Timeline for Confidence interval for geometric mean of fractions
Current License: CC BY-SA 3.0
12 events
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| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
replaced http://stats.stackexchange.com/ with https://stats.stackexchange.com/
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| Jan 29, 2015 at 3:25 | comment | added | vdi | @CristiánAntuña but it the same as $e^{\ln(X)} \in e^{\overline{\ln(X)} + b}$. I added "$exp$" into third from the last expression. May be it will bring some clarification/ | |
| Jan 29, 2015 at 3:22 | history | edited | vdi | CC BY-SA 3.0 |
add some corrections
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| Jan 28, 2015 at 21:02 | comment | added | Cristián Antuña | These events are not equivalent: $\{ ln(X_i) \in (\overline{ln(X)} + b)\}$ and $\{ e^{ln(X_i)} \in (e^{\overline{ln(X)}} + b)\}$ and, then, in general won't have the same probability. | |
| Jan 28, 2015 at 20:54 | comment | added | whuber♦ | @CristiánAntuña What, precisely, is the "wrong" statement in this answer? | |
| Jan 28, 2015 at 19:35 | comment | added | Cristián Antuña | I'll use the answer space, since it is bigger than the comments one. | |
| Jan 28, 2015 at 19:28 | comment | added | vdi | @CristiánAntuña okay, thanks, can you help to make it true? | |
| Jan 28, 2015 at 19:21 | comment | added | Cristián Antuña | If I understand what you're saying, you assume that $ln(X_i) \sim N$ and build a confidence interval for it the ususal way one would do with a Normal r.v. Your finishing expression is right, but when you say that you "exponent something" you are not being very clear on what you do and, in fact, you are stating something that is wrong. Again your last statement is right, but it seems that intuition got you there, you are not justifying it. | |
| Jan 28, 2015 at 19:05 | history | edited | vdi | CC BY-SA 3.0 |
exp functions correction
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| Jan 28, 2015 at 16:51 | vote | accept | vdi | ||
| Jan 28, 2015 at 19:26 | |||||
| Jan 24, 2015 at 18:08 | history | edited | vdi | CC BY-SA 3.0 |
added 3 characters in body
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| Jan 23, 2015 at 18:50 | history | answered | vdi | CC BY-SA 3.0 |