Timeline for Computing standard error of mean derived from multiple samples
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 22, 2015 at 23:24 | vote | accept | Delyle | ||
| Aug 22, 2015 at 0:51 | vote | accept | Delyle | ||
| Aug 22, 2015 at 0:52 | |||||
| Aug 22, 2015 at 0:51 | vote | accept | Delyle | ||
| Aug 22, 2015 at 0:51 | |||||
| Aug 21, 2015 at 23:20 | answer | added | EdM | timeline score: 1 | |
| Aug 21, 2015 at 22:12 | comment | added | Delyle | @EdM For your second question, I am not confident that the differences among variances do not arise simply because different samples have the same underlying variance. | |
| Aug 21, 2015 at 22:12 | comment | added | Delyle | @EdM I had considered weighting by the inverse variance, but I can't justify it to myself. For example, let's say you want to combine the reaction times of two people, and the first has consistent reactions times while the other has highly variable reactions. If you took the weighted average with weights$=1/\sigma_i$, the first person's reaction time would contribute more, but is that person really a better representative of the parent population just because they were more consistent? | |
| Aug 21, 2015 at 21:56 | comment | added | EdM | If you think that there are differences among the variances, have you considered weighting by the (inverse) variances in some way instead of by numbers of observations? And how confident are you that the differences among variances are real as opposed to just being from different samples having the same underlying variance? | |
| Aug 21, 2015 at 21:36 | comment | added | Delyle | @EdM Yes, it is a weighted mean based on measurements per subject. (In the practical example I'm thinking of, however, the number of measurements are all equal, i.e. $n_1=n_2=...=n_N$.) There are differences between the $\sigma^2$ values. | |
| Aug 21, 2015 at 20:38 | comment | added | EdM | So am I correct in saying that this is a weighted mean among $N$ subjects, with the weights being the number of measurements per subject? Is there any reason to think that the measurement variance would differ among individuals (that is, that there are differences among the $\sigma_i^2$ values)? | |
| Aug 21, 2015 at 20:12 | history | asked | Delyle | CC BY-SA 3.0 |