Question

A car company claims that the mean gas mileage for its luxury sedan is at least 24 miles per gallon. A random sample of 7 cars has a mean gas mileage of 23 miles per gallon and a standard deviation of 2.4 miles per gallon. At α=0.05, can you support the company’s claim assuming the population is normally distributed? Yes, since the test statistic is not in the rejection region defined by the critical value, the null is not rejected. The claim is the null, so is supported No, since the test statistic is in the rejection region defined by the critical value, the null is rejected. The claim is the null, so is not supported Yes, since the test statistic is not in the rejection region defined by the critical value, the null is not rejected. The claim is the null, so is supported No, since the test statistic is not in the rejection region defined by the critical value, the null is not rejected. The claim is the null, so is supported

Answer 1

Given that, sample size ( n ) = 7

sample mean ( M ) = 23 miles per gallon

sample standard deviation ( s ) = 2.4 miles per gallon

significance level α = 0.05

The null and the alternative hypotheses are,

Test statistic is,

t = -1.102

Since, it is left-tailed test, t-critical value at α=0.05 with degrees of freedom = 7 - 1 = 6 is, t^* = -1.943

Here, test statiatic = -1.102 > critical value = -1.943

so, we do not reject the null hypothesis.

Therefore, conclusion is,

Yes, since the test statistic is not in the rejection region defined by the critical value, the null is not rejected. The claim is the null, so is supported