Sampling Distributions for Proportions Question 7

A random sample of 300 students is selected from a large group of students who use a computer-equipped classroom on a regular basis. Occasionally, students leave their USB drive in a computer. Of the 300 students questioned, 180 said that they write their name on their USB drive. Which of the following is a 98 percent confidence interval for the proportion of all students using the classroom who write their name on their USB drive?

0.4 \pm 2.33\sqrt{\dfrac{(0.4)(0.6)}{300}}

0.4 \pm 1.96\sqrt{\dfrac{(0.4)(0.6)}{300}}

0.6 \pm 2.33\sqrt{\dfrac{(0.6)(0.4)}{300}}

0.6 \pm 1.96\sqrt{\dfrac{(0.6)(0.4)}{300}}

Approach

A 98 percent confidence interval corresponds to a $p$ -value of 0.01, which corresponds to a critical value of 2.33.

The proportion of students who write their name is

\hat{p}=\frac{180}{300}=0.6

Putting it all together:

\text{Confidence interval} = \text{statistic} \pm (\text{critical value})(\text{standard error of statistic})

\text{Confidence interval} = \hat{p} \pm (\text{critical value })\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

\text{Confidence interval} = \boxed{0.6 \pm 2.33\sqrt{\frac{(0.6)(0.4)}{300}} }