1
$\begingroup$

I'm calculating confidence intervals for data that I generated myself, so the sample size is very large (about 30,000). Using the standard formula where the sample size is taken into account, my "intervals" end up as points.

The person who told me to calculate the CIs in the first place wrote down the formula but left out the $\sqrt n$, i.e., wrote the formula as mean $\pm (z^* \times SE)$. So I'm wondering whether there could be some reason for leaving out the sample size in my case, for example because it's so large. The CIs are still narrow in this case, but not points. But I haven't been able to find any literature about this.

$\endgroup$
  • 1
    $\begingroup$ Are you aware that the formula for standard error includes sample size? $\endgroup$ – Ian_Fin Nov 10 '16 at 13:26
  • $\begingroup$ You might be want a "tolerance interval" as distinct from a confidence interval. $\endgroup$ – user20637 Nov 10 '16 at 17:08
2
$\begingroup$

The standard error already takes the sample size into account. For example, a prototypical SE formula might be $SD/\sqrt N$ (e.g., for a one-sample t-test). When you divided by $\sqrt n$ again, you double-dipped—resulting in confidence intervals that were invalid and too narrow.

$\endgroup$

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.