Timeline for How to compare 4 different normal random distributions
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 26, 2020 at 5:45 | comment | added | Dedes | @whuber - thank you for your answer. You already said we need a prior and Bayesian analysis is the way to go on your opinion. Would it be possible for you to point me an example or some reference that I can use for study and reference ? | |
| May 25, 2020 at 17:53 | comment | added | xi45 | or implicitly assumed to be uniform | |
| May 25, 2020 at 17:39 | comment | added | whuber♦ | The question is unanswerable because there cannot even be a probability until a prior is specified. | |
| May 25, 2020 at 16:57 | comment | added | xi45 | @whuber Of course this can be made more rigorous, but for this specific case I dont see why this simple solution is wrong (or better, why the question in unanswerable) since the sum of two normally distributed and independent random variables is normally distributed. | |
| May 25, 2020 at 16:46 | comment | added | whuber♦ | None of that is correct. The combination is a mixture, which will not be Normal. The general method of proceeding requires a Bayesian analysis. | |
| May 25, 2020 at 15:52 | comment | added | Dedes | @whuber, can I assume from your comment that "the prior distribution for the fours areas" is a combination for the individual distributions ? And, as the areas are all normal distributed, doesn't this mean that the distribution that combines the fours is also normal ? And, assuming I have that distribution, how do I proceed ? Thank you | |
| May 25, 2020 at 15:22 | comment | added | whuber♦ | The question is unanswerable until a prior distribution for the four areas is provided. Therefore your reply cannot be correct. Note, too, that the question appears to be asking about the mean heights per area, not individual heights. | |
| May 25, 2020 at 14:58 | history | answered | xi45 | CC BY-SA 4.0 |