Question

According to a report by the PEW Research Center, 85% of adults under 30 feel optimistic about the economy, but the optimism is shared by only 45% of those who are over 50 (Newsweek, September 13, 2010). A research analyst would like to construct 95% confidence intervals for the proportion patterns in various regions of the country. She uses the reported rates by the PEW Research Center to determine the sample size that would restrict the margin of error to within 0.05. [You may find it useful to reference the z table.) a. How large a sample is required to estimate the proportion of adults under 30 who feel optimistic about the economy? (Round intermediate calculations to at least 4 decimal places. Round "z" value to 3 decimal places. Round up your answer to the nearest whole number.) Sample size b. How large a sample is required to estimate the proportion of adults over 50 who feel optimistic about the economy? (Round intermediate calculations to at least 4 decimal places. Round " value to 3 decimal places. Round up your answer to the nearest whole number.) Sample size

Answer 1

Solution,

Given that,

a) $\hat p$ = 0.85

1 - $\hat p$ = 1 - 0.85 = 0.15

margin of error = E = 0.05

At 95% confidence level

$\alpha$ = 1 - 95%  

$\alpha$ = 1 - 0.95 =0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z0.025  = 1.96

sample size = n = (Z$\alpha$ / 2 / E )2 * $\hat p$ * (1 - $\hat p$ )

= (1.96 / 0.05 )2 * 0.85 * 0.15

= 195.9216

sample size = n = 196

b) $\hat p$ = 0.45

1 - $\hat p$ = 1 - 0.85 = 0.55

sample size = n = (Z$\alpha$ / 2 / E )2 * $\hat p$ * (1 - $\hat p$ )

= (1.96 / 0.05 )2 * 0.45 * 0.55

= 380.3184

sample size = n = 381