This question is incomplete, the complete question is;
A sociologist wants to construct an 80 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 036. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.02?
Answer:
the required or needed sample size is 946
Step-by-step explanation:
Given the data in the question;
confidence level = 80%
⇒ 100( 1 - ∝ )% = 80%
100 - 100∝ = 80
100∝ = 100 - 80
∝ = 20/100
level of significance ∝ = 0.2
[tex]Z_{\alpha /2[/tex] = [tex]Z_{0.10[/tex] = 1.2816 { from z-table }
p = 0.36
Allowable margin of error E = 0.02
let n be the required sample size.
we know that
E = √( p( 1 - p) / n )
given that margin of error E = 0.02
⇒ √( p( 1 - p) / n ) × [tex]Z_{\alpha /2[/tex] = 0.02
⇒ √( 0.36( 1 - 0.36) / n ) × 1.2816 = 0.02
⇒ √( 0.2304 / n ) × 1.2816 = 0.02
√( 0.2304 / n ) = 0.02 / 1.2816
√( 0.2304 / n ) = 0.01560549
( 0.2304 / n ) = ( 0.01560549)²
0.2304 / n = 0.000243531318
n = 0.2304 / 0.000243531318
n = 946
Therefore, the required or needed sample size is 946
