In: Statistics and Probability
A certain statistics instructor participates in triathlons. The accompanying table lists times (in seconds) he recorded while riding a bicycle. He recorded every time he rode and randomly picked five laps through each mile of a 3-mile loop. Use a .05 significance level to test the claim that it takes the same time to ride each of the miles. Mile 1 194 205 203 202 201
Mile 2 198 201 200 196 200
Mile 3 215 211 209 212 209
a) State the null and alternative hypothesis (2 points).
b) Check the requirements (3 points)
c) Carry out the hypothesis test (make sure you write everything down) (3 points)
. d) What is your conclusion? (2 points)
e) Does one of the miles appear to have a hill? (2 points)
a)
Ho: µ1=µ2=µ3
H1: not all means are equal
........
A | B | C | ||||
count, ni = | 5 | 5 | 5 | |||
mean , x̅ i = | 201.000 | 199.00 | 211.20 | |||
std. dev., si = | 4.183 | 2.000 | 2.490 | |||
sample variances, si^2 = | 17.500 | 4.000 | 6.200 | |||
total sum | 1005 | 995 | 1056 | 3056 | (grand sum) | |
grand mean , x̅̅ = | Σni*x̅i/Σni = | 203.73 | ||||
square of deviation of sample mean from grand mean,( x̅ - x̅̅)² | 7.471 | 22.404 | 55.751 | |||
TOTAL | ||||||
SS(between)= SSB = Σn( x̅ - x̅̅)² = | 37.356 | 112.022 | 278.756 | 428.1333 | ||
SS(within ) = SSW = Σ(n-1)s² = | 70.000 | 16.000 | 24.800 | 110.8000 |
no. of treatment , k = 3
df between = k-1 = 2
N = Σn = 15
df within = N-k = 12
mean square between groups , MSB = SSB/k-1 =
214.0667
mean square within groups , MSW = SSW/N-k =
9.2333
F-stat = MSB/MSW = 23.1841
anova table | ||||||
SS | df | MS | F | p-value | F-critical | |
Between: | 428.13 | 2 | 214.07 | 23.18 | 0.0001 | 3.89 |
Within: | 110.80 | 12 | 9.23 | |||
Total: | 538.93 | 14 |
..........
α = 0.05
conclusion : p-value<α , reject null hypothesis
so, means are different
...............
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