1
$\begingroup$

This is a question from a sample mid-term for a first course in Statistics

Let p denote the proportion of elderly people with artificial hips. In an earlier study, the estimated proportion of the number of elderly people with artificial hips was found to be $11 \% $.

Q.1. Using earlier study's estimate, find the number of samples we would need in order to create a $95%$ confidence interval for the proportion of elderly people with artificial which is no more than $0.05$ away from the true proportion.

Ans. I have no clue how to do this. If could get the formula on how to approach this I would appreciate it.

Q.2. What is the maximum sample size needed to create a $93%$ confidence interval for the proportion of elderly people with artificial hips which is no more than $0.04$ away from the true proportion ? Ans I would appreciate a formula for this as well. Thanks

$\endgroup$
2
  • $\begingroup$ How can a proportion be $11$? $\endgroup$ Apr 2 '13 at 9:06
  • $\begingroup$ sorry it was 11% , it was a LAtex error on my part. Thanks for the answer I am just working it out and making sure I understand it. $\endgroup$
    – Kj Tada
    Apr 2 '13 at 9:34
2
$\begingroup$

The formula for the minimum required sample size is $$ n=\frac{z^2\cdot p\cdot (1-p)}{d^2}, $$ where $z=z_{\alpha/2}$ and $p$ is the true proportion and $d$ is half the length of the $(1-\alpha)\%$ confidence interval.

So in exercises like these you

  1. determine $\alpha$, e.g. if you're looking at $95\%$ confidence intervals, then set $\alpha=0.05$. This determines $z$,

  2. find $d$ which is half the length of the confidence interval you want to obtain.

  3. and then use the formula above with $p$ being replaced by some estimate (usually obtained by earlier studies or smaller pilot studies).

Often you're not explicitly given $d$ as half the length of the confidence interval, but rather as "create a $95\%$ confidence interval such that the estimated proportion is no more than $d$ away from the true proportion".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.