0
$\begingroup$

If 1,000,000 t-shirts exist in total, and you can track 25% of them (see if they have been purchased or not), and you find that of that 25% half have been purchased, you might believe that half of the total amount of t-shirts have been purchased. Given this 25% sample size, how much deviation from your measured purchase proportion can you expect in the entire population. I'm assuming this could be quantified somehow (with a confidence interval), but don't know how to go about it. Thanks for any help!

$\endgroup$
0
$\begingroup$

Population size = 1,000,000. Sample size = population size * 25% = 250,000.

Finite population correction factor = 1- 25% = 0.75. Point estimate is half of the total amount of t-shirts have been purchased = 50%.

Variance = 0.75 * 0.5 * 0.5 /250,000 = 0.00000075.

So 95 confidence interval is $0.5 \pm 1.96\sqrt{0.00000075} =(49.83\%, 50.17\%)$.

$\endgroup$
  • $\begingroup$ So I happened to pick half, if I picked 3/5 purchased instead, would the variance be 0.75 * 0.6 * 0.4/250000 ? $\endgroup$ – davem Dec 9 '18 at 21:28
  • $\begingroup$ Yes, you are right. $\endgroup$ – user158565 Dec 9 '18 at 21:29
  • $\begingroup$ And wouldn't the finite population correction factor actually be the square root of what you did? As in ----> sqrt((N-n)/N) $\endgroup$ – davem Dec 9 '18 at 21:50
  • $\begingroup$ For variance Var = f $S^2$, SE =$\sqrt f S$. I used f in the var, so do not need sqrt. $\endgroup$ – user158565 Dec 9 '18 at 21:57

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.