Question

The owner of a local nightclub has recently surveyed a random sample of 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 population standard deviation was 5 years. What is the p-value associated with the test statistic

Answer 1

0.0416

Explanation:

Given :

Sample size, n = 300

Sample mean, x = 35.5

Population mean, m = 35

Standard deviation, s = 5

The test statistic :

Zstatistic = (x - m) / s/sqrt(n)

Zstatistic = (35.5 - 35) / 5/sqrt(300)

Zstatistic = 0.5 / 0.2886751

Zstatistic = 1.732

Using the p value calculator from Zstatistic :

One tailed P value at 95% confidence interval is : 0.0416