Question

A market research director at ABC Apparels wants to study women’s spending on apparels. A survey of the store’s credit card holders is designed in order to estimate the proportion of women who purchase their apparels primarily from ABC apparels and the mean yearly amount of money that women spend on apparels. A previous survey found that the standard deviation for the amount women spend on apparels in a year is approximately $35

a. What sample size is needed to have a 90% confidence of estimating the population proportion to within ±0.052?

B. Refer to the above. Suppose that the market research director cannot afford to survey as many women as required in part (a) of and has indicated that the maximum sample size can be 100 women. Assume that the sample size is reduced to 100:

a. Without changing the confidence level, indicate specifically what must be changed and by how much?

b. Without changing the margin of error, indicate specifically what must be changed and by how much?

Answer 1

a.

Required sample size, n = (z / E)² * p(1-p)

where z is z value for 90% confidence, E is margin of error and p is the true proportion

Since p is unknown, we will assume p = 0.5 to get the largest sample size.

z value for 90% confidence level is 1.645

n = (1.645 / 0.052)² * 0.5 * (1-0.5)

= 250 (Rounding to nearest integer)

B.

a.

Sample size, n = 100

and confidence level is 90%. Hence z = 1.645

Margin of Error, E = z *

If z is fixed, we need to increase the margin of error as the n is reduced from 250 to 100

E = 1.645 *

E = 1.645 * 0.05 = 0.08225

So, the margin of error needs to increased to ±0.08225 to get the same confidence level for sample size of 100.

b.

Margin of Error, E = z *

If E is fixed, we need to decrease confidence level as the n is reduced from 250 to 100

0.052 = z *

z = 0.052 / 0.05 = 1.04

z = 1.04 for 70% confidence level (From Z distribution)

So, the confidence level needs to decreased to 70% to get the same margin of error for sample size of 100.