Question

A researcher wants to determine whether there is a relationship between age and mode of shopping. In a random sample of 18- to 21-year-olds, 140 of 185 people shopped online more often than in stores. In a random sample of 40- to 43-year-olds, 90 of 155 shops online more often than in stores. What is the margin of error for a 95% confidence interval for the difference between the proportions of shoppers in these age groups who shop online more than in stores?

0.1229

0.0993

0.0863

0.0507

0.0026

Answer 1

Formula for margin of error for a 95% confidence interval for the difference between the proportions

Sample 1 : 18-to-21-year-olds

n₁ : Number of 18- to 21-year-olds sampled = 185

x₁ : Number of 18- to 21-year-olds shopped online more often than in stores = 140

Sample proportion of shoppers in age group 18-to-21-year-olds  who shop online more than in stores :  

= x₁/n₁ = 140/185 = 0.7568

Sample 2 : 40- to 43-year-olds

n₂ : Number of 40- to 43-year-olds sampled = 90

x₂ : Number of 40- to 43-year-olds shopped online more often than in stores = 155

Sample proportion of shoppers in age group 40-to-43-year-olds  who shop online more than in stores :

= x₂/n₂ = 90/155 = 0.5806

for 95% confidence level = (100-95)/100 =0.05

/2 = 0.05/2 =0.025

Z_/2 = Z_0.025 =1.96

margin of error for a 95% confidence interval for the difference between the proportions of shoppers in these age groups who shop online more than in stores

margin of error for a 95% confidence interval for the difference between the proportions of shoppers in these age groups who shop online more than in stores = 0.0993

Answer :

0.0993