Question

An automobile manufacturer has given its jeep a 56.1 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the actual MPG for this jeep since it is believed that the jeep performs under the manufacturer's MPG rating. After testing 200 jeeps, they found a mean MPG of 56.0. Assume the population standard deviation is known to be 2.1.
Is there sufficient evidence at the 0.02 level to support the testing firm's claim?

Answer 1

Answer:

$z=\frac{56-56.1}{\frac{2.1}{\sqrt{200}}}=-0.673$    

Since we are conducting a left tailed test the p value would be:

$p_v =P(Z$  

Since we have that the p value is higher than the significance level of 0.02 we have enough evidence to conclude that the true mean is not significantly lower than 56.1 MPG the reference value.

Step-by-step explanation:

Information given

$\bar X=56$ represent the sample mean for the MPG

$\sigma=2.1$ represent the population standard deviation

$n=200$ sample size  

$\mu_o =56.1$ represent the reference value

$\alpha=0.02$ represent the significance level

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to check if the true mean is less than 56.1 MPG the reference value, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 56.1$  

Alternative hypothesis:$\mu < 56.1$  

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

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

Replacing the info given we got:

$z=\frac{56-56.1}{\frac{2.1}{\sqrt{200}}}=-0.673$    

Since we are conducting a left tailed test the p value would be:

$p_v =P(Z$  

Since we have that the p value is higher than the significance level of 0.02 we have enough evidence to conclude that the true mean is not significantly lower than 56.1 MPG the reference value.