Question

A rental car company claims the mean time to rent a car on their website is 60 seconds with a standard deviation of 22 seconds. A random sample 36 customers attempted to rent a car on the website. The mean time to rent was 55 seconds. Is this enough evidence to say that the mean time is less than the company's claim? Consider α=0.05.

Answer 1

Answer:

Since the calculated value of Z does not fall in the critical region we accept our null hypothesis that mean = 60 seconds

Step-by-step explanation:

State the null hypothesis

H0: u = 60against the claim

Ha u≠   60   (this is a two tailed test)

Sample size n= 36

Sample mean=X`= 55

Population mean = u= 60

The significance level α = 0.05

Standard deviation= Sd = 22 seconds

Z= X`- u / Sd /√n

Z= 55- 60 / 22/√6

z= - 5/22/6

Z= -1.3635

The value of z from the table is Z∝/2= ±1.96

The critical region is less than - 1.96 and greater than 1.96.

Since the calculated value of Z does not fall in the critical region we accept our null hypothesis that mean = 60 seconds