	66.8
58	82.4
66.3	58.5
50.4	38.4
70.4	57.8
39.4	-8.3
60.2	51.8
46.2	60.4
57.8	30.6
55.1	55
58.2	39.6
53.3	41.9
52.3	65.4
57.2	39.2
74.4	93.7
58.2	25.6
67	5.5
74.4	64.2
44.8	32.2
62.7	66.2
46.2	58.3

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.)

The p-value is...

• less than (or equal to) αα
• greater than αα

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the mean difference of post-test from pre-test is not equal to 0.
• There is not sufficient evidence to warrant rejection of the claim that the mean difference of post-test from pre-test is not equal to 0.
• The sample data support the claim that the mean difference of post-test from pre-test is not equal to 0.
• There is not sufficient sample evidence to support the claim that the mean difference of post-test from pre-test is not equal to 0.

Answer 1

Test Statistic :-
S(d) = √(Σ (di - d̅)2 / n-1)
S(d) = √( 9551.585 / 23) = 20.3786
d̅ = Σdi/n = -190.2 / 24 = -7.925
t = d̅ / ( S(d) / √(n) )
t = -7.925 / ( 20.3786 / √(24) )
t = -1.9052

P - value = P ( t > 1.9052 ) = 0.0693

Reject null hypothesis if P value < α level of significance
P - value = 0.0693 > 0.05, hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis

P vlaue is greater than α

This test statistic leads to a decision to.. Fail to reject null hypothesis

There is sufficient evidence to warrant rejection of the claim that the mean difference of post-test from pre-test is not equal to 0.