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 0.05 with 99% confidence if

(a) she uses a previous estimate of 0.54?

(b)she does not use any prior estimates?

Answer 1

Solution :

Given that,

margin of error = E = 0.05

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

(a)

= 0.54

1 - = 1 - 0.54 = 0.46  

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

= (2.576 / 0.05)² * 0.54 * 0.46

= 659.33

= 660

sample size = 660

(b)

= 0.5

1 - = 1 - 0.5 = 0.5  

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

= (2.576 / 0.05)² * 0.5 * 0.5

= 663.57

= 664

sample size = 664