Question

Use the following data set – Pain Relief Study - to answer this series of questions.

Participant	Before Treatment	After Treatment
1	90	80
2	70	50
3	60	60
4	90	70
5	70	60
6	50	40
7	70	60

Patients in a drug trial rate their pain before treatment and after treatment. Higher scores equate to higher levels of reported pain with a maximum possible score of 100.

11.1 Calculate MD for the Pain Relief Study

11.2 Calculate the t-value (tobtained) for the Pain Relief Study (SHOW WORK) 1

1.3 Calculate the 95% CI for the Pain Relief Study.

11.4 What is one problem you might encounter if you use a within-subjects design compared to a between-subjects design? Explain what the problem is, explain why it is a problem, and give an example.

Answer 1

Solution;-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d = 0

Alternative hypothesis: u_d ≠ 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

11.1)

s = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 6.9007

SE = s / sqrt(n)

S.E = 2.6082

DF = n - 1 = 7 -1

D.F = 6

11.2)

t = [ (x₁ - x₂) - D ] / SE

t = 4.38

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 6 degrees of freedom is more extreme than 4.38; that is, less than - 4.38 or greater than 4.38.

Thus, the P-value = 0.005.

Interpret results. Since the P-value (0.005) is less than the significance level (0.05), we have to reject the null hypothesis.

3) 95% CI for the Pain Relief Study is C.I = (- 2.097 , 24.954).

C.I = 11.42857 + 1.96*6.9007

C.I = 11.42857 + 13.5253

C.I = (- 2.097 , 24.954)