9. A survey is planned to determine the mean annual family medical expenses of e

Question

9. A survey is planned to determine the mean annual family medical expenses of employees of a large company. The management of the company wishes to be 99% confident that the sample mean is corect to within S75 of the population mean annual family medical expenses. A previous study indicates that the standard deviation is approximately S403. a. How large a sample is necessary? b. If management wants to be correct to within ± S30, how many employees need to be selected? a. How large a sample is necessary? Round up to the nearest integer.) b. If management wants to be correct to within ± S30, how many employees need to be selected? (Round up to the nearest integer.) 1: z-table

Explanation / Answer

Question 9

(a) Here margin of error = 75

Here standard deviation = $ 403

So Here margin of error = critical test statistic * standard error of sample mean

critical test statistic (99% confidence interval) = + 2.575

standard error of sample mean = / (n) = 403/ n

so 75 = 2.575 * 403/ n

n = 2.575 * 403/ 75

n = 191.44

so sample size of 192 would be necessary

(b) Now the Margin of error is reduced to +- 30

so,

30 =  2.575 * 403/ n

n = 2.575 * 403/ 30 = 34.59

n = 1196.53 or 1197

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