A simple random sample of firms has been drawn from a large metropolitan area. T

Question

A simple random sample of firms has been drawn from a large metropolitan area. The number of firms sampled was 100. One of the variables of interest is the size of each firm in terms of employees. From the sample, you determine that the average size of a firm in the sample is 23.7 employees. The variance in firm size is 1.388. If the total number of firms in the area is 17, 579, what is the total number of employees you would estimate in the region, and what is the mean number of employees per firm? What are the 95 percent confidence limits on each of these estimates?

Explanation / Answer

For sample of size 100

n = 100

mean = 23.7

std. dev. = sqrt(1388) = 37.26

For population, mean = 23.7

std. error. = 37.26/sqrt(100) = 3.73

For 95% Confidence limits, z-score value = 1.96

Lower limit = 23.7 - 1.96*3.73 = 16.3892

Upper limit = 23.7 + 1.96*3.73 = 31.0108

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