Question

A certain vehicle emission inspection station advertises that the wait time for customers is less than 8 minutes. A local resident wants to test this claim and collects a random sample of 64 wait times for customers at the testing station. She finds that the sample mean is 7.43 ​minutes, with a standard deviation of 3.6 minutes. Does the sample evidence support the inspection​ station's claim? Use the alphaequals0.005 level of significance to test the advertised claim that the wait time is less than 8 minutes.

Answer 1

`t=(7.43-8)/((3.6)/(√(64)))=-1.27`

The degrees of freedom are given by:

`df=n-1=64-1=63`

The p value can be calculated like this:

`p_v =P(t_((63))<-1.27)=0.104`

For this case since the p value is higher than the significance level we have enough evidence to conclude that the true mean for the wait time is not significantly less than 8 minutes

Explanation:

Information given

`\bar X=7.43` represent the sampke mean in minutes

`s=3.6` represent the sample standard deviation

`n=64` sample size

`\mu_o =8` represent the value to verify

`\alpha=0.005` represent the significance level

t would represent the statistic (variable of interest)

`p_v` represent the p value

System of hypothesis

For this case we are trying to proof if inspection station advertises that the wait time for customers is less than 8 minutes, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 8`

Alternative hypothesis:
`\mu < 8`

Since the population deviation is not known the statistic can be calculated with:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

The statistic for this case is given by:

`t=(7.43-8)/((3.6)/(√(64)))=-1.27`

The degrees of freedom are given by:

`df=n-1=64-1=63`

The p value can be calculated like this:

`p_v =P(t_((63))<-1.27)=0.104`

For this case since the p value is higher than the significance level we have enough evidence to conclude that the true mean for the wait time is not significantly less than 8 minutes