Question

Question #3
Use P-Value method and 3% significance level, test the claim that the true mean weight loss produced by the three exercise programs have the same mean. Assume that the populations are normally distributed with the same variance.

Exercise A	Exercise B	Exercise C
5.6	6.8	9.3
8.1	4.9	6.2
4.3	3.1	5.8
9.1	7.8	7.1
7.1	1.2	7.9

Answer 1

The null hypothesis is : H0 : All the populations means are equal

The alternative hypothesis is : H1 : At least one mean is different from the remaining two.

The computations :-

Group totals:

	Exercise A	Exercise B	Exercise C
	5.6	6.8	9.3
	8.1	4.9	6.2
	4.3	3.1	5.8
	9.1	7.8	7.1
	7.1	1.2	7.9
Totals	34.2 = T10	23.8 = T20	36.3 = T30
Grand total = T00	94.3 = (T10  + T20 + T30 )

Now, The raw total SS is : (5.62 + 6.82 + ... + 7.92) = 662.21

Correction factor is : (T00)2/n = 94.32/15 = 592.833 and

(T10)2/5 +  (T20)2/5 +  (T30)2/5 = 610.754

So , Total SS : 662.21 - 592.833 = 69.377

SSB :  610.754 - 592.833 = 17.921

SSW : Total SS - SSB = 51.456

THE ANOVA TABLE IS GIVEN BELOW :

ANOVA
Source of Variation	SS	df	MS =SS/df	F = MSB/MSW	P-value	F crit = F0.05;2,12
Between Groups	17.921	(3-1)=2	8.960667	2.089708	0.16646	3.885294
Within Groups	51.456	(15-3)=12	4.288
Total	69.377	14 = (15-1)

Since the P value is greater than level of significance, we fail to reject the null hypothesis and conclude that the true mean weight loss produced by the three exercise programs have the same mean.