Question

A market research director at XYZ 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 XYZ 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? DO NOT USE PHSTAT4 for this part Statistics for Decision Makers, b. What sample size is needed to have 97.5% confidence of estimating the population mean within ± $5? c. In order to keep the survey costs within budgets, the market research director decided to do a pilot survey of 25 women. The pilot survey results indicate the standard deviation for the amount women spend on apparels is $20. Based on this new information, what additional sample size is needed (i.e. in addition to the pilot survey sample) to have a 97.5% confidence of estimating the population mean within ±$5? d. If the cost of surveying an additional women is $12, how much money did the market research director save by revising the estimate for sample size (as in part c of the question above) based on the results of the pilot survey compared to the sample size he would have used had he not conducted the pilot survey (as in part b of the question above)?

Answer 1

The formula for the minimum sample size needed is

n >= ( (z*σ)/ MOE )²

A) z score for two-tailed 90% CI is 1.645 ( from z table)

σ is the population standard deviation and MOE is the margin of error.

we get n >= ( 1.645*35/ 0.052)²

i.e. n >= 1225917.39

so the minimum sample size needed is 1225918

B) z score for two tailed 97.5 %CI is 2.24

MOE here is 5

therefore n >= (2.24*35/5)²

i.e. n >= 245.8

so the minimum sample size is 246

c) revised minimum sample size

n >= (2.24*20/5)²

i.e. n >= 80.28 ~ 81

D) total money saved is $ 12 * (246-81)

= $1980