Chapter 10: Q. 3.13 (page 702)

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

Expert verified

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

$n 1, n 2$ : sample 1 size, sample 2 size

Confidence interval =

role="math" localid="1654175042706" ${(x}_{1} - x_{2}) \pm t \times \sqrt{(({s_{p}}^{2} / n 1) + ({s_{p}}^{2} / n 2))} x_{1} = 0.44, x_{2} = 0.47 t = 1.96, {s_{p}}^{2} = (0.45) (0.55), n 1 = 200, n 2 = 300 {(x}_{1} - x_{2}) \pm t \times \sqrt{(({s_{p}}^{2} / n 1) + ({s_{p}}^{2} / n 2))} = (0.44 - 0.47) \pm 1.96 \sqrt{\frac{(0.45) (0.55)}{200} + \frac{(0.45) (0.55)}{300}}$