The standard deviation of a t-distribution decreases as the degrees of freedom increase, which correlates with an increase in sample size. This phenomenon occurs because larger sample sizes provide a more accurate estimate of the true population standard deviation. When calculating the sample standard deviation, using the sample mean and dividing by n leads to underestimation, while dividing by (n-1) applies Bessel's correction, which adjusts for bias. The population standard deviation, calculated as the square root of the sum of squared deviations from the mean divided by (n-1), is inherently a biased estimate due to Jensen's inequality.
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More data tends to give a more accurate estimate of the true population standard deviation.OpheliaM said:I don't understand why does the standard deviation of a t-Distribution decreases as the degree of freedom (and, thus, also the sample size) increases
The sample standard deviation underestimates the population standard deviation if you use the sample mean and divide by n. If you use the true population mean and divide by n or use the sample mean and divide by (n-1) that is not true.(CORRECTION: it is still under-estimated. See @Number Nine 's post below) For the degree of the t-distribution, you should use the n or (n-1) that you divided by.when the sample standard deviation underestimates the population standard deviation?
FactChecker said:More data tends to give a more accurate estimate of the true population standard deviation.The sample standard deviation underestimates the population standard deviation if you use the sample mean and divide by n. If you use the true population mean and divide by n or use the sample mean and divide by (n-1) that is not true. For the degree of the t-distribution, you should use the n or (n-1) that you divided by.
PS. Just to be more clear. The sample mean should always be the sum of the sample divided by n. When I say "use the sample mean and divide by (n-1)", I mean that the sum of squares of deviations from the sample mean are divided by (n-1). That is Bessel's correction. (see https://en.wikipedia.org/wiki/Bessel's_correction )
I stand corrected. Thanks. I will correct my prior post.Number Nine said:A minor point: the "population standard deviation" (i.e. the square root of the sum of squared deviations from the mean, divided by n-1) is actually a biased estimate of the standard deviation. This follows from Jensen's inequality, since the square root is a concave function. It's fairly difficult to find an unbiased estimator of a normal standard deviation, and the corrections have no closed form -- see https://en.wikipedia.org/wiki/Unbia...deviation#Results_for_the_normal_distribution