Answer to Question #199407 in Statistics and Probability for JAFFAR HUSSAIN

Question

A poll is taken among residents of a city and its suburbs to determine the feasibilities of 

a proposal to construct a civic center. If 6𝒐𝒇 𝟓𝟎 city residents favor the proposal and 

6 𝒐𝒇 𝟕𝟎 suburban residents favor it, find a 𝟗𝟓% confidence interval for the true 

difference between the proportion of city and suburban residents who favor the proposal 

to construct the civic center.

Answer 1

City Residents $(p_1):n_1=50\ ,\ x_1=6\ \ ; \ \hat p_1=\dfrac{x_1}{n_1}=\dfrac{6}{50}=0.12$

Suburban Residents $(p_2):n_2=70\ \ ;x_2=6\ \ ; \hat p_2=\dfrac{x_2}{n_2}=\dfrac{6}{70}=0.086$

The 95% confidence level $z_{\alpha/2}=1.96$

Margin of error $(E)=z_{\alpha/2}\sqrt{\dfrac{\hat p_1(1-\hat p_1)}{n_1}+\dfrac{\hat p_2(1-\hat p_2)}{n_2}}$

$E=1.96\cdot \sqrt{\dfrac{0.12\times 0.88}{50}+\dfrac{0.086\times 0.914}{70}}=1.96\times 0.0578=0.1133$

Lower limit = $(\hat p_1-\hat p_2)-E=(0.12-0.086)-0.1133=-0.0793$

Upper limit = $E-(\hat p_2-\hat p_1)=0.1133-(0.086-0.12)=0.1473$

So, confidence interval $=(-0.0793,0.1473)$