Martin Motors has in stock three cars of the same make and model. The Sales Mana

Question

Martin Motors has in stock three cars of the same make and model. The Sales Manager would like to compare the gas consumption of the three cars (labeled car A, car B, and car C) using four different types of gasoline. For each trial, a gallon of gasoline was added to an empty tank, and the car was driven until it ran out of gas. The following table shows the number of miles driven in each trial. At 0.1 significance level, state the decision rule for both treatments (Cars) and blocks (Gasoline). (Round your answers to 2 decimal places.) At .1 significance level, is there a difference in the treatments and blocks?

Explanation / Answer

there are two factors in the experiment: treatment and blocks

so the variation in the data is due to two things : the effect of the treatments and the effect of the blocks

here treatment is the type of cars and block is the gasoline.

number of treatments : r=3 number of blocks: v=4

let the model be

yij=u+bi+tj+eij i=1,2,3 j=1,2,3,4

where yij=value of response variable=the number of miles driven in each trial

u=common overall effect

bi=additional effect due to ith treatment

tj=additional effect due to jth block

eij=error associated with yij

we are interested in testing for the effect of treatments and blocks

so there are actually two set of hypothesis

H01: b1=b2=b3=0 vs H11: not H01

H02: t1=t2=t3=t4=0 vs H12: not H02

we have the anova table as follows (obtained from MINITAB)

Source DF SS MS F P
treatment 2 1.6517 0.82583 4.39 0.067
block 3 7.3492 2.44972 13.03 0.005
Error 6 1.1283 0.18806
Total 11 10.1292

we have the test statistic for H01 as F=MST/MSE where MST is the mean square treatment and MSE is the mean square error

F follows an F distribution with dfs 2 and 6

level of significance alpha=0.1

hence H01 is rejected iff F>F0.1:2,6   where F0.1:2,6 is the upper 0.1 point of an F distribution with dfs 2 and 6

using MINITAB F0.1:2,6=3.46

for H02 the test statistic F=MSB/MSE where MSB is the mean square due to block

F follows an F distribution with dfs 3 and 6

level of significance alpha=0.1

hence H01 is rejected iff F>F0.1:3,6   where F0.1:3,6 is the upper 0.1 point of an F distribution with dfs 3 and 6

using MINITAB F0.1:3,6=3.29

hence

for treatments (cars) reject H0 if F>3.46

for block reject H0 if F>3.29

now we have MST/MSE=4.39

so for H01 F>3.46

hence H01 is rejected

we have MSB/MSE=13.03

so for H02 F>3.29

hence H02 is also rejected

reject H0. there is significant difference between the four types of gasoline

reject H0: there is significant difference in the cars

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