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 95% confidence interval in a margin of error within plus or -6 dollars. Assuming a standard deviation for spending on Mother’s Day is $36. The required sample is

Answer 1

Solution

standard deviation =   =$36

Margin of error = E = $6

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

sample size = n = [Z/2* / E] 2

n = ( 1.96* 36 / 6 )2

n =138.29

Sample size = n =139