Question

An automobile manufacturer has given its car a 52.6 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the actual MPG for this car since it is believed that the car has an incorrect manufacturer's MPG rating. After testing 250 cars, they found a mean MPG of 52.8. Assume the population standard deviation is known to be 1.6. A level of significance of 0.05 will be used. State the null and alternative hypotheses.

Answer 1

Answer:

Null hypothesis:$\mu = 52.6$  

Alternative hypothesis:$\mu \neq 52.6$  

$z=\frac{52.8-52.6}{\frac{1.6}{\sqrt{250}}}=1.976$    

$p_v =2*P(z>1.976)=0.0482$  

Since the p value is lower than the significance level wedon't have enough evidence to conclude that the true mean is significantly different from 52.6 MPG.

Step-by-step explanation:

Information provided

$\bar X=52.8$ represent the sample mean  for the MPG of the cars

$\sigma=1.6$ represent the population standard deviation

$n=250$ sample size  of cars

$\mu_o =52.6$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test

Hypothesis

We need to conduct a hypothesis in order to check if the true mean of MPG is different from 52.6 MPG, the system of hypothesis would be:  

Null hypothesis:$\mu = 52.6$  

Alternative hypothesis:$\mu \neq 52.6$  

Since we know the population deviation the statistic is given by:

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

Calculate the statistic

Replacing we have this:

$z=\frac{52.8-52.6}{\frac{1.6}{\sqrt{250}}}=1.976$    

Decision

Since is a two tailed test the p value would be:  

$p_v =2*P(z>1.976)=0.0482$  

Since the p value is lower than the significance level wedon't have enough evidence to conclude that the true mean is significantly different from 52.6 MPG.