Question

According to the U.S. Federal Highway Administration, the mean number of miles driven annually is 12,200 with a standard deviation of 3800 miles. A resident of the state of Montana believes the drivers in Montana drive more than the national average. She obtains a random sample of 35 drivers from a list of registered drivers in the state and finds the mean number of miles driven annually for these drivers to be 12,895.90. Is there sufficient evidence to show that residents of the state of Montana drive more than the national average?

What is the p-value for this hypothesis test?
What is the test statistic for this hypothesis test?
What is the critical value?
What is the decision?

Options:

A.	Reject Ho because the test statistic is in the rejection region and the p-value is less than alpha.
B.	2.21
C.	-0.2724
D.	2.575
E.	1.08
F.	1.96
G.	0.1393
H.	(1.6223, 1.6331)
I.	1.645
J.	-0.5364
K.	(1.6116, 1.6166)
L.	0.2146
M.	Fail to reject Ho because the test statistic is not in the critical region and the p-value is greater than alpha.
N.	1.96
O.	0.7854
P.	Reject Ho because the test statistic is not in the rejection region and the p-value is less than alpha.
Q.	Fail to reject Ho because the test statistic is not in the rejection region and the p-value is less than alpha.
R.	(1.6113, 1.6168)

Answer 1

given data :-

sample mean () = 12895.9

population sd () = 3800

sample size (n) = 35

hypothesis:-

the test statistic be:-

the p value of the study is:-

[ for more accurate calculation i am using z=1.0834 i spite of 1.08 ]

[ in any blank cell of excel type =NORMSDIST(1.0834) ]

the critical value is = 1.645 (E)

[ this is the Z critical value for alpha=0.05, right tailed test]

decision:-

Fail to reject Ho because the test statistic is not in the critical region and the p-value is greater than alpha (M).

[ p value = 0.1393 >0.05 (alpha)..so, we fail to reject the null hypothesis

in critical value approach:-

..so, we fail to reject the null hypothesis .]

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