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A survey is planned to determine the mean annual family medical expenses of SVSU staff and...

A survey is planned to determine the mean annual family medical expenses of SVSU staff and faculty. The staff conducting this research wish to be 95% confident that the sample mean is correct to within $50 of the population mean. A previous study indicates the standard deviation is approximately $294.

How large of a sample is necessary?

Math   Statistics-And-Probability   
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Answer #1

Solution :

Given that,

standard deviation = = 294

margin of error = E = 50

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

Sample size = n = ((Z/2 * ) / E)2

= ((1.960* 294) / 50)2

= 132.82

= 133

Sample size = 133

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