For Exercises 3 through 8, the null hypothesis was rejected. Use the Scheffé test when sample sizes are unequal or the Tukey test when sample sizes are equal, to test the differences between the pairs of means. Assume all variables are normally distributed, samples are independent, and the population variances are equal.
Exercise 9 in Section 12−1

There is a significant difference in the mean cost to drive 25 miles between hybrid cars and hybrid trucks and between hybrid SUVs and hybrid trucks.

Here we first need to test the hypotheses for differences to do the Scheffe and Tukey test

The sample mean and variance for:-

Hybrid Cars = 1.89 , 0.29

Hybrid SUVs = 2.22 , 0.02

Hybrid trucks = 3.53 , 0.02

Hence the total mean will be

(2.10+2.70+1.67+⋯+3.62+3.4)/12

= 2.301

Hence we generate the following Hypotheses

Let H₀be the hypotheses that the means obtained are equal i.e,

H₀ : μ₁ = μ₂ = μ₃.

Let

H₁ : At least one of the means differs from the other means

Here we can see that the number of samples is 12 and we have 3 groups of data,

hence N = 12 and k = 3

hence we get degrees of freedom as

d.f.N = k - 1

= 2

d.f.D = N - k

= 9

Hence to get the critical value we need to look up the F distribution with a significance level of 0.05 and a fraction of 2/9

hence critical value is 4.26

To calculate the between-group variance we use the formula

(∑n(sample mean - total mean)²)/(k-1)

= [5(1.89−2.301)² +5(2.22−2.301)² + 2(3.53−2.301)²]/2

= 1.95

The within-group variance will be

0.14

Hence the test statistic

F = between group variance/within group variance

= 1.95/0.14

= 13.93

Now, the critical value of F is less than the test value

Therefore, the value of the test statistic is in the rejection region

Hence the null Hypotheses H₀ is rejected

Hence there is a difference in the mean cost to drive 2525 miles for different types of hybrid vehicles which are concluded by sufficient evidence.

The critical value for the Scheffe test

F' = critical value X (k - 1)

= 4.26 X 2

= 8.52

For Hybrid Cars vs Hybrid SUVs, we have the Scheffe test as

F1 = (sample mean of cars - sample mean of SUVs)²/(withing group variance(1/n + 1/n))

= 1.945

Since F' > F1

the difference between the mean of Hybrid cars and SUVs is insignificant

Similarly, for SUVs and Truck, we get

F2 = 17.51

Here F'<F2

Hence the difference in the mean is significant

and for Cars and Trucks

F3 = 27.44

F1<F3

the difference is significant

Hence,

The evidence is enough to conclude a significant difference in mean cost to drive 25 miles between hybrid cars and hybrid trucks and between hybrid SUVs and hybrid trucks.

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