Question

Q1. The quantitative reasoning scores on the GRE were recorded for students admitted to three different graduate programs as below.

(a): Construct an ANOVA table for the data using Completely Randomized Design?

(b): If you find a significant difference in the average scores for the three programs, Use Tukey’smethod for paired comparisons to determine which means differ significantly. Use α = .05.

Life	Physical	Social
630	660	440
640	640	330
470	720	670
600	690	570
580	530	590
660	760	540
660	670	4 50
480	700	570
650	710	530
710	450	630

Answer 1

	Life	Physical	Social	Total
Sum	6080	6530	5320	17930
Count	10	10	10	30
Mean, Sum/n	608	653	532
Sum of square, Ʃ(xᵢ-x̅)²	55360	79610	90960

a) Null and Alternative Hypothesis:

Ho: µ1 = µ2 = µ3

H1: At least one mean is different.

Number of treatment, k = 3

Total sample Size, N = 30

df(between) = k-1 = 2

df(within) = N-k = 27

df(total) = N-1 = 29

SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 - (Grand Sum)²/ N = 74806.6667

SS(within) = SS1 + SS2 + SS3 = 225930

SS(total) = SS(between) + SS(within) = 300736.6667

MS(between) = SS(between)/df(between) = 37403.3333

MS(within) = SS(within)/df(within) = 8367.7778

F = MS(between)/MS(within) = 4.4699

p-value = F.DIST.RT(4.4699, 2, 27) = 0.0210

Decision:

P-value < α, Reject the null hypothesis.

ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	74806.6667	2	37403.3333	4.4699	0.0210
Within Groups	225930.0000	27	8367.7778
Total	300736.6667	29

b) At α = 0.05, k = 3, N-K = 27, Q value = 3.51

HSD = 3.51*√(8367.7778/10) = 101.5342

Comparison	Diff. = (xi - xj)	HSD	Results
x̅1-x̅2	-45	101.5342	Means are not different
x̅1-x̅3	76	101.5342	Means are not different
x̅2-x̅3	121	101.5342	Means are different