Question

ch 23* in 2012, respondents to the general social survey were asked: \there are always some people whose ideas are considered bad or dangerous by other people. for instance, somebody who is against churches and religion. if such a person wanted to make a speech in your (city/town/community) against churches and religion, should he be allowed to speak, or not?\ among the 573 male respondents, 459 said, \allow,\ whereas among the 719 female respondents, 552 said, \allow.\ take ( p_m ) and ( p_f ) to be the proportions of all males and females who would allow an anti - religionist to speak. the 90% large - sample confidence interval for the difference ( p_m - p_f ) in the proportions of males and females who would allow an anti - religionist to speak is about ( 0.033pm0.045 ). ( 0.033pm0.002 ). ( 0.033pm0.038 ). question 12 1 pts ch 23 when comparing proportions from two populations, we make inference about: ( p1 - p2 ). ( p1 + p2 ). ( hat{p}1-hat{p}2 ). ( hat{p}1+hat{p}2 ).

Explanation:

Step1: Calculate the sample proportions

For males: $\hat{p}_M=\frac{459}{573}\approx0.801$
For females: $\hat{p}_F=\frac{552}{719}\approx0.768$
The difference in sample proportions is $\hat{p}_M-\hat{p}_F = 0.801 - 0.768=0.033$

Step2: Calculate the standard error

The formula for the standard error of the difference in proportions is $SE=\sqrt{\frac{\hat{p}_M(1 - \hat{p}_M)}{n_M}+\frac{\hat{p}_F(1 - \hat{p}_F)}{n_F}}$
Substitute the values:
$\hat{p}_M = 0.801$, $n_M = 573$, $\hat{p}_F = 0.768$, $n_F = 719$
$SE=\sqrt{\frac{0.801\times(1 - 0.801)}{573}+\frac{0.768\times(1 - 0.768)}{719}}$
$=\sqrt{\frac{0.801\times0.199}{573}+\frac{0.768\times0.232}{719}}$
$=\sqrt{\frac{0.159399}{573}+\frac{0.178176}{719}}$
$=\sqrt{0.0002782 + 0.0002478}$
$=\sqrt{0.000526}\approx0.0229$

Step3: Find the critical value

For a 90% confidence interval, the critical value $z$ (from the standard normal distribution) is $z_{0.05}=1.645$

Step4: Calculate the margin of error

Margin of error $ME = z\times SE$
$ME=1.645\times0.0229\approx0.038$

Answer:

The 90% large - sample confidence interval for the difference $p_M - p_F$ is about $0.033\pm0.038$