Question
In: Statistics and Probability
You may need to use the appropriate appendix table or technology to answer this question. In...
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.)
Solutions
Expert Solution
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
orchestra answered 2 years agoRelated Solutions
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