Question

In a​ poll, 37​% of the people polled answered yes to the question​ "Are you in favor of the death penalty for a person convicted of​ murder?" The margin of error in the poll was 5​%, and the estimate was made with 94​% confidence. At least how many people were​ surveyed?

Answer 1

Answer:

The number of people ​surveyed was 330.

Step-by-step explanation:

The (1 - α)% confidence interval for the population proportion is:

$CI=\hat p\pm z_{\alpha/2}\cdot \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The margin of error for this interval is:

$MOE= z_{\alpha/2}\cdot \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The information provided is:

$\hat p=0.37\\MOE=0.05\\\text{Confidence level}=0.94\\\Rightarrow \alpha=0.06$

The critical value of z for 94​% confidence level is, z = 1.88.

*Use a z-table.

Compute the value of n as follows:

$MOE= z_{\alpha/2}\cdot \sqrt{\frac{\hat p(1-\hat p)}{n}}$

      $n=[\frac{z_{\alpha/2}\times \sqrt{\hat p(1-\hat p)}}{MOE}]^{2}$

         $=[\frac{1.88\times \sqrt{0.37(1-0.37)}}{0.05}]^{2}\\\\=(18.153442)^{2}\\\\=329.5475\\\\\approx 330$

Thus, the number of people ​surveyed was 330.

Answer 2

Answer:

329  people were​ surveyed

Step-by-step explanation:

Percent of people polling yes to the question​ "Are you in favor of the death penalty for a person convicted of​ murder?"= 37 %

Margin error in the poll=5%

Confidence Interval=94%

The alpha value =1-0.94= 0.06

The Z ( critical value) for Confidence Interval of 94% =1.88

The sample size is given by

$n=pq(\frac{z}{e})^2$

where, p=0.37, q=0.63,  e= 5/100= 0.05, z=1.88

therefore,

$n=0.37\times0.63(\frac{1.88}{0.05})^2$

=329.54745

=329  people were​ surveyed