I was reading on the computation of the unbiased estimation of standard deviation and the source I read stated
(...) except in some important situations, the task has little relevance to applications of statistics since its need is avoided by standard procedures, such as the use of significance tests and confidence intervals, or by using Bayesian analysis.
I was wondering if anyone could elucidate the reasoning behind this statement, for example doesn't the confidence interval use the standard deviation as part of the calculation? Therefore, wouldn't the confidence intervals be affected by a biased standard deviation?
Thanks for the answers so far, but I'm not quite sure I follow some of the reasoning for them so I'll add a very simple example. The point is that if the source is correct, then then something is wrong from my conclusion to the example and I would like someone to point how how the p-value doesn't depend on the standard deviation.
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 with a significance level of 0.05. 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. This is below our significance level of 0.05 so we reject the null hypothesis.
So in this example, wouldn't the p-value (and maybe your conclusion) change depending on how you estimated your sample standard deviation?