A one-way ANOVA was conducted at a = 0.05. The study involved three (3) different groups with 12 subjects in each. The means for each group were as follows: M1 = 10.92, M2=13.83, and M3 = 14.92. The sum of squares values is shown in the table below.
Source	SS	df	MS	F	p
Between-treatments	102.72				<.05
Within-treatments	229.5
Total	332.22
a. Assuming = 0.05, complete the rest of the table
	M1	M2	M3	Total
	10.92	13.83	14.92	39.67
n	12	12	12	36
Mean	0.91	1.15	1.24	1.10
df total = 3*12 - 1 =	35
DfC = C - 1 =	2
Dfe= N - C =	33
F critical value=	3.28
MS(between) =	SS(between)/df(between)
	102.72/2
	51.36
MS(within) =	SS(within)/df(within)
	229.5/33
	6.95
F =	MS(between)/MS(within)
	51.36/6.95
	7.39
Source	SS	df	MS	F	p
Between-treatments	102.72	2	51.36	7.39	<.05
Within-treatments	229.5	33	6.95
Total	332.22
The decision to be made is to reject the null hypothesis because the observed value of 7.39 is greater than the critical F-value which is 3.28. It means that there is a significant difference between the means of three groups.
b. Conduct a Tukey's HSD and indicate which means differ significantly.		(I did the answering on A I just need the B.) thank you .