Question

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 ?

I need help how to solve in Excel

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 mean time to rent a car on their website is 60 seconds.

Alternative hypothesis: Ha: the mean time to rent a car on their website is not 60 seconds.

H0: µ = 60 versus Ha: 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

α = 0.05

Critical value = -1.96 and 1.96

(by using z-table or excel command =NORMSINV(0.05/2))

Z = (75 – 60)/[30/sqrt(36)]

Z = 3

P-value = 0.0027

(by using Z-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is not sufficient evidence to conclude that the mean time to rent a car on their website is 60 seconds.