Question

13) 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 0.28 (b) there is no preliminary estimate for p

Answer 1

Solution :

Given that,

= 0.28

1 - = 1 - 0.28 = 0.72

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 * 0.28 * 0.72

= 77.44

Sample size =78

(B)

Solution :

Given that,

= 0.5 ( assume 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