Question

A study found that 73% of elementary students buy lunch at school at least once a month. How large a sample would be necessary to estimate the true proportion within 3 percentage points with 95% confidence?

Answer 1

Solution :

Given that,

= 73% = 0.73

1 - = 1 - 0.73 = 0.27

Margin of error = E = 3% = 0.03

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Sample size = (  Z_/2 / E)² * * (1 - )

= (1.96 / 0.03)² * 0.73 * 0.27

= 841.31 = 842

Sample size = n = 842