In: Statistics and Probability
Flight 1 | -2 | -1 | -2 | 2 | -2 | 0 | -2 | -3 |
Flight 19 | 19 | -4 | -5 | -1 | -4 | 73 | 0 | 1 |
Flight 21 | 18 | 60 | 142 | -1 | -11 | -1 | 47 | 13 |
Listed below are departure delay times (minutes) for american Airline flights from New York to Los Angeles. Negative values correspond to flights that departed early.
Use a 0.05 significance level to test the claim that the different flights have the same mean departure delay time.What is the critical value (F-value)? [Round to 4 decimal places]
treatment | Flight 1 | Flight 19 | Flight 21 | |||||
count, ni = | 8 | 8 | 8 | |||||
mean , x̅ i = | -1.250 | 9.88 | 33.38 | |||||
std. dev., si = | 1.6 | 26.6 | 50.3 | |||||
sample variances, si^2 = | 2.500 | 709.839 | 2525.411 | |||||
total sum | -10 | 79 | 267 | 336 | (grand sum) | |||
grand mean , x̅̅ = | Σni*x̅i/Σni = | 14.00 | ||||||
( x̅ - x̅̅ )² | 232.563 | 17.016 | 375.391 | |||||
TOTAL | ||||||||
SS(between)= SSB = Σn( x̅ - x̅̅)² = | 1860.500 | 136.125 | 3003.125 | 4999.75 | ||||
SS(within ) = SSW = Σ(n-1)s² = | 17.500 | 4968.875 | 17677.875 | 22664.2500 |
no. of treatment , k = 3
df between = k-1 = 2
N = Σn = 24
df within = N-k = 21
mean square between groups , MSB = SSB/k-1 =
4999.75/2= 2499.8750
mean square within groups , MSW = SSW/N-k =
22664.25/21= 1079.2500
F-stat = MSB/MSW = 2499.875/1079.25=
2.32
ANOVA | ||||||
SS | df | MS | F | p-value | F-critical | |
Between: | 4999.75 | 2 | 2499.88 | 2.316 | 0.1233 | 3.467 |
Within: | 22664.25 | 21 | 1079.25 | |||
Total: | 27664.00 | 23 | ||||
α = | 0.05 |
critical F value= 3.4668
there is insufficient evidence to reject the claim that the different flights have the same mean departure delay time.