The personnel director of a large corporation wishes to stud

The personnel director of a large corporation wishes to study absenteeism among clerical workers at the corporation\'s central office during the year. A random sample of 25 clerical workers reveals the following:

- Absenteeism: X (with line over top) = 9.7, S = 4.0 days.

- 12 clerical workers were absent more than 10 days.

a. Construct a 95% confidence interval estimate for the mean number of absences for clerical workers during the year.

b. Construct a 95% confidence interval estimate for the population proportion of clerical workers absent more than 10 days during the year.

Suppose that the personnel directior also wishes to take a surver in a branch office. Answer these questions.

c. What sample size is needed to have 95% confidence in estimating the popularion mean absenteeism to within + - 1.5 days if the population standard is estimated to be 4.5 days?

d. How many clerical workers need to be selected to have 90% confidence in estimating the population proportion to within + - 0.075 if no prevois estimate is available?

e. Based on c and d, what sample size is needed if a single survey is being conducted?

Solution

(a) Given a=0.05, Z(0.025) = 1.96 (from standard normal table)

So 95% confidence interval is

xbar +/- Z*s/vn

--> 9.7 +/- 1.96*4/sqrt(25)

--> (8.132, 11.268)

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(b) p=12/25 = 0.48

So 95% confidence interval is

p +/- Z*sqrt(p*(1-p)/n)

--> 0.48 +/- 1.96*sqrt(0.48*(1-0.48)/25)

--> (0.2841569, 0.6758431)

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(c) n= (Z*s/E)^2

=(1.96*4.5/1.5)^2

=34.5744

Take n=35

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(d) Given a=0.1, Z(0.05)=1.645 (from standard normal table)

We use p=0.5 as estimated.

So n=(Z/E)^2*p*(1-p)

=(1.645/0.075)^2*0.5*0.5

=120.2678

Take n=121

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(e) sample size 35