In: Statistics and Probability
A statisitcs instructor participates in triathalons. the table lists the times in minutes and seconds, he recorded while riding five laps through each mile of a 3-mile loop. Use a .05 significance level to test the claim that is takes the same time to ride each of the miles.
Mile 1 | 3:15 | 3:25 | 3:23 | 3:22 | 3:22 |
Mile 2 | 3:19 | 3:23 | 3:20 | 3:17 | 3:19 |
Mile 3 | 3:34 | 3:30 | 3:29 | 3:30 | 3:29 |
Determine the null and alternate hypotheses.
Find the F statistic
P VALUE
What is the conlusion for this hypothesis test?
(fail to reject, reject) There is (sufficient, insufficient) evidence to warrant the rejection of the claime that the three different miles have the same mean ridetime.
Does one of the miles appear to have a hill?
Here we are given time in minutes and seconds. First we convert time in secconds.
e.g : 3:15 = 3 minutes and 15 seconds.
60 seconds = 1 minute
so 3 minutes = 180 seconds.
Hence 3:15 = 180+15 = 195 seconds.
So we have table as :
Mile1 | Mile2 | Mile3 |
195 | 199 | 214 |
205 | 203 | 210 |
203 | 200 | 209 |
202 | 197 | 210 |
202 | 199 | 209 |
We run one way ANOVA in excel :
Data tab > data analysis > ANOVA:Single factor
We get output as :
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
Mile1 | 5 | 1007 | 201.4 | 14.3 | ||
Mile2 | 5 | 998 | 199.6 | 4.8 | ||
Mile3 | 5 | 1052 | 210.4 | 4.3 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 334.8 | 2 | 167.4 | 21.46154 | 0.000109 | 3.885294 |
Within Groups | 93.6 | 12 | 7.8 | |||
Total | 428.4 | 14 |
The hypothesis are :
H0: The three different miles have the same mean ridetime.
H1: The three different miles have different mean ridetime.
The test statistic is
F = 21.462
P value : 0.0001
Here p value <
Hence we reject null hypothesis.
There is sufficient evidence to warrant the rejection of the claim that the three different miles have the same mean ridetime.
This means one of the miles appear to have a hill.