Question

Independent random samples of voters from two voting districts, G and H, were selected to investigate the proportion of all voters who favor a proposal to widen a road that runs through both districts. The difference between the sample proportions (G minus H) was used to create the 95 percent confidence interval (0.13,0.47)
for the population difference between districts.

Which of the following is the best interpretation of the interval?

It is likely that the majority of voters in both districts favor the proposal, because all values in the interval are positive.
A

It is likely that more voters in district H favor the proposal than in district G, because all values in the interval are positive.
B

It is likely that more voters in district G favor the proposal than in district H, because all values in the interval are positive.
C

It is likely that less than half the voters in both districts favor the proposal, because all values in the interval are less than 0.5.
D

It is likely that there is no difference between the districts in voters who favor the proposal, because 0 is not contained in the interval.

Answer 1

Answer:

$0.13 \leq p_G - p_H \leq 0.47$

And since the lower value for the confidence interval of the difference is higher than 0 we can conclude that the proportion of voters who favor a proposal in district G is significantly higher than the proportion for the district H and the best solution for this case is:

C. It is likely that more voters in district G favor the proposal than in district H, because all values in the interval are positive.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$p_G$ represent the real population proportion for district G  

$\hat p_G$ represent the estimated proportion for district G

$n_B$ is the sample size required for district G

$p_H$ represent the real population proportion for district H

$\hat p_H$ represent the estimated proportion for district H

$n_H$ is the sample size required for district H

$z$ represent the critical value for the margin of error  

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$  

Solution to the problem

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

$(\hat p_G -\hat p_H) \pm z_{\alpha/2} \sqrt{\frac{\hat p_G(1-\hat p_G)}{n_G} +\frac{\hat p_H (1-\hat p_H)}{n_H}}$  

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 required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

The confidence interval for this case is given by:

$0.13 \leq p_G -p_H \leq 0.47$

And since the lower value for the confidence interval of the difference is higher than 0 we can conclude that the proportion of voters who favor a proposal in district G is significantly higher than the proportion for the district H and the best solution for this case is:

C. It is likely that more voters in district G favor the proposal than in district H, because all values in the interval are positive.