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).

Answer 1

The difference between the observed sample mean (x_bar) and the population mean (mu) should be $100.

The population standard deviation (sigma) = 656
So, standard error = standard deviation of the sample means = sigma / sqrt(sample size) = 656/sqrt(n) where n is the sample size.

Z = NORMSINV(1 - (0.05/2)) = 1.96

x + Z*standard error = x + 100
or, standard error = 100/Z
or, 656/sqrt(n) = 100/Z
or, n = (656*Z/100)² = (656*1.96/100)² = 165.3 or 166 (rounded off)

When the confidence level is 99%, Z will be = NORMSINV(1 - (0.01/2)) = 2.58

n = (656*2.58/100)² = 285.5 or 286 (rounded off)

Using Excel sheet will give us result as -