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 0.15


(b) there is no preliminary estimate for p

Answer 1

Solution:

Given that,

(a) a preliminary estimate for p is 0.15

= 0.15

1 - = 1 - 0.15 = 0.85

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 = Z_0.025 = 1.960

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

= (1.960 / 0.1)² * 0.15 * 0.85

= 48.9804

= 49

n = sample size = 49

(b) there is no preliminary estimate for p

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 4% = 0.1

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

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

= (1.960 / 0.1)² * 0.5 * 0.5

= 96.04

n = sample size = 96