Comments on: Ignorance doesn’t change reality: statistics pitfall http://www.johndcook.com/blog/2009/02/16/statistics-pitfall/ The blog of John D. Cook Sat, 11 Feb 2012 01:10:06 -0500 http://wordpress.org/?v=2.8.4 hourly 1 By: Mike Kowalski http://www.johndcook.com/blog/2009/02/16/statistics-pitfall/comment-page-1/#comment-25656 Mike Kowalski Fri, 09 Oct 2009 02:39:13 +0000 http://www.johndcook.com/blog/?p=1538#comment-25656 Must add to John's comment. If the sample mean has a N(μ, σ²) distribution then the standardized sample mean (x-bar-μ)/(σ/√n) also has a normal distribution. The t-distribution comes in to play when we must estimate σ with the sample standard deviation, 's'. The standardized sample mean, stadardized by using the sample standard deviation (x-bar-μ)/(s/√n) follows a t-distribution. No (correct) theory says that the sample mean follows a t-distribution! Must add to John’s comment. If the sample mean has a N(μ, σ²) distribution then the standardized sample mean (x-bar-μ)/(σ/√n) also has a normal distribution. The t-distribution comes in to play when we must estimate σ with the sample standard deviation, ’s’. The standardized sample mean, stadardized by using the sample standard deviation (x-bar-μ)/(s/√n) follows a t-distribution. No (correct) theory says that the sample mean follows a t-distribution!

]]> By: John Johnson http://www.johndcook.com/blog/2009/02/16/statistics-pitfall/comment-page-1/#comment-13634 John Johnson Fri, 20 Feb 2009 14:51:52 +0000 http://www.johndcook.com/blog/?p=1538#comment-13634 Er, The sample mean has a N(μ, σ²/n) distribution. Er, The sample mean has a N(μ, σ²/n) distribution.

]]> By: John Johnson http://www.johndcook.com/blog/2009/02/16/statistics-pitfall/comment-page-1/#comment-13585 John Johnson Thu, 19 Feb 2009 12:36:33 +0000 http://www.johndcook.com/blog/?p=1538#comment-13585 The sample mean has a N(μ, σ²) distribution. The z-score (x-bar-μ)/(σ/√n) has the t. Student didn't say anything about the sample mean, but rather the standardized mean under the null hypothesis. All this assumes you're not modeling (μ, σ²) themselves with probability distributions. The sample mean has a N(μ, σ²) distribution. The z-score (x-bar-μ)/(σ/√n) has the t. Student didn’t say anything about the sample mean, but rather the standardized mean under the null hypothesis.

All this assumes you’re not modeling (μ, σ²) themselves with probability distributions.

]]> By: John Venier http://www.johndcook.com/blog/2009/02/16/statistics-pitfall/comment-page-1/#comment-13476 John Venier Tue, 17 Feb 2009 18:39:42 +0000 http://www.johndcook.com/blog/?p=1538#comment-13476 Interesting observation -- I had not known folks thought that way about it. That might be important to keep in mind when teaching this stuff, that students may draw the erroneous conclusion. I think this all comes from teaching the case of estimating the population mean when the population variance is known (and the population distribution is known to be normal). Has anyone ever encountered this case in real life? Even disregarding the normality assumption? Of course it is a nice mathematical introduction to the case when the population variance is unknown, but still. Anyway, another illustration is that even when the population variance is known, if you construct your estimator using the sample variance you get exactly the same result as when it is unknown. I guess it looks more obvious that way around. By the way, people generally object to corrections for multiple looks on the grounds that there is no spooky quantum effect of having looked at the data already. Same thing for multiple testing. Interesting observation — I had not known folks thought that way about it. That might be important to keep in mind when teaching this stuff, that students may draw the erroneous conclusion.

I think this all comes from teaching the case of estimating the population mean when the population variance is known (and the population distribution is known to be normal). Has anyone ever encountered this case in real life? Even disregarding the normality assumption?

Of course it is a nice mathematical introduction to the case when the population variance is unknown, but still.

Anyway, another illustration is that even when the population variance is known, if you construct your estimator using the sample variance you get exactly the same result as when it is unknown. I guess it looks more obvious that way around.

By the way, people generally object to corrections for multiple looks on the grounds that there is no spooky quantum effect of having looked at the data already. Same thing for multiple testing.

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