The owner of a local nightclub has recently surveyed a random sample of n = 300 customers of the club.She would now like to determine whether or not the mean age of her customers is over 35.If so,she plans to alter the entertainment to appeal to an older crowd.If not,no entertainment changes will be made.Suppose she found that the sample mean was 35.5 years and the population standard deviation was 5 years.What is the p-value associated with the test statistic?
A)0.9582
B)1.7300
C)0.0418
D)0.0836

Question

Quizplus · Accepted Answer

We need to conduct a one-sample t-test, where the null and alternative hypotheses are:

$H_0: \mu \leq 35$ (the mean age is not over 35)
$H_1: \mu > 35$ (the mean age is over 35)

The test statistic can be calculated as:

$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{35.5 - 35}{\frac{5}{\sqrt{300}}} \approx 2.23$

The degrees of freedom for the t-distribution is $n-1 = 299$, and using a t-table or calculator, we can find the p-value associated with this test statistic to be approximately 0.0418.

Therefore, the best choice is C, with the p-value of 0.0418.

The Owner of a Local Nightclub Has Recently Surveyed a Random