Question

The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is greater than 30. 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 30.45 years and the sample standard deviation was 5 years. What is the p-value associated with the test statistic?

Answer 1

Answer:

0.077994

Step-by-step explanation:

Given that the owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club.

$H_0: Mean = 30\\H_a: Mean >30$

(Right tailed test)

Sample size = $n=250$

Sample mean = 30.45

Mean difference =$30.45-30=0.45$

Sample std dev s = 5

Sample std error = $\frac{5}{\sqrt{n} } =\frac{5}{\sqrt{250} } \\=0.3163$

Test statistic = Mean diff/std error = $=\frac{0.45}{0.3163} =1.423$

Since population std deviation is not know we use t test

p value = $0.077994$