Question

Call me: A sociologist wants to construct a 98% confidence interval for the proportion of children aged 8-10 living in New York who own a cell phone.

(a) A survey by the National Consumers League taken in

2012 estimated the nationwide proportion to be 0.28. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.03?

Answer 1

Solution :

Given that,

= 0.28

1 - = 1 - 0.28=0.72

margin of error = E = 0.03

At 98% confidence level the z is,

= 1 - 98%

= 1 - 0.98 = 0.02

/2 = 0.01

Z/2 = 2.33 ( Using z table )

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

= (2.33 / 0.03)2 * 0.28 * 0.72

=1216.07

Sample size =1216 rounded