Question

A random sample of $200$New York State voters included $88$Republicans, while a random sample of $300$California voters produced$141$Republicans. Which of the following represents the $95\%$confidence interval for the true difference in the proportion of Republicans in New York State and California?

a. $(0.44-0.47)\pm 1.96\left(\right(0.44\left)\right(0.56)+(0.47\left)\right(0.53)200+300)$

b. $(0.44-0.47)\pm 1.96\left(\right(0.44\left)\right(0.56)200+(0.47\left)\right(0.53\left)300\right)$

c. ($0.44-0.47)\pm 1.96(0.44\left)\right(0.56)200+(0.47\left)\right(0.53)300$

d. $(0.44-0.47)\pm 1.96(0.44)(0.56)+(0.47)(0.53)200+300$

e. $(0.44-0.47)\pm 1.96(0.45)(0.55)200+(0.45)(0.55)300$

Short answer

d.$(0.44-0.47)\pm 1.96\sqrt{\frac{(0.45)(0.55)}{200}+\frac{(0.45)(0.55)}{300}}$ represent the 95% confidence interval for the true difference in the proportion of Republicans in New York State and California

Step-by-step solution

Given information

We need to find that which statement represent the $95\%$ confidence interval for the true difference in the proportion of Republicans in New York State and California

Explanation

To 95% confidence interval , we will use following formula ;

Here $x_{1,}{x}_2$are sample mean$1$and $2$

$t$: the $t-$critical value based on the confidence level , ${s_p}^2$: pooled variance

$n1,n2$: sample 1 size, sample 2 size

Confidence interval =

role="math" localid="1654175042706" $$\begin{gathered} {(x}_1–x_2)\pm t\times \sqrt{\left(\right({s_p}^2/n1)+({s_p}^2/n2\left)\right)} \\ {x}_1=0.44,x_2=0.47 \\ t=1.96,{s_p}^2=(0.45\left)\right(0.55),n1=200,n2=300 \\ {(x}_1–x_2)\pm t\times \sqrt{\left(\right({s_p}^2/n1)+({s_p}^2/n2\left)\right)}=(0.44-0.47)\pm 1.96\sqrt{\frac{(0.45)(0.55)}{200}+\frac{(0.45)(0.55)}{300}} \end{gathered}$$