The number of the sample that is necessary is 171. The number of employees that would be selected if E is 30 = 476
The standard deviation = 398
error E = 50
Alpha is = 1-0.90 = 0.10
a. We are to find the value of the size of the sample
Z∝/2 = Z0.05
= 1.645
sample size,n >= [Za/2 *(sigma/E)]^2
n >= [1.645 * (398/50)] ^2
n >= 171.46
n = 171
b.
error,E = 30
n >= [1.645 * (398/30)] ^2
n >= 476.27
n = 476
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 90% confident that the sample mean is correct to within ±$50 of the population mean annual family medical expenses. A previous study indicates that the standard deviation is approximately $398. a. How large a sample is necessary?
b. If management wants to be correct to within ±$30, how many employees need to be selected?
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