Question

(17 points) A random sample of ?=80n=80 in-coming college freshmen showed that 32 were planning to bring a car with them to campus next fall. Suppose that we are interested in forming a confidence interval at 99 percent for the proportion of all in-coming college freshmen who are planning to bring a car to campus next fall.

Where appropriate, express your answer as a proportion (not a percentage). Round answers to three decimal places.

(a) The estimate is:  .

(b) The standard error is:

(c) The multiplier is:

Answer 1

a. The estimate is $p=\frac{32}{80}=0.4$

b. Standard error is $SE=\sqrt{\frac{pq}{n}}=\sqrt{\frac{0.4*0.6}{80}}=0.055$

c. z value for 99% CI is 2.576 as P(-2.576<z<2.576)=0.99

So Margin of Error is

E = 2 *SE= 2.576 * 0.055 = 0.142

Hence CI is $p \pm E=0.4 \pm 0.142=(0.258, 0.542)$