Question

The following data represent the results from an independent-measures experiment comparing three treatment conditions. Use an analysis of variance with α = .05 to determine whether these data are sufficient to conclude that there are significant differences between the treatments (Hint: start by filling in the missing values on the table below).

Be sure to show all formulas with symbols (and plug in numbers), steps, processes and calculations for all steps and all parts of all answers.

Treatment 1	Treatment 2	Treatment 3
2	4	8
3	1	4
5	3	6
3	2	5	N = 12
M = 3.25	M = 2.5	M = 5.75	G =
SS =	SS =	SS =	Ex2 =

Answer 1

	Treatment 1	Treatment 2	Treatment 3
Sum	13	10	23
Count	4	4	4
Mean, Sum/n	3.25	2.5	5.75
Sum of square, Ʃ(xᵢ-x̅)²	4.75	5	8.75

N = 12

G = (M1 + M2 + M3)/3 = (3.25+2.5+5.75)/3= 3.8333

∑x² =2²+3²+ 5²+...… +6²+5² = 218

Number of treatment, k = 3

Total sample Size, N = 12

df(between) = k-1 = 2

df(within) = N-k = 9

df(total) = N-1 = 11

SS(between) = (M1)²*n1 + (M2)²*n2 + (M3)²*n3 - (G)²*N = 23.1667

SS(within) = SS1 + SS2 + SS3 = 4.75+5+8.75 =18.5

SS(total) = ∑x² - (∑x)²/n = 218 - (13+10+23)²/12 = 41.6667

MS(between) = SS(between)/df(between) = 11.5833

MS(within) = SS(within)/df(within) = 2.0556

F = MS(between)/MS(within) = 5.6351

p-value = F.DIST.RT(5.6351, 2, 9) = 0.0259

ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	23.1667	2	11.5833	5.6351	0.0259
Within Groups	18.5000	9	2.0556
Total	41.6667	11

Null and Alternative Hypothesis:

Ho: µ1 = µ2 = µ3

H1: At least one mean is different.

Test statistic:

F =5.6351

p-value = 0.0259

Decision:

P-value < α, Reject the null hypothesis.

There is sufficient evidence to conclude that there are significant differences between the treatments.

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