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.
model. A comprehensive statistics examination is given to 16 students in order to determine whether or...
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