Question

Consider the following data from a repeated-measures design. You want to use a repeated-measures t test to test the null hypothesis H_0: mu_D = 0 (the null 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: The mean difference score is M_D = For a repeated-measures t test, you need to calculate the t statistic, which requires you to calculate s and s_MD. What is the estimated standard deviation of the difference scores? s = Squareroot = Squareroot 6.80 = What is the estimated standard error of the mean difference scores? s_MD = s = What is the t statistic for the repeated-measures t test to test the null hypothesis Ho: mu_D = 0? t = M_D- = 1.37 You conduct a two-tailed test at alpha = .05. To use the Distributions tool to find the critical values, you first need to set the degrees of freedom in the tool. The degrees of freedom are The critical values (the values for t-scores that separate the tails from the main body of the distribution, forming the critical region) are. Finally, since the t statistic in the critical region, you the null hypothesis.

Answer 1

Answer

The completed table is as follows

Observation	A	B	D=B-A	D^2
1	1	2	1	1
2	3	3	0	0
3	4	6	2	4
4	6	5	-1	1
5	6	8	2	4
Total			4	10

Hence the mean difference score is $M_D=4/5=0.8$

Now estimated standard deviation of the different scores is

$s=\sqrt{\frac{\sum D^2-n.M_D^2}{n-1}}=\sqrt{\frac{6.8}{4}}=1.30384$

Now the estimated standard error of the mean difference scores,

$s_{MD}=\frac{s}{\sqrt{n}}=0.5831$

The t statistic to test the null hypothesis $H_0:\mu_D=0$

$t=\frac{M_D-\mu_0}{s_{MD}}$ here $\mu_0$ is the value of $\mu_D$ under Ho , hence 0

  $=\frac{0.8}{0.5831}=1.37$

Now to conduct a two tailed test the degrees of freedom are n-1=5-1=4

At level of significance $\alpha=0.05$ the critical values are -2.78 and 2.78

Finally, since the t-statistic does not belong to the critical region (since -2.78<1.37<2.78) Hence we accept null hypothesis based on the data.

Hnce the answer...

Thank you......