Question

2. At a computer manufacturing company, the actual size of computer chips is uormally^ distributed with a mean of 10 centimeter and a standard deviation of 2.1 centimeter. A random sample of 24 computer chips is taken. What is the probability that the sample mean will be between 10.95 and 11.05 centimeters? 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: [20 points] Absenteeism: X -9.7 days, S- 4.0 days. 12 clerical workers were absent more than 10 days. Set up a 95 % confidence interval estimate of the average number of absences for clerical workers last vear a. b. Set up a 95% confidence interval estimate of the population proportion of clerical workers absent more than 10 days last year 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 t 1.5 days and the population standard deviation is assumed to be 4.5 days? c. 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 tharn 10 days if no previous estimate is available?

Answer 1

2.
Standard error of sample mean = s / sqrt(n) = 2.1 / sqrt(24) = 0.4286607

Probability that the sample mean lie between 10.95 and 11.05 = P(10.95 < X < 11.05)
= P(X < 11.05) - P(X < 10.95)
= P[Z < (11.05 - 10) / 0.4286607] - P[Z < (10.95 - 10) / 0.4286607]
= P[Z < 2.45] - P[Z < 2.22]
= 0.9929 - 0.9868
= 0.0061

a.
Standard error of sample mean = s / sqrt(n) = 4 / sqrt(25) = 0.8
Degree of freedom = n - 1 = 25 - 1 = 24
Critical value of t at 95% confidence interval is 2.06
95% confidence interval is
(9.7 - 2.06 * 0.8, 9.7 + 2.06 * 0.8)
(8.052, 11.348)

b.
Sample proportion = 12/ 25 = 0.48
Standard error of sample proportion = sqrt(p(1-p)/n) = sqrt(0.48 * (1- 0.48)/25) = 0.1
z value for 95% confidence interval is 1.96
95% confidence interval is,
(0.48 - 1.96 * 0.1, 0.48 + 1.96 * 0.1)
(0.284, 0.676)

c.
Margin of error E = 1.5
Std deviation, s = 4.5
Sample size = (z * s / E)²

= (1.96 * 4.5 / 1.5)²

= 35 (Rounding to nearest integer)

d.

If no previous estimate is available, we assume p = 0.5

Sample size, n = (z / E)² * p(1-p)

= (1.96 / 0.075)² * 0.5 * (1 - 0.5)

= 171    (Rounding to nearest integer)