Answer to Question #109971 in Statistics and Probability for Diana

Question

Some say Hillary Clinton is unstoppable for 2016. A recent poll indicates that 586/1000 voters will favor her in the next election.

A) Create a 90% confidence interval for P ( the population proportion) and interpret.

B) Based off of that interval is there strong evidence to say she will win?

C) How large of a sample is needed for a desired margin of error of 2% using 90% confidence and 586/1000 as a point estimate for P.

Answer 1

A) We need to construct the 90% confidence interval for the population proportion. We have been provided with the following information about the number of favorable cases:

$Favorable\ Cases \ X=586$

$Sample\ Size \ N=1000$

The sample proportion is computed as follows, based on the sample size $N=1000$ and the number of favorable cases$X=586.$

$$\hat{p}={X \over N}={586 \over 1000}=0.586$$

The critical value for $\alpha=0.1$  is $z_c=z_{1-\alpha/2}=1.645.$ The corresponding confidence interval is computed as shown below:

$$CI(Proportion)=\big(\hat{p}-z_c\sqrt{{\hat{p}(1-\hat{p}) \over N}},\hat{p}+z_c\sqrt{{\hat{p}(1-\hat{p}) \over N}}\big)$$

$$=\big(0.586-1.645\sqrt{{0.586(1-0.586) \over 1000}},0.586+1.645\sqrt{{0.586(1-0.586) \over 1000}}\big)$$

$$=(0.560378,0.611622)$$

Therefore, based on the data provided, the 90% confidence interval for the population proportion is $0.560378<p<0.611622,$ which indicates that we are 90% confident that the true population proportion $p$ is contained by the interval $(0.560378,0.611622).$

B) Based off of that interval there is strong evidence to say she will win $(p>0.5).$

C) Margin of error $E$

$$E=z_c\sqrt{{\hat{p}(1-\hat{p}) \over N}}\leq0.02$$

$$N\geq{(1.645)^2 (0.586)(1-0.586) \over (0.02)^2}$$

$$N\geq1642$$

$1642$