Question

An employee of a small software company in Minneapolis bikes to work during the summer months. He can travel to work using one of three routes and wonders whether the average commute times (in minutes) differ between the three routes. He obtains the following data after traveling each route for one week.

Route 1 34 35 29 36 29

Route 2 21 28 25 30 24

Route 3 22 27 25 30 26

a-1. Construct an ANOVA table.

a-2. At the 1% significance level, do the average commute times differ significantly between the three routes. Assume that commute times are normally distributed.

b. Use Tukey’s HSD method at the 1% significance level to determine which routes' average times differ.

Answer 1

	Route 1	Route 2	Route 3	Total
Sum	163	128	130	421
Count	5	5	5	15
Mean, Sum/n	32.6	25.6	26
Sum of square, Ʃ(xᵢ-x̅)²	45.2	49.2	34

Number of treatment, k = 3

Total sample Size, N = 15

df(between) = k-1 = 2

df(within) = N-k = 12

df(total) = N-1 = 14

SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 - (Grand Sum)²/ N = 154.5333

SS(within) = SS1 + SS2 + SS3 = 128.4

SS(total) = SS(between) + SS(within) = 282.9333

MS(between) = SS(between)/df(between) = 77.26667

MS(within) = SS(within)/df(within) = 10.7

F = MS(between)/MS(within) = 7.2212

p-value = F.DIST.RT(7.2212, 2, 12) = 0.0087

Critical value Fc = F.INV.RT(0.01, 2, 12) = 6.927

a-1) ANOVA table:

ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	154.5333	2	77.2667	7.2212	0.0087
Within Groups	128.4000	12	10.7000
Total	282.9333	14

a-2)

Null and Alternative Hypothesis:

Ho: µ1 = µ2 = µ3

H1: At least one mean is different.

Test statistic:

F = 7.22

p-value = F.DIST.RT(7.2212, 2, 12) = 0.0087

Decision:

P-value < α, Reject the null hypothesis.

There is enough evidence to conclude that at the 1% significance level, the average commute times differ significantly between the three routes.

b) At α = 0.01, k = 3, N-K = 12, Q value = 5.04

Critical Range, CV = Q*√(MSW/n) = 5.04*√(10.7/5) = 7.37

Comparison	Diff. = (xi - xj)	Critical Range	Results
x̅1-x̅2	7	7.37	Means are not different
x̅1-x̅3	6.6	7.37	Means are not different
x̅2-x̅3	-0.4	7.37	Means are not different