Question

The Retail Advertising and Marketing Association would like to estimate the average amount of money that a person spends for Mother’s Day with a 99% confidence interval and a margin of error within plus or minus $12. Assuming the standard deviation for spending on Mother’s Day is $36, the required sample size is ________. a.) 60 b.)72 c.)97 d.)108

Answer 1

a.) 60

Explanation:

Margin of Error formula = z × s/√n

z = z score of confidence interval

We are given z score of 99% = 2.58

s = Standard deviation = $36

Margin of Error= ± 12

n= unknown

= 12 = 2.58 × 36/√n

12 × √n = 92.88

√n = 92.88/12

√n = 7.74

Square both sides

(√n)² = 7.74²

n = 59.9076

Approximately n ≈ 60

Therefore, the required sample size = 60