Please help with all parts of this question Consider the following data from a repeated-measures design. You want to use a repeated-measures t test to test the null hypothesis Ho: PD = 0 (the%3Dnull hypothesis states that the mean difference for the general population is zero). The data consist of five observations, each with two measurements,A and B, taken before and after a treatment. Assume the population of the differences in these measurements are normally distributed.Complete the following table by calculating the differences and the squared differences:Difference ScoreSquared Difference ScoreObservationA(D = B - A)(D2)233.4.6.4.6.5.6.8.The mean difference score is Mp =For a repeated-measures t test, you need to calculate the t statistic, which requires you to calculate s and sMDWhat is the estimated standard deviation of the difference scores?S = V6.80/What is the estimated standard error of the mean difference scores? (Note: For best results, retain at least six decimal places from your calculation ofs.)SMDs/%3DWhat is the t statistic for the repeated-measures t test to test the null hypothesis Ho: Pp = 07Mp1.37t DistributionDegrees of Fedom 7pe here to searchF1DIIESF2F3F4F5F6PrtScnHomeF7F8F9@#32$&23 The mean difference score is Mp =For a repeated-measures t test, you need to calculate the t statistic, which requires you to calculate s and syD.What is the estimated standard deviation of the difference scores?6.80/S =What is the estimated standard error of the mean difference scores? (Note: For best results, retain at least six decimal places from your calculation ofs.)SMDs/%3DWhat is the t statistic for the repeated-measures t test to test the null hypothesis Ho: Pp = 0?t =Mp%3D1.37t DistributionDegrees of Freedom = 750002500S004.0-3.0-2.0-1.00.01.00.7112.03.04.00.711You conduct a two-tailed test at a = .05. To use the Distributions tool to find the critical values, you first need to set the degrees of freedom in thetool. The degrees of freedom areThe critical values (the values for t scores that separate the tails from the main body of the distribution, forming the critical region) areFinally, since the t statisticin the critical region, youthe null hypothesis.

The data shows the five observations of each with two measurements A and B.The differences and the squared differences is, Observation A B D=B-A D2 1 1 2 2-1=1 12=1 2 3 3 3-3=0 02=0 3 4 6 6-4=2 22=4 4 6 5 5-6=-1 -12=1 5 6 8 8-6=2 22=4 The mean difference score is, MD=∑Dn=1+0+2-1+25=45=0.8 Therefore, the mean difference score is 0.8.The estimated standard deviation of the difference scores is, s=∑D2-∑D2nn-1=1+0+4+1+4-1+0+2-1+2255-1=10-1654=6.84=1.7=1.3038 Therefore, the estimated standard deviation of the difference scores is 1.3038.The estimated standard error of the mean difference scores is, sMD=sn=1.30385=0.5831 Therefore, the estimated standard error of the mean difference scores is 0.5831. The t statistic is, t=MD-μDsMD=0.8-00.5831=1.37 Therefore, the t statistic is 1.37.The degrees of freedom is, df=n-1=5-1=4 Therefore, the degrees of freedom is 4. The distribution plot is obtained by changing the degrees of freedom to 4. The critical values are ±2.776.Conclusion: The test statistic is less than the critical value. Based on the critical region, fail to reject the null hypothesis. Finally, since the t statistic is not in the critical region, you fail to reject the null hypothesis.