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> in 95% of runs (in the long run) the true value would be in the interval [a,b] obtained on that run So 95% of the runs produce an interval that includes the
by Hermel 11y ago
> in 95% of runs (in the long run) the true value would be in the interval [a,b] obtained on that run
So 95% of the runs produce an interval that includes the true value. Doesn't that imply that when doing one run, there is a 95% chance of the resulting interval including the true value? I.e. the probability is 95% that the true value is in the interval?
- kgwgk 11y agoNot really. For example, following a procedure which behaves well in the long run and provides 95% coverage you could get a confidence interval that contains only impossible values. Consider also the following example: you want to estimate x from two random numbers taken uniformly from the interval[x-1,x+1]. The interval defined by the two random numbers is a 50% confidence interval (half of the times x will be in the middle of the two). However, for some specific realizations (when the distance between the numbers is grater than one) you will be able to say with certainty that the interval contains x. See also https://github.com/richarddmorey/ConfidenceIntervalsFallacy https://github.com/richarddmorey/ConfidenceIntervalsFallacy
- Hermel 11y agoThanks, you are right. I thought I understood confidence intervals, but apparently I did not.
- gjm11 11y agokgwgk's explained why the answer is no, but let me offer a slightly different perspective. The situation is similar to the one we discussed earlier where you do the same experiment twice. Before you do the experiment, your estimate of Pr(true value is in confidence interval) is 95%. But after you do the experiment, you have extra information: you know how the experiment turned out. And this may lead you to a different estimate of Pr(true value is in the confidence interval).