Question

Question 15 of 30 (1 point) View problem in a pop-up 7.3 Section Exercise 27 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 Part 1 out of 3 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? A sample of children aged 8-10 living in New York is needed to obtain a 98% confidence interval with a margin of error of 0.03 using the estimate 0.28 forp ll A Sa

Answer 1

standard error of proportion, se =    where n is the sample size

z value for 98% confidence interval = 2.33

Now,

Margin of error, E = z * se

E = z * $\sqrt{p(1-p)/n}$

=> n = (z / E)² * p(1-p)

n = (2.33 / 0.03)² * 0.28 * (1-0.28) = 1216.074

n = 1216 (Rounding to nearest integer)