Timeline for How is the confidence interval calculated for the ACF function?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 8, 2022 at 14:21 | history | edited | User1865345 | CC BY-SA 4.0 |
added 27 characters in body
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| Mar 31, 2017 at 10:46 | history | edited | Robert de Graaf | CC BY-SA 3.0 |
Removed word repetition ('we want we want')
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| May 13, 2016 at 11:39 | comment | added | Robert de Graaf | Hi Glen_b, I have tried to update my answer accordingly - I have also borrowed some of your words, which I hope is okay. | |
| May 13, 2016 at 11:38 | history | edited | Robert de Graaf | CC BY-SA 3.0 |
Updated to incorporate Glen_B's suggestions to completely answer the original question, appropriating some of the words in Glen_B's comment, hopefully with permission
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| May 11, 2016 at 2:35 | comment | added | Glen_b | Good answer but this doesn't discuss the part of the question about why (0.95+1)/2 (or whatever other value ci takes); that's not really enough for a separate answer, I think, so I'll mention it here: That's simply because we want $\alpha/2$ in both tails, so we want the $1-\alpha/2$ quantile. and then see that $(1+1-\alpha)/2=1-\alpha/2$. (Robert: if you want to incorporate something along these lines into your answer please go ahead) | |
| May 10, 2016 at 21:26 | history | edited | Robert de Graaf | CC BY-SA 3.0 |
Added year for Kendall/Stuart
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| May 10, 2016 at 15:13 | vote | accept | Nick Nikolaev | ||
| May 10, 2016 at 12:00 | history | answered | Robert de Graaf | CC BY-SA 3.0 |