Question

Question 3 (25 points) A study is being conducted to determine whether a particular exercise program is effective at reducing the mean blood cholesterol level on males between the ages of 35 and 50. Six subjects participated in the program, and their cholesterol levels before and after the program are as follows. 1 Blood Cholesterol Level Subject Before After 265 229 2 240 231 3 258 227 4 295 240 5 251 238 6 225 226 Using this data, we wish to test the following hypothesis using a level of significance of 0.05, where My denotes the mean blood cholesterol level before the program, and 42 denotes the mean blood cholesterol level after the program. Ho:M1 - U2 = 0 H1:41 - U2 > 0 Answer questions a through e assuming that you have chosen to use a paired t-test. Part a (4 points): State any assumptions that are required to perform this test. Part b (6 points): Identify the test statistic and critical region for the test. State the conclusion for this test. Part c (5 points): What is the P-value for this test? State the conclusion for this test. Part d (5 points): Construct a confidence interval that is relevant to this problem. Show that this confidence interval verifies your conclusion from part c. Part e (5 points): If the null hypothesis is true, what is the probability you conclude that H1 > Hz?

Answer 1

	Before	After	Difference =before - After
	265	229	36
	240	231	9
	258	227	31
	295	240	55
	251	238	13
	225	226	-1
Average	255.667	231.833	23.833
St. Dev.	23.864	5.845	20.615
n	6	6	6

sample means are:

Xˉ1​=255.667

Xˉ2​=231.833

Also, the provided sample standard deviations are:

s1​=23.864

s2​=5.845

and the sample size is n = 6. For the score differences we have

Dˉ=23.833

sD​=20.615

null and alternative hypotheses need to be tested:

Ho: μD​ = 0

Ha: μD​ > 0

significance level is α=0.05, and the degrees of freedom are df=5.

Hence, it is found that the critical value for this right-tailed test is tc​=2.015, for α=0.05 and df=5.

The rejection region for this right-tailed test is R=t:t>2.015.

Test Statistics

D t= 23.833 20.615/16 = 2.832 8p/vn

Since it is observed that 5t=2.832>tc​=2.015, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0183, and since p=0.0183<0.05, it is concluded that the null hypothesis is rejected.

Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is greater than μ2​, at the 0.05 significance level.

(nı – 1)si + (n2 – 1)s ni + n2 – 2 (6 – 1) 23.8642 + (6 – 1) x 5.8452 6 + 6 – 2 17.373 Since we assume that the population variances are equal, the standard error is computed as follows: 1 se — SpV ni n2 1 1 — 17.373 x + 6 6 = 10.03 Now, we finally compute the confidence interval: CI = (X – X – te X se, Xi - X +te X se) (255.667 – 231.833 – 2.228 x 10.03, 255.667 – 231.833 + 2.228 x 10.03) = (1.485, 46.183)