Question

A researcher wishes to estimate the proportion of adults who have​ high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within

0.03 with 90% confidence if

​(a) she uses a previous estimate of

0.36?

​(b) she does not use any prior​ estimates?

Answer 1

Solution :

Given that,

margin of error = E = 0.03

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

(a)

= 0.36

1 - = 1 - 0.36 = 0.64

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.645 / 0.03)² * 0.36 * 0.64

= 692.74 = 693

sample size = 693

(b)

= 0.5

1 - = 1 - 0.5 = 0.5

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.645 / 0.03)² * 0.5 * 0.5

= 751.67 = 752

sample size = 752