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

0.050.05

with

9999​%

confidence if

​(a) she uses a previous estimate of

0.560.56​?

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

Answer 1

Solution :

Given that,

= 0.56

1 - = 1 - 0.56= 0.44

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 = 2.58

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

= (2.58 / 0.05)² * 0.56 * 0.44

= 656

Sample size = 656

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5 ( estimate is not given than use 0.5)

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 = 2.58

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

= (2.58 / 0.05)² * 0.5 * 0.5

= 665.64

Sample size = 666