Chapter 10, Problem 13PT3

To determine

To identify the 95% confidence interval that would estimate the true difference in the proportion of Republicans in New York and California:

Answer to Problem 13PT3

  $\text{(c) }(0.44-0.47)\pm 1.96\sqrt{\frac{(0.44)(0.56)}{200}+\frac{(0.47)(0.53)}{300}}$

Explanation of Solution

Given information:

  $\begin{array}{l}{n}_1=200x_1=88\\ n_2=300x_2=141\end{array}$

Where $n_1\&n_2$ are number of voters in New York City and California City respectively

Also $x_1\&x_2$ are number if Republicans in New York City and California City

respectively

The formula for 95% confidence interval for true difference in the population proportion $(P_1-P_2)$ is as follows:

  $(p_1-p_2)\pm Z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{\frac{(p_1)(1-p_1)}{n_1}+\frac{(p_2)(1-p_2)}{n_2}}$

Where $p_1\&p_2$ are sample proportions.

  $p_1=\frac{x_1}{n_1}$ and $p_2=\frac{x_2}{n_2}$

Calculation:

For the given problem $p_1\&p_2$ are obtained as below:

  $\begin{array}{l}{p}_1=\frac{88}{200}=0.44\\ p_2=\frac{141}{300}=0.47\end{array}$

The critical value of Z at 5% level of significance is $Z_{\alpha /2}=Z_{0.025}=1.96$

Therefore from the formula the confidence interval will be,

  $(0.44-0.47)\pm 1.96\sqrt{\frac{(0.44)(0.56)}{200}+\frac{(0.47)(0.53)}{300}}$

Hence 95% confidence interval to estimate the true differences in the proportion of Republicans in the two cities is $(0.44-0.47)\pm 1.96\sqrt{\frac{(0.44)(0.56)}{200}+\frac{(0.47)(0.53)}{300}}$