The personnel director of a large corporation wants to study absenteeism among c

Question

The personnel director of a large corporation wants 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= 9.7 days, S = 4.0 days.
12 clerical workers were absent more than 10 days.

a. Set up a 95 % confidence interval estimate of the average number of absences for clerical workers last year.
b. Set up a 95% confidence interval estimate of the population proportion of clerical workers absent more than 10 days last year.
c. If the personnel director also wants to take a survey in a branch office, what sample size is needed if the director wants to be 95% confident of being correct to within ± 1.5 days and the population standard deviation is assumed to be 4.5 days?
d. If the personnel director also wants to take a survey in a branch office, what sample size is needed if the director wants to be 90% confident of being correct to within ± 0.075 of the population proportion of workers who are absent more than 10 days if no previous estimate is available?

Explanation / Answer

a) here std error=std deviation/(n)1/2 =0.8

for 95% CI and 24 degree if freedom t=2.0639

hence 95% confidence interval =sample mean -/+ t*std error =8.0489 ; 11.3511

b) estimated proportion =12/25=0.48

std error =(p(1-p)/n)1/2 =0.0999

for 95% CI, z=1.96

hence confidence interval =sample proportion -/+ z*std error =0.2842 ; 0.6758

c) margin of error E =1.5

for 95% CI, z=1.96

std deviation =4.5

hence sample size n=(z*std deviation/E)2 =~35

d) margin of error E=0.075

for no previous estimate p=0.5

for 90% CI, z=1.645

hence sample size n=~p(1-p)*(z/E)2 =~121

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.