Bastien, Inc. has been manufacturing small automobiles that have averaged 50 miles per gallon of gasoline in highway driving. The company has developed a more efficient engine for its small cars and now advertises that its new small cars average more than 50 miles per gallon in highway driving. 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. You conduct a test with a 0.05 level of significance to determine whether or not the manufacturer's advertising campaign is legitimate. What is the p-value associated with the sample results (2 decimals)?

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.

Let [tex]\mu[/tex] = population average gasoline of new small cars in highway driving.

SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\leq[/tex] 50 miles per gallon {means that the manufacturer's advertising campaign is not correct}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 50 miles per gallon {means that the manufacturer's advertising campaign is correct}

The test statistics that will be used here is One-sample z test statistics as we know about the population standard deviation;

T.S. = [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)

where, [tex]\bar X[/tex] = sample average gasoline = 51.5 miles per gallon

[tex]\sigma[/tex] = population standard deviation = 6 miles per gallon

n = sample of automobiles = 36

So, test statistics = [tex]\frac{51.5-50}{\frac{6}{\sqrt{36} } }[/tex]

= 1.50

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.

Also, P-value is given by the following formula;

P-value = P(Z > 1.50) = 1 - P(Z [tex]\leq[/tex] 1.50)

= 1 - 0.93319 = 0.067 or 6.7%

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.