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Answer:
[tex] 0.13 \leq p_G - p_H \leq 0.47[/tex]
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". Â
The margin of error is the range of values below and above the sample statistic in a confidence interval. Â
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". Â
[tex]p_G[/tex] represent the real population proportion for district G Â
[tex]\hat p_G [/tex] represent the estimated proportion for district G
[tex]n_B[/tex] is the sample size required for district G
[tex]p_H[/tex] represent the real population proportion for district H
[tex]\hat p_H [/tex] represent the estimated proportion for district H
[tex]n_H[/tex] is the sample size required for district H
[tex]z[/tex] represent the critical value for the margin of error Â
The population proportion have the following distribution Â
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex] Â
Solution to the problem
The confidence interval for the difference of two proportions would be given by this formula Â
[tex](\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}}[/tex] Â
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution. Â
[tex]z_{\alpha/2}=1.96[/tex] Â
The confidence interval for this case is given by:
[tex] 0.13 \leq p_G -p_H \leq 0.47[/tex]
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.
An independent random sample is a sequence of observations which are not dependent on any other sample or data. Random samples chosen randomly not according to the sequences. Thus, the option C is correct
The followings are the key points that are mentioned below:
- Because according to the Question the difference between the sample proportions G minus H more population. Â It is likely that more voters in district G favor the proposal than in district H, because all values in the interval are positive.
- Confidence interval  a confidence interval (CI) is a range of estimates, defined by a lower bound and upper bound, for an unknown parameter. The interval is computed at a designated confidence level.
- A 95% confidence level is most common, but other levels, such as 90% or 99%, are sometimes used.
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