Question

Help with AP Stats??? I will award medals!!!

2. You're interested in the proportion of students in your school who favor the elimination of a local curfew on teenagers. Assume your school has 1,000 students.

A. To generate a 95% confidence interval for the proportion of students who favor the elimination of a curfew, what's the minimum number of students you need if you want the margin of error to be 5%?

Answer 1

For a 95% C.I. the critical value is something like 1.96, if I remember correctly, and the margin of error is given by the product of this critical value and the standard error for the estimator. In this case, since you're estimating the true proportion, the standard error will have the form
\[\sqrt{\frac{p(1-p)}{n}}\]
where \(p\) is the estimated proportion.

So the margin of error formula will look like
\[1.96\sqrt{\frac{p(1-p)}{n}}=0.05\]
Presumably, you're given a certain value for \(p\), which will allow you to solve for \(n\).

Answer 2

Correction: It *is* 1.96.

Answer 3

THANK YOU!!! @SithsAndGiggles

Answer 4

yw

Answer 5

What about this???

Suppose your principal, who never took a statistics class, says that a 5% margin of error just isn't accurate enough. He wants you to produce a confidence interval with a margin of error of only 2.5%. How big does your sample have to be now? What do you think of the principal's proposal?

Answer 6

It's the same equation with the exception of replacing 0.05 with 0.025

Answer 7

Ok thank you so much again!! Do you know how to do these two?


C. In general, if you wanted to cut the margin of error for a given confidence level in half, by what amount would you have to increase the sample size? Use the formula for minimum sample size to explain your answer.


D. In general, if you wanted to cut the margin of error for a given confidence level in half, by what amount would you have to increase the sample size? Use the formula for minimum sample size to explain your answer.

That's all after this :)

Answer 8

Uhhh, those are the same question...

Answer 9

Sorry I copied and pasted them twice in the word document. Then that question is the last one! Do you know how to do that one?

Answer 10

I'm not familiar with the "minimums sample size formula" name, but it looks like a simple algebra question.

Let the margin of error be \(M\), the critical value for the CI be \(Z\), the standard error be \(s_p\), the estimated proportion be \(p\), and the sample size be \(n\). Then
\[M=Z\sqrt{\frac{p(1-p)}{n}}\]
You want to cut \(M\) in half, so to find the appropriate sample size, you would rearrange the above equation so that
\[\frac{M}{2}=Z\sqrt{\frac{p(1-p)}{n}}~~\iff~~M=\cdots\]
Particularly,
\[M=2Z\sqrt{\frac{p(1-p)}{n}}=Z\sqrt{4\cdot\frac{p(1-p)}{n}}=Z\sqrt{\frac{p(1-p)}{n/4}}\]
Suppose you have a margin of error \(M\) with a sample size of \(\dfrac{n}{4}=100\). Then, from the above work, you will get \(\dfrac{M}{2}\) if you use a sample size of \(4\cdot\dfrac{n}{4}=n=400\).

Thus in general, you should quadruple the sample size in order to halve the error margin.

Answer 11

Thank you so much!! You helped a ton :)

Answer 12

You're welcome!