A fire insurance company thought that the mean distance from a home to the neare
ID: 3020093 • Letter: A
Question
A fire insurance company thought that the mean distance from a home to the nearest fire department in a suburb of Chicago was at least 5.9 miles. It set its fire insurance rates accordingly. Members of the community set out to show that the mean distance was less than 5.9 miles. This, they thought, would convince the insurance company to lower its rates. They randomly indentified 56 homes and measured the distance to the nearest fire department from each. The resulting sample mean was 5. If = 1.9 miles, does the sample show sufficient evidence to support the community's claim at the = .05 level of significance?
(a) Find z. (Give your answer correct to two decimal places.)
(ii) Find the p-value. (Give your answer correct to four decimal places.)
(b) State the appropriate conclusion. Reject the null hypothesis, there is not significant evidence that the mean distance is less than 5.9 miles.Reject the null hypothesis, there is significant evidence that the mean distance is less than 5.9 miles. Fail to reject the null hypothesis, there is significant evidence that the mean distance is less than 5.9 miles.Fail to reject the null hypothesis, there is not significant evidence that the mean distance is less than 5.9 miles.
Explanation / Answer
Here hypothesis for the test is,
H0 : mu 5.9 Vs H1 : mu < 5.9
number of indentified homes (n) = 56
sample mean (Xbar) = 5
= 1.9 miles
alpha = 0.05
The test statistic for testing mean is,
Z = (Xbar - mu) / ( sd/sqrt(n) )
Z = (5 - 5.9) / (1.9/sqrt(56))
Z = -3.54
We can find P-value by using EXCEL.
syntax is,
=NORMSDIST(z)
where z is test statistic value.
P-value = 0.0002
P-value < alpha
Reject H0 at 5% level of significance.
Conclusion : There is significant evidence that the mean distance is less than 5.9 miles.
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