Question

A leasing firm claims that the mean number of miles driven annually, u , in its leased cars is less than 12800 miles. A random sample of 50 cars leased from this firm had a mean of 12499 annual miles driven. It is known that the population standard deviation of the number of miles driven in cars from this firm is 3140 miles. Is there support for the firm's claim at the 0.05 level of significance? Perform a one-tailed test. a. null hypothesis? b. alternative hypothesis? c. type of test statistic? d. value of the test statistic? e. the p-value?

Answer 1

The mean number of miles driven annually is greater than or equal 12800 miles.

Explanation:

We are given the following in the question:

Population mean, μ = 12800 miles

Sample mean,
`\bar{x}` = 12499 miles

Sample size, n = 50

Alpha, α = 0.05

Population standard deviation, σ = 3140 miles

First, we design the null and the alternate hypothesis

`H_(0): \mu \geq 12800\text{ miles}\\H_A: \mu < 12800\text{ miles}`

We use One-tailed(left) z test to perform this hypothesis.

Formula:

`z_(stat) = \displaystyle\frac{\bar{x} - \mu}{(\sigma)/(√(n)) }`

Putting all the values, we have

`z_(stat) = \displaystyle(12499 - 12800)/((3140)/(√(50)) ) = -0.6778`

We calculate the p-value from the z table.

P-value at 0.05 significance level = 0.2492

Since,

P-value > 0.05

P-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.

Thus, the mean number of miles driven annually is greater than or equal 12800 miles.