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model. A comprehensive statistics examination is given to 16 students in order to determine whether or not there is a signifi



At a 0.05 level of significance, test to see if there is a significant difference in the performance of the students in the t

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Answer #1
Accounting Management Marketing Total
Sum 386 485 395 1266
Count 5 6 5 16
Mean, Sum/n 77.2 80.83333 79
Sum of square, Ʃ(xᵢ-x̅)² 606.8 950.8333 820

Null and Alternative Hypothesis:

Ho: µ1 = µ2 = µ3

H1: At least one mean is different.

Number of treatment, k = 3

Total sample Size, N = 16

df(between) = k-1 = 2

df(within) = N-k = 13

df(total) = N-1 = 15

SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 - (Grand Sum)²/ N = 36.1167

SS(within) = SS1 + SS2 + SS3 = 2377.6333

SS(total) = SS(between) + SS(within) = 2413.75

MS(between) = SS(between)/df(between) = 18.0583

MS(within) = SS(within)/df(within) = 182.8949

F = MS(between)/MS(within) = 0.0987

p-value = F.DIST.RT(0.0987, 2, 13) = 0.9067

Critical value Fc = F.INV.RT(0.05, 2, 13) = 3.80556525

Decision:

P-value > α, Do not reject the null hypothesis.

ANOVA
Source of Variation SS df MS F P-value
Between Groups 36.1167 2 18.0583 0.0987 0.9067
Within Groups 2377.6333 13 182.8949
Total 2413.7500 15

There is not enough evidence to conclude that there is a significant difference in the performance of the student in the three majors at 0.05 significance level.

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