The Bureau of Labor Statistics is planning the next yearly survey to determine t

Question

The Bureau of Labor Statistics is planning the next yearly survey to determine the average cost of a summer vacation for a US family. The following standards have been set a confidence level of 95% and an error of less than $100. Past research has indicated that the standard deviation should be $656. What is the required sample size? What would happen to your results if you use a sample smaller than the required sample size? What would be the required sample size if the BLS standards specify a 99% confidence level instead? You can either do the calculations on your own, or use the "Calculating required sample size" Excel template).

Explanation / Answer

a)

let sample size be n

standard deviation = 656

standard deviation of average cost of a summer vacation = 656/n

margin of error for 95% confindence interval is 1.96*(standard deviation of average cost) i.e 1.96*(656/n)

1.96*(656/n) = 100

n = 1.96*656/100 = 12.8576

n = 12.85762 = 165.32 ~166

required sample size is 166

b)

if you use a sample smaller than the required sample size i.e 166 the error will be greater than $100

c)

margin of error for 99% confindence interval is 2.575*(standard deviation of average cost) i.e 2.575*(656/n)

2.575*(656/n) = 100

n = 2.575*656/100 = 16.892

n = 16.8922 = 285.339 ~ 286

required sample size is 286

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