In a sample survey of 68 high-tech companies, Pelham Associates reported that the average training budget per employee for companies in the sample was $894.Assume the sample was selected from a population of 1320 companies, and that the standard deviation for the sample was $446.Build a 95% confidence interval estimate of the average training budget per employee for the population of high-tech firms represented here.Now suppose you want to reduce the margin of error for the interval you produced to $20, without sacrificing confidence.How big a sample size would you recommend?
A)781
B)1824
C)1911
D)924
E)1135

Question

Quizplus · Accepted Answer

The correct answer is A.The margin of error for a 95% confidence interval is given by:ME = z * (s / sqrt(n))where z is the z-score corresponding to the desired confidence level, s is the sample standard deviation, and n is the sample size.For a 95% confidence level, z = 1.96. Substituting the given values into the formula, we get:ME = 1.96 * (446 / sqrt(68)) = 123.47The 95% confidence interval is then:(894 - 123.47, 894 + 123.47) = (770.53, 1017.47)To reduce the margin of error to $20, we need to solve the following equation for n:20 = 1.96 * (446 / sqrt(n))Solving for n, we get:n = (1.96 * 446 / 20)^2 = 781Therefore, we would recommend a sample size of 781 to reduce the margin of error to $20.

In a Sample Survey of 68 High-Tech Companies, Pelham Associates