Question

An article a newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the results of a pol, showed that 46% would like more discussion about the family's financial situation, 37% would like to talk about school, and 30% would like to talk about religion. These and other percentages were based on a national sampling of 523 teenagers. Estimate the proportion of a teenagers who want more family discussions about school. 90% confidence level. Express the answer in the form and round to the nearest thousandth O A 0.63 0.035 OB. 0.37 0.035 OC. 0.63 +0.002 OD. 0.37 0.002

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = 0.37

1 - $\hat p$ = 1 - 0.37 = 0.63

Z$\alpha$/2 = Z0.05 = 1.645

Margin of error = E = Z$\alpha$ / 2 * (($\hat p$ * (1 - $\hat p$ ))$\sqrt$ / n)

= 1.645 ($\sqrt$((0.37 * 0.63) / 523)

= 0.035

A 90% confidence interval for population proportion p is ,

$\hat p$ ± E   

= 0.37 ± 0.035

correct option is = B