Question

What is the minimal sample size needed for a 95% confidence interval to have a maximal margin of error of 0.1 in the following scenarios? (Round your answers up the nearest whole number.)

(a) a preliminary estimate for p is 0.39


(b) there is no preliminary estimate for p

Answer 1

Solution :

Given that,

= 0.39

1 - = 1 - 0.39 = 0.61

margin of error = E = 0.1

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

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

= (1.96 / 0.1)2 * 0.39* 0.61

= 91.39

Sample size = 92

(B)

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 0.1

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

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

= (1.96 / 0.1)2 * 0.5* 0.5

= 96.04

Sample size =96