Question

Dr. P is trying to determine increase flexibility of patients struggling with arthritis in their hips. He is using water therapy to see if flexibility in patients improves over time, specifically with hamstring flexibility. 0 AND Ha: μ Set up the null and alternative hypothesis for this t test. Ho: μ The change in flexibility for each patient in each program is below (flexibility measured by having patients do the "sit and reach test" and the distance they could reach past their toes is measured in inches. if they couldn't reach their toes the inches are shown as negative Patient 1 2 3 45678 9 10 before -2 3 -4 01-12-3-1 0 after 1 -3 -2 1 00 0-2 01 (0 Determine the test statistic& P-value: t =-0.7589 Fill in the blanks for the conclusion of the test: We (reject/fail to reject) fail to reject evidence that the mean flexibility of the patients improved. p= 0.9673 the null hypothesis at the .o5 level. We (do/do not) do have significant

Answer 1

Below are the null and alternative hypothesis
Null Hypothesis, H0: μ <= 0
Alternate Hypothesis, Ha: μ > 0

Before	After	dbar
-2	-1	-1
-3	-3	0
-4	-2	-2
0	1	-1
1	0	1
-1	0	-1
2	0	2
-3	-2	-1
-1	0	-1
0	1	-1

dbar = μ(after) - μ(before)	0.5
s(dbar)	1.1785
SE = s(dbar)/sqrt(n)	0.3727
Test Statisitcs, t = dbar/SE	1.3416
degrees of freedom = n-1	9
p-value	0.1063

We fail to reject the null hypothesis at 0.05 level. We do not have significant evidence that the mean flexibility of the patients improved.