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

Answer:

$t=\frac{7.43-8}{\frac{3.6}{\sqrt{64}}}=-1.27$    

The degrees of freedom are given by:

$df=n-1=64-1=63$  

The p value can be calculated like this:

$p_v =P(t_{(63)}$  

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

Step-by-step 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:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

The statistic for this case is given by:

$t=\frac{7.43-8}{\frac{3.6}{\sqrt{64}}}=-1.27$    

The degrees of freedom are given by:

$df=n-1=64-1=63$  

The p value can be calculated like this:

$p_v =P(t_{(63)}$  

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