Wage earners in a large firm are stratified into management and clerical classes

Question

Wage earners in a large firm are stratified into management and clerical classes, the first having 300 and the second having 500 employees. To assess attitude on sick-leave policy, independent random samples of 100 workers each were selected, one sample from each of the classes. After the sample data were collected, the responses were divided according to gender. In the table of results, a= Number who like the policy, b = Number who dislike the policy: and c = Number who have no opinion. Find an estimate and an estimated variance of that estimate for each parameter listed below: Proportion of managers who like the policy. Proportion of clerical workers who like the policy. Difference between the proportion of managers who like the policy and the proportion of clerical workers who like the policy. Is this difference significant? Proportion of wage earners who like the policy. Total number of clerical wage earners who dislike the policy Total number of clerical wage earners who have no opinion on the policy.

Explanation / Answer

Solution

Back-up Theory

1. An estimate of the population proportion is the sample proportion, say p.

2. Variance of the above estimate = p(1 - p)/n, where n = sample size.

3. An estimate of the number who fall into one category in the population is corresponding number in the sample.

4. Variance of the above estimate = np(1 - p), where n = sample size.

Part (a)

Estimate of proportion of managers who like the policy,

paM = number of managers in the sample who like the policy/total number of employees surveyed

= {60(M) + 10(F)}/100 = 0.7 ANSWER1

Estimate of variance of paM = (0.7)(0.3)/100 = 0.0021 ANSWER2

Part (b)

Estimate of proportion of clerical staff who like the policy,

paC = number of clerical staff in the sample who like the policy/total number of employees surveyed

= {24(M) + 42(F)}/100 = 0.66 ANSWER1

Estimate of variance of paC = (0.66)(0.34)/100 = 0.002234 ANSWER2

Part (c)

Estimate of the difference in proportion of managers who like the policy and proportion of clerical staff who like the policy = paM - paC = 0.7 – 0.66 = 0.04 ANSWER 1

Estimate of variance of this estimate = variance of paM + variance of paC

= 0.0021 + 0.002234 = 0.004334 ANSWER 2

To test if this difference is significant,

the test statistic is Z = (difference between two proportions)/{p(1 - p)(2/n)}, where p = (paM + paC)/2 and n = 100.

So, Z = 0.04/(0.68x0.32x0.02) = 0.04/0.004352 = 0.04/0.0660 = 0.606

Z has Standard Normal Distribution and so we conclude the difference is not significant if the above value of Z is greater than 1.645(i.e., upper 5% point of Standard Normal Distribution obtained from Statistical Tables.)

Since 0.606 < 1.645, we conclude that the difference is not significant. ANSWER 3

Part (d)

Estimate of proportion of wage earners who like the policy,

paW = number of wage earners in the sample who like the policy/total number of employees surveyed

= (number of managers + number of clerical staff) who like the policy/200 = (70 + 66)/200 = 0.68 ANSWER1

Estimate of variance = (0.68)(0.32)/200 = 0.0011 ANSWER2

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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