For the homework what is Test statistic and P value? You wish to test the following claim (\(H_a\)) at a significance level of \(\alpha = 0.02\). For the context of this problem, \(\mu_d = \text{PostTest} - \text{PreTest}\) 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 \(\mu_1\) and \(\mu_2\) are so that the differences are computed correctly.)\[H_0 : \mu_d = 0\]\[H_a : \mu_d \neq 0\]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 ||----------|-----------|| 42.6 | 60.3 || 35.9 | 30.9 || 35.5 | 38.2 || 42.3 | 36.5 || 38.5 | 36.2 || 47.1 | 67.9 || 35.9 | 39.3 || 48.2 | 60.5 || 50.4 | 45.7 || 50.4 | 56.1 || 35.5 | 39.4 || 48.8 | 45.2 || 45.1 | 56.4 || 36.9 | 24.5 || 35.5 | 27.8 |What is the test statistic for this sample?\[ \text{test statistic} = \_\_\_\_\_ \](Report answer accurate to 4 decimal places.)What is the p-value for this sample?\[ \text{p-value} = \_\_\_\_\_ \](Report answer accurate to 4 decimal places.)

Suppose μd is the difference between mean scores of PostTest and PreTest.That is, μd = μPostTest – μPreTest.The values are considered for the same observations in two different scenarios. One for pre-test and another for post-test. The observations are related. Thus, it is suitable to use paired t test.The claim is that there is a significant difference between the mean scores of PostTest and PreTest.The null and alternative hypotheses are:H0: μd = 0Ha: μd ≠ 0The test is both tailed.The sample size is n = 15.Consider the below table.The mean difference is,The sample standard deviation of the difference is,The test statistic is: The test statistic is 0.8819.The degrees of freedom is, df = n – 1 = 15 – 1 = 14.Now, using the EXCEL formula =T.DIST.2T(0.8819,14) the p-value is obtained as approximately 0.3927.The excel output is, Decision rule:If p-value is less than or equal to the level of significance, reject the null hypothesis. Otherwise, failed to reject the null hypothesis.Conclusion:The p-value of 0.3927 is greater than the level of significance of α = 0.02.Fail to reject H0. The data do not provide convincing evidence to support the claim that there is a significant difference between the mean scores of PostTest and PreTest, at 2% level of significance.Test statistic = 0.8819p-value = 0.3927Fail to reject H0. The data do not provide convincing evidence to support the claim that there is a significant difference between the mean scores of PostTest and PreTest, at 2% level of significance.