Question

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.

Table: HDL mean and standard deviation results for participants of the clinical trial by exercise group. (N=60)

Exercise Group	N	Mean	Std Dev
Group 1: Weight Training	20	49.7	10.2
Group 2: Aerobic Exercise	20	43.1	11.1
Group 3: Stretching/Yoga	20	57.0	12.5

Source	Sums of Squares	Df	Mean Squares	F	P value
Between	1933.8
Within	21860
Total

Write out the Null Hypothesis of no difference in means between the groups.

Write out the Alternate Hypothesis?

What is the F statistic?

What is the approximate critical F value for this problem at an a=.05?

Is the p value greater than or less than .05?

Do you reject or accept the null at an a=.05?

Answer 1

Null Hypothesis

Alternative Hypothesis H1: Atleast two means are different

Source	Sum of Squares	Df	Mean Squares	F	P Value
Between	1933.8	k-1 =3-1 =2	1933.8/2=966.9	2.521	0.089
Within	21860	N-k=60-3=57	21860/57 =383.51
Total	23793.8	N-1 =60-1=59

• The F statistic is =966.9/383.51 = 2.521
• Degrees of Freedom , (numerator, denominator) = (2,57)  
• Level of significance,
• The critical value of F is 3.159
• Since p value is greater than the significance level , i.e. (0.08>0.05)
• We accept Null hypothesis at 5% significance level.