Question

A candidate for one of Ohio's two U.S. Senate seats wishes to compare her support among registered voters in the northern half of the state with her support among registered voters in the southern half of the state. A random sample of 2000 registered voters in the northern half of the state is selected, of which 1062 support the candidate. Additionally, a random sample of 2000 registered voters in the southern half of the state is selected, of which 900 support the candidate.
A 95% confidence interval for the difference in proportion of registered voters that support this candidate between the northern and southern halves of the state is


A. 0.050 to 0.112.
B. 0.035 to 0.127.
C. 0.040 to 0.122.
D. 0.037 to 0.119.

Answer 1

Answer:

$(0.531-0.45) - 1.96 \sqrt{\frac{0.531(1-0.531)}{2000} +\frac{0.45(1-0.45)}{2000}}=0.050$  

$(0.531-0.45) + 1.96 \sqrt{\frac{0.531(1-0.531)}{2000} +\frac{0.45(1-0.45)}{2000}}=0.112$  

And the best option for this case would be:

A. 0.050 to 0.112.

Step-by-step explanation:

Data given

$p_A$ represent the real population proportion of who support the cnadite for the northern half state  

$\hat p_A =\frac{1062}{2000}=0.531$ represent the estimated proportion of who support the candidate for the northern half state  

$n_A=2000$ is the sample size required the northern half state  

$p_B$ represent the real population proportion of who support the candidate for the southern half state  

$\hat p_B =\frac{900}{2000}=0.45$ represent the estimated proportion of people who support the candidate for the southern half state  

$n_B=2000$ is the sample size for the northern half state  

$z$ represent the critical value

The confidence interval for the difference of two proportions would be given by this formula  

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$  

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile in the normal standard distribution and we got.  

$z_{\alpha/2}=1.96$  

Replacing into the formula we got:

$(0.531-0.45) - 1.96 \sqrt{\frac{0.531(1-0.531)}{2000} +\frac{0.45(1-0.45)}{2000}}=0.050$  

$(0.531-0.45) + 1.96 \sqrt{\frac{0.531(1-0.531)}{2000} +\frac{0.45(1-0.45)}{2000}}=0.112$  

And the best option for this case would be:

A. 0.050 to 0.112.