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 to the nearest whole number)

(a) a preliminary estimate for p is 0.34

(b) there is no preliminary estimate for p

Answer 1

a)

Solution :

Given that,

= 0.34

1 - = 1 - 0. 34= 0.66

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 = 1.96

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

= (1.96 / 0.1)² * 0.34 * 0.66

= 86.205504

Sample size =87

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 = Z_0.025 = 1.96

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

= (1.96 / 0.1)² * 0.5 * 0.5

= 96.04

Sample size =96