Chapter 10: Q. 13 (page 668)

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 that should be used to estimate the true difference in the proportions of Republicans in New York State and California?
(a) $(0.44 - 0.47) \pm 1.96 \frac{(0.44) (0.56) + (0.47) (0.53)}{\sqrt{200 + 300}}$
(b) $(0.44 - 0.47) \pm 1.96 \frac{(0.44) (0.56)}{\sqrt{200}} + \frac{(0.47) (0.53)}{\sqrt{300}}$
(c) $(0.44 - 0.47) \pm 1.96 \sqrt{\frac{(0.44) (0.56)}{200} + \frac{(0.47) (0.53)}{300}}$
(d) $(0.44 - 0.47) \pm 1.96 \sqrt{\frac{(0.44) (0.56) + (0.47) (0.53)}{200 + 300}}$
(e) $(0.44 - 0.47) \pm 1.96 \sqrt{(0.45) (0.55) (\frac{1}{200} + \frac{1}{300})}$

Short Answer

Expert verified

The true difference in the proportions of Republicans in New York State and California is (c) $(0.44 - 0.47) \pm 1.96 \sqrt{\frac{(0.44) (0.56)}{200} + \frac{(0.47) (0.53)}{300}}$

Step by step solution

Given information

We are given that a random sample of New York voters and California voters include republicans. We have to estimate the true difference in the proportions of Republicans in New York State and California.

Simplify

According to question

Sample of New York $n_{1} = 200$

Republicans in New York $x_{1} = 88$

Sample of California $n_{2} = 300$

Republicans in $x_{2} = 141$

For difference of proportions:

$(\hat{p_{1}} - \hat{p_{2}}) - z_{α / 2} \times \sqrt{\frac{\hat{p_{1}} (1 - \hat{p_{1}})}{n_{1}} + \frac{\hat{p_{2}} (1 - \hat{p_{2}})}{n_{2}}} \hat{p_{1}} = \frac{x_{1}}{n_{1}} = \frac{88}{200} = 0.44 \hat{p_{2}} = \frac{x_{2}}{n_{2}} = \frac{141}{300} = 0.47$

Critical value $z_{α / 2} = z_{0.025} = 1.96$

Now , from above formula

$= (0.44 - 0.47) \pm 1.96 \sqrt{\frac{0.44 (1 - 0.44)}{200} + \frac{0.47 (1 - 0.47)}{300}} = (0.44 - 0.47) \pm 1.96 \sqrt{\frac{0.44 (0.56)}{200} + \frac{0.47 (0.53)}{300}}$