Question

Check my wor The following output summarizes the results for a one-way analysis of variance experiment in which the treatments were three different hybrid cars and the variable measured was the miles per gallon (mpg) obtained while driving the same route. (You may find it useful to reference the table.) Hybrid 1: 2 - 27, n = 20 Hybrid 2: = 41, n2 = 15 Hybrid 3 = 34, n = 18 df Source of Variation Between Groups Within Groups Total MS 588.61 29.36 D-value 20.050 .000e 1.177.21 1,468.03 2,645.24 se S2 a. At the 5% significance level, can we conclude that average mpg differs between the hybrids? Yes. since the p-value is less than the significance level Yes. since the p value is not less than the significance level No. since the pvalue is less than the significance level

Answer 1

Solution

Back –up Theory

A hypothesis on difference between/among means is significant if

the test statistic value > the critical value or equivalently, if p-value < significance level............................. (1)  

Now, to work out the solution,

Part (a)

Vide (1), First Option Answer 1

Part (b)

Tuckey’s HSD Method

The Tukey-Kramer confidence limits for all pair-wise comparisons with confidence coefficient of at least 1−α are:

(yi.bar – yj.bar) ± [(1/√2)q_{α,r, N - r} σ_hat√{(1/n_i) + (1/n_j)}

where

yi.bar and yj.bar are the means of two treatments under analysis

q_{α,r, N - r} = studentized range, which are available in standard text books or can accessed from the net

1 - α = confidence coefficient,

r = number of treatments in ANOVA,

N = total number of observations in ANOVA

σ_hat = standard error of the residual = sqrt(MSE of ANOVA)

n_i, n_j = the number of observations per treatment i and j respectively in ANOVA

Here

r = 3; N = (20 + 15 + 18) = 53, α = 0.05 and hence 1 - α = 0.95, σ_hat = √29.36 = 5.42, q_{α,r, N – r} = q_{0.5,3, 50} = 3.42

Thus, f = (1/√2)q_{α,r, N - r} σ_hat = 13.1072  

Mean Difference	|yibar - yjbar|	√{(1/ni) + (1/nj)}	MoE	Lower Limit	Upper Limit
µ1 - µ2	14	0.2478	3.2486	10.7514	17.2486
µ1 - µ3	7	0.2578	3.3787	3.6213	10.3787
µ2 - µ3	7	0.2320	3.0402	3.9598	10.0402

Hybrid’s mean differ if the confidence interval does not hold zero.

Finally, answer in the desired format

Mean Difference	Lower Limit	Upper Limit	Conclusion
µ1 - µ2	10.7514	17.2486	Hybrid’s means differ
µ1 - µ3	3.6213	10.3787	Hybrid’s means differ
µ2 - µ3	3.9598	10.0402	Hybrid’s means differ

Answer 2

DONE