Question

Q1. A clinical trial is run to assess the effects of different forms of regular exercise on HDL levels in persons between the ages of 18 and 29. Participants in the study are randomly assigned to one of three exercise groups - Weight training, Aerobic exercise, or Stretching/Yoga – and instructed to follow the program for 8 weeks. Their HDL levels are measured after 8 weeks and are summarized below.

Weight Training

48
49
51
53
45
47
54
49
50
47
48
45
44
55
52
55
50

Aerobic Exercise

38
41
42
37
39
44
43
42
36
34
33
35
38
40
34
32
35

Stretching/Yoga

56
57
60
55
58
64
61
62
60
56
55
59
61
62
63
67
64

Is there a significant difference in mean HDL levels among the exercise groups? Run the test at a 5% level of significance. (10 points)  

REMEMBER: SHOW ALL WORK. NO WORK=NO CREDIT

a) Hypotheses (1 point) HO: vs. HA

b) Summary statstics (2 points)

Group	Exercise Group	N	Mean	Std Dev
1	Weight Training
2	Aerobic Exercise
3	Stretching/Yoga

c) MSB (1 point) MSB=

d) MSW (1 point) MSW=

e) Compute F-stat (3 points) F-stat=

f) P-value (1 point)

g) Conclusion (1 point)(circle one ) Accept H0 Reject H0 REMEMBER: SHOW ALL WORK. NO WORK=NO CREDIT

Answer 1

a)

Null and Alternative Hypothesis:

Ho: µ1 = µ2 = µ3

H1: At least one mean is different.

b)

For Weight Training :

∑x = 842, ∑x² = 41894, n1 = 17

Mean , x̅1 = Ʃx/n = 842/17 = 49.5294

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(41894-(842)²/17)/(17-1)] = 3.4481

For Aerobic Exercise :

∑x = 643, ∑x² = 24543, n2 = 17

Mean , x̅2 = Ʃx/n = 643/17 = 37.8235

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(24543-(643)²/17)/(17-1)] = 3.7289

For Stretching/Yoga :

∑x = 1020, ∑x² = 61396, n2 = 17

Mean , x̅3 = Ʃx/n = 1020/17 = 60.0000

Standard deviation, s3 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(61396-(1020)²/17)/(17-1)] = 3.5000

	Weight Training	Aerobic Exercise	Stretching/Yoga
Count	17	17	17
Mean, Sum/n	49.5294	37.8235	60
Standard deviation	3.4481	3.7289	3.5

c)

Number of treatment, k = 3

Total sample Size, N = 51

df(between) = k-1 = 2

df(within) = N-k = 48

df(total) = N-1 = 50

SS(between) = (x̅1)²*n1 + (x̅2)²*n2 + (x̅3)²*n3+(x̅4)²*n4 - (Grand Mean)²*N = 4184.588

SS(within) = (n1-1)*s1² + (n2-1)*s2² + (n3-1)*s3² + (n4-1)*s4² = 608.7059

SS(total) = SS(between) + SS(within) = 4793.294

MSB = MS(between) = SS(between)/df(between) = 2092.294

d)

MSW = MS(within) = SS(within)/df(within) = 12.68137

e)

F = MS(between)/MS(within) = 164.9896

f)

p-value = F.DIST.RT(164.9896, 2, 48) = 0.0000

Decision:

P-value < α, Reject the null hypothesis.