AP Statistics 3.3: Standard and Margin Error

Calculate standard error and margin of error for a proportion interval, then rearrange the margin formula to find a required sample size.

An environmental science teacher at a high school with a large population of students wanted to estimate the proportion of students at the school who regularly recycle plastic bottles. The teacher selected a random sample of students at the school to survey. Each selected student went into the teacher's office, one at a time, and was asked to respond yes or no to the following question.
Do you regularly recycle plastic bottles?
Based on the responses, a 95 percent confidence interval for the proportion of all students at the school who would respond yes to the question was calculated as (0.584, 0.816).

How many students were in the sample selected by the environmental science teacher?

Part (a):
Using the standard formula for a confidence interval for one proportion, the interval ( 0.584 to 0.816 ) is found as follows. $\hat{p} \pm z^{*} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ where $\hat{p}=\frac{0.584+0.816}{2}=0.7$ , the margin of error is 0.816-0.7=0.116, and $z^{*}=1.96$ .
Solving $1.96 \sqrt{\frac{0.7(1-0.7)}{n}}=0.116$ yields $n=\frac{(1.96)^{2}(0.7)(1-0.7)}{(0.116)^{2}} \approx 59.95$ . The sample size was 60.

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3.3.D—Calculate the standard error and margin of error of a sample statistic for a confidence interval for a… question 1