Question

computing z scores

A rental car company claims the mean time to rent a car on their website is 60 seconds with a standard deviation of 30 seconds. A random sample of 36 customers attempted to rent a car on the website. The mean time to rent was 75 seconds. Is this enough evidence to contradict the company's claim?

Answer 1

Solution:

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

The null and alternative hypotheses are given as below:

H₀: µ = 60 versus H_a: µ ≠ 60

This is a two tailed test.

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

µ = 60

Xbar = 75

σ = 30

n = 36

Assume α = 0.05

Critical value = - 1.96 and 1.96

(by using z-table or excel)

Z = (75 – 60)/[30/sqrt936)]

Z = 3.00

P-value = 0.0027

(by using Z-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is enough evidence to contradict the company’s claim that the mean time to rent a car on their website is 60 seconds.