Answer:
(A) Null Hypothesis, :
{means that there is no significant difference in the variances of the amounts spent in the two counties}
Alternate Hypothesis, :
{means that there is a significant difference in the variances of the amounts spent in the two counties}
(B) The F-table gives critical values of 0.359 and 2.781 for (14,17) degrees of freedom for the two-tailed test.
(C) The value of the F-test statistic is 0.611.
(D) We conclude that there is no significant difference in the variances of the amounts spent in the two counties.
Step-by-step explanation:
We are given that a politician wishes to compare the variances of the amount of money spent on road repair in two different counties.
The data are given here below;
County A County B
s1 = $11,596 s2 = $14,837
n1 = 15 n2 = 18
Let = variances of the amounts spent in County A.
= variances of the amounts spent in County B.
(A) So, Null Hypothesis, :
{means that there is no significant difference in the variances of the amounts spent in the two counties}
Alternate Hypothesis, :
{means that there is a significant difference in the variances of the amounts spent in the two counties}
The test statistics that would be used here <u>Two-sample F-test statistics </u>distribution;
T.S. = ~
where, = sample standard deviation for County A = $11,596
= sample standard deviation for County B = $14,837
= sample size for County A = 15
= sample size for County B = 18
(B) Now at 0.05 level of significance, the F-table gives critical values of 0.359 and 2.781 for (14,17) degrees of freedom for the two-tailed test.
(C) So, <u><em>the test statistics</em></u> = ~
= 0.611
The value of the F-test statistic is 0.611.
Now, as we can see that our test statistics lie within the range of critical values of F, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.
(D) Therefore, we conclude that there is no significant difference in the variances of the amounts spent in the two counties.