In: Statistics and Probability
| P. Turnpike | MF. Expressway | BV. Expressway | 
| 7 | 10 | 1 | 
| 14 | 1 | 12 | 
| 32 | 1 | 1 | 
| 19 | 0 | 9 | 
| 10 | 11 | 1 | 
| 11 | 1 | 11 | 
A state employee wishes to see if there is a significant difference in the number of employees at the interchanges of three state toll roads. The data are shown. At α=0.10, test the claim that there is no difference in the average number of employees at each interchange.
| P. Turnpike | MF. Expressway | BV. Expressway | Total | |
| Sum | 93 | 24 | 35 | 152 | 
| Count | 6 | 6 | 6 | 18 | 
| Mean, Sum/n | 15.5 | 4 | 5.8333 | |
| Sum of square, Ʃ(xᵢ-x̅)² | 409.5 | 128 | 144.8333 | 
Null and Alternative Hypothesis:
Ho: µ1 = µ2 = µ3
H1: At least one mean is different.
Number of treatment, k = 3
Total sample Size, N = 18
df(between) = k-1 = 2
df(within) = N-k = 15
df(total) = N-1 = 17
SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 - (Grand Sum)²/ N = 458.1111
SS(within) = SS1 + SS2 + SS3 = 682.3333
SS(total) = SS(between) + SS(within) = 1140.4444
MS(between) = SS(between)/df(between) = 229.0556
MS(within) = SS(within)/df(within) = 45.4889
F = MS(between)/MS(within) = 5.0354
Critical value Fc = F.INV.RT(0.1, 2, 15) = 2.695
Decision:
As F = 5.0354 > 2.695, Reject the null hypothesis.
There is enough evidence to reject the claim that there is no difference in the average number of employees at each interchange.
| ANOVA | ||||
| Source of Variation | SS | df | MS | F | 
| Between Groups | 458.1111 | 2 | 229.0556 | 5.0354 | 
| Within Groups | 682.3333 | 15 | 45.4889 | |
| Total | 1140.4444 | 17 |