Question

You want to study the variance in gas mileage of the Ford fusion compared with the Ford F150 truck. You randomly select 8 fusions and 7 trucks and test them on the same track. The data below are the results in miles per gallon (mpg.)

Fusion:27, 25.2, 27, 25.7, 24.5, 26, 25, 25.9

F150:21, 21.3, 22.5, 23.1, 21.9, 23.1, 21.8

Is there evidence to conclude that the variances are different

Answer 1

Following table shows the output of descriptive statistics:

Descriptive statistics
	Fusion, X1	F150, X2
count	8	7
mean	25.788	22.100
sample standard deviation	0.897	0.831
sample variance	0.804	0.690
minimum	24.5	21
maximum	27	23.1
range	2.5	2.1

Here we have n-8, s0.804 n27, s20.69 Hypotheses are: 0 H o Level of significance: a 0.05 Test is two tailed. Degrees of freedom: df1=n1-1=7, df2=n2-1=6 Test Statistics: (0.804)/(0.69) = 1.17 Critical value(CVs): 0.2,5.7 Rejection Region: If F< 0.2 or F>5.7, Reject HO Decision: Since test statistics does not lie in rejection region so we fail to reject the null hypothesis P-value: P-value 0.864 Decision: Since p-value is not less than level of significance so we fail to reject the null hypothesis. Excel function for critical value: -FINV((1-(0.05/2)), 7,6) and -FINV(0.05/2),7,6) Excel function for p-value: 2*MIN(1-FDIST(1.17,7,6)),FDIST(1.17,7,6) Conclusion: That is on the basis of sample evidence we cannot conclude that population variances are not equal