Question

13. A large software development firm recently relocated its facilities. Top management is interested in fostering good relations with its new local community and has encouraged its professional employees to engage in local service activities. The company believes that its professionals volunteer an average of more than 15 hours per month. If this is not the case, it will institute an incentive program to increase community involvement.

(a) A random sample of 24 professionals reported the sample mean and standard deviation are 16.6 hours and 2.22 hours respectively. For a more accurate determination, top management wants to estimate the average number of hours volunteered per month by their professional staff to within half an hour with 99% confidence. How many randomly selected professional employees would they need to sample?

(b) Now, suppose 40 professional employees are randomly selected. This sample yields a mean of 15.2 hours and a standard deviation of 1.8 hours. Construct the 95% confidence interval. (Hint: 1.8 is the sample standard deviation, not the population standard deviation)

Answer 1

(a). We know that the Confidence interval for population mean is where

is the Standard Normal variable corresponding to the probability 0.99. We are given that =16.6 hrs and SD(s)=2.22 hours. Here, for 99% the is 2.5758. We are asked to find out the n so that the Upper-Lower=0.5 hours.

ie , This has to 0.25 because it is the two sided interval.

Therefore we have  

  

  

  

(b). We are given that hours   hours and

We need to construct a 95% confidence interval for the mean.

The Confidence interval as before is given by  . We shall look into the tables or EXCEL to find the value of

The 95% confidence interval is   

  

Therefore the 95% confidence interval is