The discussion revolves around the relationship between standard deviation, sample size, and the t-distribution. Participants explore the implications of sample size on the accuracy of standard deviation estimates and the conditions under which sample standard deviation may underestimate population standard deviation.
Participants do not reach a consensus on the implications of the sample standard deviation and its relationship to the population standard deviation. Multiple viewpoints and clarifications are presented, indicating ongoing debate and exploration of the topic.
Limitations include the dependence on definitions of standard deviation and the conditions under which estimators are considered biased or unbiased. The discussion does not resolve the complexities surrounding these definitions and their implications.
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