Question

A sociologist wants to construct a 90% confidence interval for the proportion of children aged 8-10 living in New York who own a cell phone. A survey by the National Consumers League taken in 2012 estimates the nationwide proportion to be 0.41. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.02?

Answer 1

Solution :

Given that,

= 0.41

1 - = 1 - 0.41 = 0.59

margin of error = E =0.02

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.645 / 0.02)2 * 0.41 * 0.59

=1636.46

Sample size = 1637