Question

A large software development firm recently relocated its facilities. Top management is interested in fostering good relations with their new local community and has encouraged their professional employees to engage in local service activities. They believe that the​ firm's professionals volunteer an average of more than 21 hours per month. If this is not the​ case, they will institute an incentive program to increase community involvement. A random sample of 24 professionals yields a mean of 22 hours and a standard deviation of 2.22 hours.
12 13 14 14 15 15 15 16 16 16 16 16
17 17 17 18 18 18 18 19 19 19 20 21

The sample has a mean of 16.6 hours and a standard deviation of 2.22 credit hours.

At α = 0.05,

I. we reject the null hypothesis.
II. we fail to reject the null hypothesis.
III. the firm shouldn't need to institute an incentive program because the evidence indicates that professional employees volunteer an average of more than 15 hours per month in their local community.

a. II and III
b. I and III
c. II only
d. I only
e. III only

Answer 1

Answer:

B. i and iii

Step-by-step explanation:

There are mistakes in your question. But I solve this problem using the average mean of 15 because that is what is contained in the solution

Hypothesis

H0: u <= 15

H1: u > 15

Mean of x = 16.6

S = 2.22

Test statistic used is t test

16.6 - 15 / 2.22/√24

= 3.531

We find the critical value

Degree of freedom = 24-1 = 23

This is a one tailed test

Critical value = 1.714

We find p value using Ms excel T distribution function

= 0.000894

To make the decision

3.531 > 1.714 so we conclude that

I.) Reject null hypothesis

Iii) conclude an average if more than 15 hours is volunteered per month. So no need to institute the program.

The answer option Is B