Question

An automobile manufacturer has given its car a 46.7 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 150 cars, they found a mean MPG of 46.5. Assume the population standard deviation is known to be 1.1. A level of significance of 0.05 will be used. Find the value of the test statistic. Round your answer to two decimal places.

Answer 1

`z=(46.5-46.7)/((1.1)/(√(150)))=-2.23`

The p value would be given by:

`p_v =2*P(z<-2.23)=0.0257`

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

Explanation:

Information given

`\bar X=46.5` represent the mean

`\sigma=1.1` represent the population standard deviation

`n=150` sample size

`\mu_o =46.7` represent the value to verify

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

z would represent the statistic

`p_v` represent the p value

Hypothesis to test

We want to test if the true mean for this case is 46.7, the system of hypothesis would be:

Null hypothesis:
`\mu = 46.7`

Alternative hypothesis:
`\mu \\eq 46.7`

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

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

Replacing we got:

`z=(46.5-46.7)/((1.1)/(√(150)))=-2.23`

The p value would be given by:

`p_v =2*P(z<-2.23)=0.0257`

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG