Question

Please answer all parts in detail :)

8. 15 adult males between the ages of 35 and 50 participated in a study to evaluate the effect of diet and exercise on blood cholesterol levels. The total cholesterol was measured in each subject initially and then three months after participating in an aerobic exercise program and switching to a low-fat diet. The data are shown in the following table. Blood Cholesterol Level Subject Before After 1 265 229 2 240 231 3 258 227 4 295 240 5 251 238 6 245 241 7 287 234 8 314 256 9 260 247 10 279 239 11 283 246 12 240 218 13 238 219 14 225 226 15 247 233 (a) (2 points) Calculate the sample mean and standard deviation of the differences between before and after cholesterol levels. (b) (3 points) Find P-value. (It can be a range.) (c) (5 points] Do the data support the claim that low-fat diet and aerobic exercise are of value in producing a mean reduction in blood cholesterol levels? Use a = 0.05. (Hint: use paired t-test) (d) [5 points] Calculate a one-sided confidence limit using a = 0.05. Do the data support the claim that low-fat diet and aerobic exercise are of value in producing a mean reduction in blood cholesterol levels?

Answer 1

Solution

We will use paired t test and the procedure is as follows

The following table is obtained:

	Before	After	Difference = Before - After
	265	229	36
	240	231	9
	258	227	31
	295	240	55
	251	238	13
	245	241	4
	287	234	53
	314	256	58
	260	247	13
	279	239	40
	283	246	37
	240	218	22
	238	219	19
	225	226	-1
	247	233	14
Average	261.8	234.93	26.867
St. Dev.	24.963	10.484	19.037
n	15	15	15

From the sample data, it is found that the corresponding sample means are: X1 = 261.8 X = 234.933 Also, the provided sample standard deviations are: 81 = 24.963 82 = 10.484 and the sample size is n = 15. For the score differences we have D = 26.867 8D = 19.037 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: ud = 0 Ha: up > 0 This corresponds to a right-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 14. Hence, it is found that the critical value for this right-tailed test is te = 1.761, for a= 0.05 and df = 14. The rejection region for this right-tailed test is R=t:t > 1.761. (3) Test Statistics The t-statistic is computed as shown in the following formula: D SD/n 26.867 19.037/15 = 5.466 (4) Decision about the null hypothesis Since it is observed that t = 5.466 > t = 1.761, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p= 0, and since p=0 < 0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that low-fat diet and aerobic exercise are of value in producing mean reduction in blood cholesterol levels​, at the 0.05 significance level.

Confidence Interval

The 95% confidence interval is 16.324<μD​<37.409.