Question

You may need to use the appropriate appendix table or technology to answer this question. In the past, 19% of all homes with a stay-at-home parent had the father as the stay-at-home parent. An independent research firm has been charged with conducting a sample survey to obtain more current information. (a) What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.03? Use a 95% confidence level. (Round your answer up to the nearest whole number.) . (b) Repeat part (a) using a 99% confidence level. (Round your answer up to the nearest whole number.)

Answer 1

Solution,

Given that,

= 0.19

1 - = 1 - 0.19 = 0.81

margin of error = E = 0.03

a) At 95% confidence level

= 1 - 95%

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025  = 1.96

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

= (1.96 / 0.03)2 * 0.19 * 0.81

= 656.91

sample size = n = 657

b) At 99% confidence level

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = Z0.005  = 2.576

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

= (2.576 / 0.03)2 * 0.19 * 0.81

= 1134.71

sample size = n = 1135