Question

Listed below are times(in minutes and seconds) while an athlete rode his bicycle through each mile of a 3 mile loop. Use a 0.05 significance level to test the claim that it takes the same time to ride each of the miles.

Mile 1	3:14	3:24	3:23	3:22	3:21
Mile 2	3:19	3:22	3:21	3:17	3:19
Mile 3	3:34	3:31	3:29	3:31	3:29

Answer 1

The data converted into decimals is as below

Mile 1	Mile 2	Mile 3
3.23	3.32	3.57
3.4	3.37	3.52
3.38	3.35	3.48
3.37	3.28	3.52
3.35	3.32	3.48

The summary statistics is as below

	Mile 1	Mile 2	Mile 3
Total	16.73	16.64	17.57
n	5	5	5
Mean	3.35	3.33	3.51
Sum Of Squares	0.02	0.00	0.01
Variance	0.0046	0.0012	0.0014

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The Hypothesis:

H0: There is no difference between the mean time taken for the three miles.

Ha: The mean time of at least one mile is different from the others.

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The ANOVA table is as below.

Source	SS	DF	Mean Square	F	Fcv	p
Between	0.105	2	0.053	21.98	3.89	0.0001
Within/Error	0.03	12	0.002
Total	0.13	14

The p value is calculated for F = 21.98 for df1 = 2 and df2 = 12

The F critical is calculated at = 0.05 for df1 = 2 and df2 = 12

The Decision Rule:

If F test is > F critical, Then Reject H0.

Also if p-value is < , Then reject H0.

The Decision:

Since F test (21.98) is > F critical (3.89), We Reject H0.

Also since p-value (0.0001) is < (0.05), We Reject H0.

The Conclusion: There is sufficient evidence at the 95% level of significance to conclude that the mean time of at least one mile is different from the others.

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Calculations For the ANOVA Table:

Overall Mean = [3.35 +3.33 +3.51) / 3 = 3.40

SS treatment = SUM [n* ( Individual Mean - overall mean)²] = 5 * (3.35 - 3.4)² + 5 * (3.33 - 3.4)² + 5 * (3.51 - 3.4)² = 0.105

df1 = k - 1 = 3 - 1 = 2

MSTR = SS treatment/df1 = 0.105 / 2 = 0.053

SS error = SUM (Sum of Squares) = 0.02 + 0.0 + 0.01 = 0.03

df2 = N - k = 15 - 3 = 12

Therefore MS error = SS error/df2 = 0.03 / 12 = 0.002

F = MSTR / MSE = 0.053 / 0.002 = 21.98

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