2 ms·

I was a bit confused by the article initially: > Perhaps most simply, with a t-statistic of 2, your 95% confidence intervals will nearly touch 0. Your 95% CI

by FabHK 3y ago

I was a bit confused by the article initially: > Perhaps most simply, with a t-statistic of 2, your 95% confidence intervals will nearly touch 0. Your 95% CI will include 0, unless you have more than 50 or so data points, in which case there's no point in using Student's t-distribution, might as well use the Gaussian, which the author seems to assume, and which I thought gave rise to the z-score (in my mind, t-statistic = t-distribution, z-score = normal distribution). But then looking things up, it turns out that difference is that the z-score is computed with population mean and sd, while the t-statistic is computed with sample mean and sd. So, yeah, practically you'll use the t-statistic (and it will be t-distributed if the population is normally distributed), unless you already know population mean and sd, in which case you can compute the z-score (which will approach the normal distribution by CLT under certain conditions with large enough samples, but is otherwise not predicated on normality in any way). Then all the author was pointing out is that if we take a +/- 2 standard error CI, then if your statistic is 2, the CI goes from 0 to 4, giving rise to a 100% "half-width" of the CI, while if your statistic is 4, say, the CI goes from 2 to 6, giving rise to just 50% half-width.

nerdponx 3y ago
The T distribution arises as the ratio between a Gaussian r.v. and the square root of a Chi-square r.v.: X ~ Gaussian(m, v) S ~ ChiSquare(n) T = X / sqrt(S / n) The sampling distribution of the sample mean is Gaussian whenever the data is Gaussian, or whenever the CLT applies. The sampling distribution of the sample variance is Chi-square whenever the data is Gaussian. But the CLT does not have any effect here. In general there isn't much else that we can say about the sampling distribution of the sample variance. Thus if we want to compute a "sample Z statistic" using those estimated quantities, and we know the data is Gaussian, then we know the sampling distribution of that "sample Z statistic". It's the T distribution. But the assumption of underlying Gaussian data is important here. The CLT doesn't help us derive a T distribution. But it is true that our "sample Z statistic" is asymptotically Gaussian, in which case the T distribution itself is approximately Gaussian. [0] So a "T test" (meaning "test of a difference in means"), using the actual T distribution as the null distribution of the test statistic, is basically never valid on non-Gaussian data. But it's valid (asymptotically) using the Gaussian distribution as the null distribution of the test statistic. That's a lot of reasoning to say: `avg(y) / sqrt(var(y) / n)` could be either T or Gaussian, depending on the assumptions and context. But I would push back on conflating that with `avg(y) / sqrt(σ^2 / n)`. Even if they have the same distribution, they are not the same thing. [0]: If you want a good writeup, see https://stats.stackexchange.com/a/253318/36229 https://stats.stackexchange.com/a/253318/36229. Or look into a good statistics textbook.

by FabHK 3y ago

nerdponx 3y ago

The T distribution arises as the ratio between a Gaussian r.v. and the square root of a Chi-square r.v.: X ~ Gaussian(m, v) S ~ ChiSquare(n) T = X / sqrt(S / n) The sampling distribution of the sample mean is Gaussian whenever the data is Gaussian, or whenever the CLT applies. The sampling distribution of the sample variance is Chi-square whenever the data is Gaussian. But the CLT does not have any effect here. In general there isn't much else that we can say about the sampling distribution of the sample variance. Thus if we want to compute a "sample Z statistic" using those estimated quantities, and we know the data is Gaussian, then we know the sampling distribution of that "sample Z statistic". It's the T distribution. But the assumption of underlying Gaussian data is important here. The CLT doesn't help us derive a T distribution. But it is true that our "sample Z statistic" is asymptotically Gaussian, in which case the T distribution itself is approximately Gaussian. [0] So a "T test" (meaning "test of a difference in means"), using the actual T distribution as the null distribution of the test statistic, is basically never valid on non-Gaussian data. But it's valid (asymptotically) using the Gaussian distribution as the null distribution of the test statistic. That's a lot of reasoning to say: `avg(y) / sqrt(var(y) / n)` could be either T or Gaussian, depending on the assumptions and context. But I would push back on conflating that with `avg(y) / sqrt(σ^2 / n)`. Even if they have the same distribution, they are not the same thing. [0]: If you want a good writeup, see https://stats.stackexchange.com/a/253318/36229 https://stats.stackexchange.com/a/253318/36229. Or look into a good statistics textbook.