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?

Answer 1

Here, we have to use one sample z test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H₀: The residents of the state of Montana drive the same as the national average of 12,200 miles.

Alternative hypothesis: H_a: The residents of the state of Montana drive more than the national average of 12,200 miles.

H₀: µ = 12,200 versus H_a: µ > 12,200

This is an upper tailed test.

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

µ = 12200

Xbar = 12895.9

σ = 3800

n = 35

α = 0.05

Critical value = 1.6449

(by using z-table or excel)

Z = (12895.9 - 12200)/[ 3800/sqrt(35)]

Z = 1.0834

P-value = 0.1393

(by using Z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that the residents of the state of Montana drive more than the national average of 12,200 miles.