Timeline for Why does this excerpt say that unbiased estimation of standard deviation usually isn't relevant?
Current License: CC BY-SA 4.0
16 events
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| Dec 24, 2021 at 4:03 | history | edited | Galen | CC BY-SA 4.0 |
This answer doesn't make sense without knowing what a pivotal quantity is in precise terms. I've added a link to help readers.
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| Dec 9, 2016 at 11:42 | history | edited | Scortchi♦ | CC BY-SA 3.0 |
fixed typos
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| Jul 29, 2012 at 13:58 | vote | accept | BYS2 | ||
| Jul 29, 2012 at 13:53 | comment | added | BYS2 | @MichaelChernick, ok thank you very much! I really appreciate the effort you guys put in answering these questions :D | |
| Jul 29, 2012 at 13:04 | comment | added | Michael R. Chernick | @BYS2 Yes that's right. | |
| Jul 29, 2012 at 7:18 | comment | added | BYS2 | @MichaelChernick, ah so what you are saying is that if I use a different estimate for standard deviation (say by multiplying it by a constant), then I must multiply the t statistic by that same constant for it to be applicable/coherent with my modified SD? I cannot just multiply the SD by a constant and use the normal t statistic right? | |
| Jul 28, 2012 at 19:35 | comment | added | Michael R. Chernick | No what Cardinal is correctly pointing out is that it is possible to multiply the t statistic by a constant to essentially use a different estimate of standard deviation. The test statistic no longer has the t distribution. It is a slightly different distribution because of the constant. The mean changes by a factor of b and so does the standard deviation. When you go about computing the critical value for the test statistic it changes appropriately as he demonstrates above. | |
| Jul 28, 2012 at 15:26 | comment | added | BYS2 | @cardinal sorry I deleted my previous comment to rewrite it, so if you were replying to that then this may be confusing :S | |
| Jul 28, 2012 at 15:25 | comment | added | BYS2 | @cardinal, thank you for your reply :D. But I am still a bit fuzzy regarding your reasoning. For example say I multiply my SD by a scale factor of 1.5 (to make it unbiased), then the t-value would change (t would equal 1.63), which would affect my p-value and my CI (since endpoints are +/- sample mean + (target_t_value * SD)/sqrt(N). This may change your conclusion if it was on the borderline of the significance level could it not? Sorry, I am trying to learn these things as a hobby so I've just using online tutorials and wikipedia which aren't as rigorous as they could be sometimes. | |
| Jul 28, 2012 at 15:24 | comment | added | cardinal | @BYS2: In the normal model case using the $t$-statistic, there is a nice correspondence between CIs and $p$-value. So, the $p$-value will not change if you "rescale" the sample standard deviation by a known constant. For example: Let $\tilde T_b = (\bar X - \mu)/(b\cdot \hat \sigma) = T / b$ for fixed $b > 0$. Then, $$\mathbb P(\tilde T_b > u) = \mathbb P(T > b u)$$ and so the critical value $\tilde t_{b,\alpha} = b t_\alpha$, i.e., there is a one-to-one correspondence between them. Does that make sense? | |
| Jul 28, 2012 at 14:37 | comment | added | cardinal | @BYS2: Note that in terms of the interval constructed in the example you give, nothing changes by multiplying the sample standard deviation by a scale factor (e.g., to make it unbiased). The distribution of the test statistic would change (slightly) in this case, but the CI constructed would end up being exactly the same! Now, if you did some "correction" that depended on the data themselves, that would yield something different (in general). See my comment under Glen's answer. | |
| Jul 28, 2012 at 14:02 | comment | added | BYS2 | @MichaelChernick, hmm ok so what you are saying is that the t distribution is only valid as a test statistic when we are estimating the standard deviation/variance using the formula with Bessel's correction (multiplying the standard sample variance by n(n−1)). So essentially the t-distribution as a test statistic intrinsically compensates for this and if we used another estimator for SD/variance then it would be invalid? | |
| Jul 28, 2012 at 13:37 | comment | added | gung - Reinstate Monica | +1, this is a really great answer; clear & w/ sufficient background info to make the answer make sense. | |
| Jul 28, 2012 at 13:36 | comment | added | Michael R. Chernick | It does depend on the estimate but there is only one estimate for which the t with 19 degrees of freedom applies and that estimate is the square root of the usual estimate of the sample variance. If you use a different estimate of the standard deviation you have a different reference distribution for the test statistic under the null hypothesis. It is not the t. | |
| Jul 28, 2012 at 12:40 | comment | added | BYS2 | Ok so suppose a researcher wished to test whether the mean score of fifth graders on a test in his or her city differed from the national mean of 76. The researcher randomly sampled the scores of 20 students. The sample mean was 80.85 with a sample standard deviation of 8.87. This means: t = (80.85-76)/(8.87/sqrt(20)) = 2.44. A t-table is then used to calculate that the two-tailed probability value of a t of 2.44 with 19 df is 0.025. So in this example, wouldn't your p-value (and maybe your conclusion) change depending on how you estimated your sample standard deviation? | |
| Jul 28, 2012 at 9:50 | history | answered | Michael R. Chernick | CC BY-SA 3.0 |