Question

11 of 18 pact that octane has on gas mileage. The accompanying table shows the gas mileag 87, 89, and 93 octane. The gas mileage was recorded for each octane level. Each More Info Car A Gas Mileage 87 Octane 89 Octane 31.1 32.6 21.6 30.8 31.1 31.8 25.3 295 33.4 30.7 26.1 28 2 C 93 Octane 27.5 29.8 25.3 32.5 25.6 26.4 E F Print Done

Answer 1

# In R, I use the car as A=1, B=2, C=3, D=4, E=5, F=6.

## R command

Mileage=c(31.1,21.6,31.1,25.3,33.4,26.1,32.6,30.8,31.8,29.5,30.7,28.2,
27.5,29.8,25.3,32.5,25.6,26.4)
Octane=c(87,87,87,87,87,87,89,89,89,89,89,89,93,93,93,93,93,93)
Car=c(1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6)
## ANOVA model
Model=aov(Mileage~factor(Octane)+factor(Car))
summary(Model)

## End

> Mileage=c(31.1,21.6,31.1, 25.3, 33.4,26.1, 32.6, 30.8, 31.8, 29.5, 30.7,28.2, + 27.5, 29.8,25.3, 32.5,25.6,26.4) > Octane=c(87, 87, 87, 87, 87, 87, 89, 89, 89, 89, 89, 89,93,93,93,93,93,93) > Car=c(1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6) > ## ANOVA model > Model=aov (Mileage-factor (Octane) +factor (Car)) > summary (Model) DE Sum Sq Mean Sq F value Pr(>F) factor (Octane) 2 27.75 13.875 1.132 0.360 factor (Car) 5 29.32 5.865 0.479 0.785 Residuals 10 122.53 12.253

a) The correct hypothesis to test the for differences in the mileage octane levels is for the B) Hos M M 93 H,: Mot all the Mis are equal. The test statistic F = 1.32 The p-valu 0 360 Conabussion about the population means. fail to reject Ho. conclude that there is no sufficient significant difference between the means. The correct hypothesis to test for differences in the block means is E = ME HoiM =MBF Me - M EM H.: Not all the M's are equal. The list statistic The FBL = 0.479 p-value = 0.785 Conclusion: fail to reject Ho. There is no evidence suggesting that the blocking was effective.

c) The mileages for all the three octance are insignificant. Hence, filling this blank does not exist.

The mileage for ______ were significantly different.