Timeline for Difference in Means vs. Mean Difference
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 12, 2017 at 1:26 | vote | accept | user84756 | ||
| Dec 16, 2015 at 15:09 | comment | added | bers | I posted a follow-up question to clear the confusion: stats.stackexchange.com/questions/187067/… | |
| Dec 16, 2015 at 13:45 | comment | added | shadowtalker | Which mean paired difference is "the" mean paired difference you're talking about? | |
| Dec 16, 2015 at 13:40 | comment | added | shadowtalker | @bers I think you're confused, but I'm confused as to what you're confused about. | |
| Dec 16, 2015 at 13:27 | comment | added | bers | Assume $C=A$. Then $A-C$ and $A-B$ are two different sequences. The confidence interval for the mean paired difference will certainly be different in both cases. But the difference of the means, and so it's confidence interval, will be indentical both for $A-C$ and $A-B$. Or am I wrong? | |
| Dec 16, 2015 at 13:23 | comment | added | shadowtalker | @bers what does $A-A$ have to do with it? | |
| Dec 16, 2015 at 13:07 | comment | added | bers | Can't edit my previous post any longer. The 3rd sentence should begin "A sequence of paired 'mean differences' ..." | |
| Dec 15, 2015 at 19:37 | comment | added | bers | But two confidence intervals calculated for "the difference of the means" and "the mean difference" will be different, right? This can be seen by looking at $A = [1, 2, 3, 4, 5, ...]$ and $B = [..., 5, 4, 3, 2, 1]$. A paired "mean difference" will be different for $A - A$ (which is all zero) versus $A - B$ (which is not all zero); the difference of the means is unaffected by the order of the elements. | |
| Apr 29, 2015 at 15:54 | history | edited | shadowtalker | CC BY-SA 3.0 |
added 102 characters in body
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| Apr 29, 2015 at 15:14 | history | edited | whuber♦ | CC BY-SA 3.0 |
added 7 characters in body
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| Apr 29, 2015 at 14:22 | history | answered | shadowtalker | CC BY-SA 3.0 |