Question

# Q1. The quantitative reasoning scores on the GRE were recorded for students admitted to three different...

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

 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

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