Question

You wish to test the following claim (HaHa) at a significance level of α=0.10α=0.10. 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
56.5	69.8
47.8	50.2
51.2	41.4
49.7	65.9
45.6	28.5
54.8	52.9
62.3	50.9
52.3	52.7
52.3	46.5
56	58.7
38.7	47.7
37.7	44.5
52.8	56.2
51.8	76.4
44.7	34.1
58.9	55.7
46	29.6

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

Here, the null hypothesis is H0: d = 0 and the alternate hypothesis Ha: d ≠ 0

where d is the mean difference in scores between pre - test and post - test

Answer

The Excel Output of the above data is given below -

and the formulas used to calculate the measures are given below -

(where C3, D3 etc. are cell numbers)

Therefore,

Number of observations, n = 17

Mean of Differences, = 0.1529

Standard Deviation of differences, s' = 11.5740

and already it is given that the population of differences follows a Normal Distribution

The test statistic t = (n^0.5)( - d) / s' follows t - distribution with (n - 1) degrees of freedom under the null hypothesis

Substituting all values, t = 0.0545

The P - Value is 0.9572

Generally if the P - Value is lesser than the Significance level, we reject the null hypothesis else we fail to reject the null. We accept the null

In this case,

The P - Value is > Significance Level (= 0.1)

The test statistic leads to a decision to accept the null

As such the final conclusion is -

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.