Question

An automobile manufacturer claims that its van has a 36.9 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 240 vans, they found a mean MPG of 37.3. Assume the standard deviation is known to be 2.2. A level of significance of 0.01 will be used. Find the value of the test statistic.

Answer 1

`z=(37.3-36.9)/((2.2)/(√(240)))=2.817`

`p_v =2P(z>2.817)=0.005`

since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is different from 36.9 mpg at 1% of significance.

Explanation:

Data given and notation

`\bar X=37.3` represent the sample mean

`\sigma=2.2` represent the sample population deviation

`n=240` sample size

`\mu_o =36.9` represent the value that we want to test

`\alpha=0.01` 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 (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is different from 36.9 mpg, the system of hypothesis are:

Null hypothesis:
`\mu = 36.9`

Alternative hypothesis:
`\mu \\eq 36.9`

The statistic is given by:

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

Calculate the statistic

We can replace in formula (1) the info given like this:

`z=(37.3-36.9)/((2.2)/(√(240)))=2.817`

P-value

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

`p_v =2P(z>2.817)=0.005`

Conclusion

since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is different from 36.9 mpg at 1% of significance.