Timeline for Is normality testing 'essentially useless'?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 6, 2024 at 15:30 | history | edited | Nick Cox | CC BY-SA 4.0 | added 17 characters in body |
| Mar 31, 2023 at 17:55 | comment | added | Frank Harrell | You're missing something big. You have to actually compute the two confidence tail non-coverages to get the point. "Looking normal" isn't nearly good enough. | |
| Mar 27, 2023 at 9:58 | comment | added | Singularity24601 | @FrankHarrell Really? I simulated the mean of N iid log-normal distributed variables for >5000 reps and the distribution of these means looked reasonably normal for values of N that were much lower than 50000. Eg, I got sample skewness = 0.17 for N = 1000. Am I missing something? | |
| Oct 19, 2022 at 18:05 | history | wiki removed | kjetil b halvorsen♦ | ||
| Mar 19, 2021 at 10:34 | comment | added | Dave | @FrankHarrell Do you mean that, in a simulation of $10,000$ iterations, of the $\sim 500$ $95\%$ confidence intervals that should fail to contain the mean, we should be concerned that the split is $\sim 250$ with upper endpoints below the mean and $\sim 250$ with lower endpoints above the mean, rather than most of the intervals missing high (or low)? | |
| Nov 23, 2020 at 22:09 | comment | added | Frank Harrell | Beg to disagree. It is not appropriate to consider confidence coverage devoid of directionality IMHO, and if a method works poorly under likely-to-occur situations, the method is usually not good enough for field use where asymptotics are completely irrelevant. | |
| Nov 23, 2020 at 16:19 | comment | added | AdamO | @FrankHarrell truly it's impossible for the "truth" to be entirely one way or the other when we are speaking in broad terms. Agree that for a skewed distribution using a symmetric confidence interval will be inefficient (too broad), but on the other hand it is correct in that it "covers the replicates" 95% of the time. | |
| Nov 22, 2020 at 12:23 | comment | added | Frank Harrell | This is simply not the case, and such opinions usually fail to consider the non-coverage probabilities in both tails for a confidence interval. For example, n=50,000 from a log-normal distribution is inadequate for the central limit theorem to work well enough when computing the CI for a mean. | |
| Jun 11, 2020 at 14:32 | history | edited | CommunityBot | Commonmark migration | |
| S Mar 12, 2018 at 17:59 | history | answered | AdamO | CC BY-SA 3.0 | |
| S Mar 12, 2018 at 17:59 | history | made wiki | Post Made Community Wiki by AdamO |