Question

An automobile manufacturer claims that its jeep has a 36.1 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the MPG for this jeep since it is believed that the jeep has an incorrect manufacturer's MPG rating. After testing 150 jeeps, they found a mean MPG of 35.8. Assume the standard deviation is known to be 2.4. A level of significance of 0.02 will be used. Make a decision to reject or fail to reject the null hypothesis. Make a decision.

Answer 1

Answer:

$z=\frac{35.8-36.1}{\frac{2.4}{\sqrt{150}}}=-1.531$    

$p_v =2*P(z$  

Since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean is different from 36.1 MPG (the claim by the manufacturer)

Step-by-step explanation:

Information given  

$\bar X=35.8$ represent the sample mean for the MPG

$\sigma=2.4$ represent the population deviation

$n=150$ sample size  

$\mu_o =36.1$ represent the value to check

$\alpha=0.02$ represent the significance level for the hypothesis test.  

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We need to conduct a hypothesis in order to check if manufacturer is incorrect, the null and alternative hypothesis are:  

Null hypothesis:$\mu = 36.1$  

Alternative hypothesis:$\mu \neq 36.1$  

The statistic is given by:

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

Replacing we got:

$z=\frac{35.8-36.1}{\frac{2.4}{\sqrt{150}}}=-1.531$    

P value

Since we are conducting a bilateral test we can find the p value like this:

$p_v =2*P(z$  

Conclusion

Since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean is different from 36.1 MPG (the claim by the manufacturer)