Question

A group of adults with elevated cholesterol levels participated in a diet and exercise program to determine the effect of this program on their cholesterol. The resulting data follows:

Subject	Before	After
1	259	244
2	260	250
3	272	246
4	260	235
5	281	273
6	254	245

Test the hypothesis that the diet and exercise program lowered the subject's total chloresterol by at least
15 points. alpha = .05.
a. Find the mean of the differences.  
b. Find the standard deviation of the differences.  
c. Find the lower confidence limit of the differences.  
d. Find the upper confidence limit of the differences.  
e. According to the confidence interval, the null hypotesis that the difference is equal to 20 cannot be rejected.

TRUE

FALSE   

Please show work

Answer 1

Subject	Before	After	Difference
1	259	244	15
2	260	250	10
3	272	246	26
4	260	235	25
5	281	273	8
6	254	245	9
		Total	93

Sample size = n = 6

a) Sample mean of difference =    = 15.5

b) Standard deviation of difference = = 8.1179

c) We have to construct 95% confidence interval.

Formula is

Here E is a margin of error.

Degrees of freedom = n - 1 = 5 - 1 = 4

Level of significance = 0.05

tc = 2.571 ( Using t table)

So confidence interval is ( 15.5 - 8.5192 , 15.5 + 8.5192) = > ( 6.9808 , 24.0192)

c) Lower confidence limit = 6.9808

d) Upper confidence limit = 24.0192

e) TRUE because 20 is lies into interval.