Question

A sample of holiday shoppers is taken randomly from a local mall to determine the average daily spending on gifts. From a preselected sample, the standard deviation was determined to be $26. You would like to construct a 95% confidence interval for the mean daily spending on all holiday spending.

1. Find the appropriate sample size necessary to achieve a margin of error of $5.

2. Find the appropriate sample size necessary to achieve a margin of error of $8.

Answer 1

The sample size is calculated as

$n=\left (\frac{Z\sigma}{M.E} \right )^2$

For 95% Confidence Interval, Z = 1.96, given  $\sigma=26$

Therefore, the required sample size to achieve

a. margin of error of $5 is

$n=\left (\frac{1.96\times 26}{5} \right )^2=103.9\approx 104$

a. margin of error of $8 is

$n=\left (\frac{1.96\times 26}{8} \right )^2=40.6\approx 41$