Question

In 1987, the General Social Survey asked, "Have you ever been active in a veteran's group? " For this question, 52 people said that they did out of 98 randomly selected people. We will make a 95% confidence interval for p. What is the margin of error of the confidence interval? Group of answer choices

Answer 1

Answer:

$0.531 - 1.96 \sqrt{\frac{0.531(1-0.531)}{98}}=0.432$

$0.531 + 1.96 \sqrt{\frac{0.531(1-0.531)}{98}}=0.630$

And the 95% confidence interval would be given (0.432;0.630).

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".  

The population proportion have the following distribution  

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

Solution to the problem

The estimated proportion for this case is:

$\hat p = \frac{52}{98} =0.531$

The confidence interval would be given by this formula

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

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.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.531 - 1.96 \sqrt{\frac{0.531(1-0.531)}{98}}=0.432$

$0.531 + 1.96 \sqrt{\frac{0.531(1-0.531)}{98}}=0.630$

And the 95% confidence interval would be given (0.432;0.630).