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SAMPLE SIZES WHEN RANDOM SAMPLING FINITE POPULATIONS a Estimating m and t
Consider randomly selecting a sample of n measurements without replacement from a finite population consisting of N measurements and having variance s2. Also consider the sample size given by the formula n = Ns2/[(N - 1)D + s2]. It can be shown that this sample size makes the margin of error in a 100(1 - a) percent confidence interval for m equal to E if we set D equal to (E/za/2) . It can also be shown that this sample size makes the margin of error in a 100(1 - a) percent confidence interval for t equal to E if we set D equal to [E/(za/2N)] . Now consider Exercise 8.52. Using s2 = (1.26)2 = 1.5876 as an estimate of s2, determine the sample size that makes the margin of error in a 95 percent confidence interval for the total number of person-days lost to unexcused absences last year equal to 100 days.
Estimating p and t.
Consider randomly selecting a sample of n units without replacement from a finite population consisting of N units and having a proportion p of these units fall into a particular category.
Also, consider the sample size given by the formula n = Np(1 - p)/[(N - 1)D + p(1 - p)]. It can be shown that this sample size makes the margin of error in a 100(1 - a) percent confi- dence interval for p equal to E if we set D equal to (E/z )2. It can also be shown that this sample size makes the margin of error in a 100(1 - a) percent confidence interval for t equal to E if we set D equal to [E/(z N)]2. Now consider Exercise 8.53. Using pˆ = .31 as an estimate of p, determine the sample size that makes the margin of error in a 95 percent confidence interval for the proportion of the 1,323 vouchers that were filled out incorrectly equal to .04.
Text Book: Business Statics in Practice By BOWERMAN.
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