Question

How do I even begin to solve this using R statistical software?
A random sample of eight pairs of twins was randomly assigned to treatment A or treatment B. The data are given in the following table:

Twins	1	2	3	4	5	6	7	8
Treatment A	48.3	44.6	49.7	40.5	54.3	55.6	45.8	35.4
Treatment B	43.5	43.8	53.7	43.9	54.4	54.7	45.2	34.4

What is the p-value of the Wilcoxon signed-rank test?

Is there any significant evidence that the two treatments differ using an α = 0.05 and Wilcoxon-signed rank test.

Compare the p-value of the Wilcoxon-signed rank test and paired t-test.

Answer 1

> treatment_A=c(48.3,44.6,49.7,40.5,54.3,55.6,45.8,35.4)
> treatment_B=c(43.5,43.8,53.7,43.9,54.4,54.7,45.2,34.4)
> wilcox.test(treatment_A, treatment_B, paired = TRUE, alternative = "two.sided")

Wilcoxon signed rank test

data: treatment_A and treatment_B
V = 22, p-value = 0.6406
alternative hypothesis: true location shift is not equal to 0

As p-value = 0.6406 >0.05 we accept the null hypothesis and conclude there is no significant difference between two treatments.

# let's perfom paired t-test

> t.test(treatment_A, treatment_B, paired = TRUE, alternative = "two.sided")

Paired t-test

data: treatment_A and treatment_B
t = 0.076822, df = 7, p-value = 0.9409
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-2.233538 2.383538
sample estimates:
mean of the differences
0.075

Here p-value = 0.9409 >0.05 so we accept the null hypothesis and conclude that their is no significant difference between two treatments.

Compare the p-value of the Wilcoxon-signed rank test and paired t-test.

p-value for paired t-test is 0.9409 and Wilcoxon-signed rank test p-value = 0.6406 so p-value for paired t-test is high as compare to Wilcoxon-signed rank test.

And also indicates that paired t-test has strong evidence for accepting null hypothesis.