Question

An automobile manufacturer claims that its van has a 29.1 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the MPG for this van since it is believed that the van has an incorrect manufacturer's MPG rating. After testing 200 vans, they found a mean MPG of 28.8. Assume the standard deviation is known to be 2.4. A level of significance of 0.01 will be used. State the hypotheses. Enter the hypotheses:

Answer 1

Answer:

$z=\frac{28.8-29.1}{\frac{2.4}{\sqrt{200}}}=-1.768$    

The p value would be given by this probability:

$p_v =2* P(z<-1.768) =0.0771$ 

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and then we can conclude that the true mean is not significantly different from 29.1 MPG

Step-by-step explanation:

Information given

$\bar X=28.8$ represent the sample mean

$\sigma=2.4$ represent the population standard deviation

$n=200$ sample size  

$\mu_o =29.1$ represent the value to verify

$\alpha=0.01$ represent the significance level

t would represent the statistic  

$p_v$ represent the p value

Hypothesis to test

We want to verify if the true mean is equal to 29.1 MPG, the system of hypothesis would be:  

Null hypothesis:$\mu = 29.1$  

Alternative hypothesis:$\mu \neq 29.1$  

The statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (1)  

Replacing the info:

$z=\frac{28.8-29.1}{\frac{2.4}{\sqrt{200}}}=-1.768$    

The p value would be given by this probability:

$p_v =2* P(z<-1.768) =0.0771$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and then we can conclude that the true mean is not significantly different from 29.1 MPG