Question

A recent survey showed that among 100 randomly selected college seniors, 20 plan to attend graduate school and 80 do not. determine a 95 % confidence interval for the population proportion of college seniors who plan to attend graduate school. (enter each answer rounded to three decimal places.)

Answer 1

The confidence interval would be (0.122, 0.278).

We first find our z-score.  We want a 95% confidence interval:

0.95/2 = 0.475

Looking this up in the z-table, (http://www.statisticshowto.com/tables/z-table/) we see the z-score is 1.96.

The formula we will use is:

$p \pm z\sqrt{\frac{p(1-p)}{n}}$

In this problem, p = 20/100 = 0.2, and n=100:

$0.2 \pm 1.96\sqrt{\frac{(0.2)(0.8)}{100}} \\ \\0.2\pm 0.0784 \\ \\(0.2-0.0784, 0.2+0.0784) \\ \\(0.1216,0.2784)$