Answer:
We conclude that the the manufacturer's advertising campaign is not correct.
P-value of test = 0.067.
Step-by-step explanation:
We are given that an independent testing service road-tested 36 of the automobiles. The sample showed an average of 51.5 miles per gallon. The population standard deviation is 6 miles per gallon.
We have to conduct a test to determine whether or not the manufacturer's advertising campaign is legitimate.
<u><em>Let </em></u><u><em> = population average gasoline of new small cars in highway driving.</em></u>
SO, <u>Null Hypothesis</u>, :
50 miles per gallon {means that the manufacturer's advertising campaign is not correct}
<u>Alternate Hypothesis</u>, :
> 50 miles per gallon {means that the manufacturer's advertising campaign is correct}
The test statistics that will be used here is <u>One-sample z test</u> <u>statistics</u> as we know about the population standard deviation;
T.S. = ~ N(0,1)
where, = sample average gasoline = 51.5 miles per gallon
= population standard deviation = 6 miles per gallon
n = sample of automobiles = 36
So, <em><u>test statistics</u></em> =
= 1.50
<em>Now at 0.05 significance level, the z table gives critical value of 1.6449 for right-tailed test. Since our test statistics is less than the critical values of z as 1.50 < 1.6449, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.</em>
<u>Also, P-value is given by the following formula;</u>
P-value = P(Z > 1.50) = 1 - P(Z 1.50)
= 1 - 0.93319 = <u>0.067 or 6.7%</u>
Therefore, we conclude that the the manufacturer's advertising campaign is not correct as its new small cars average less than or equal to 50 miles per gallon in highway driving.