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 $\mu$ = population average gasoline of new small cars in highway driving.

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

Alternate Hypothesis, $H_A$ : $\mu$ > 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. = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average gasoline = 51.5 miles per gallon

$\sigma$ = population standard deviation = 6 miles per gallon

n = sample of automobiles = 36

So, test statistics = $\frac{51.5-50}{\frac{6}{\sqrt{36} } }$

= 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 $\leq$ 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.

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4 years ago