Short premise: this question is very simple and I expected the answer to be uncontroversial. Contrary to my expectations, however, I did meet some disagreement, so let's hear your opinion.

For simplicity, let's consider a normal population with known variance $\sigma^2$. We can form an exact 95% confidence interval for the mean $\mu$, using the sample mean $\bar{X}_N$ and $\sigma$. The standard interpretation of the confidence interval is that it's a random interval which has a probability $p=0.95$ of containing $\mu$. In other words, if you repeatedly get $N$ samples from your population and each time form the corresponding confidence interval, then, as the number of repetitions $ M \to \infty$, the proportion of confidence intervals which contain $\mu$ approaches 0.95. That's ok.

However, I could also describe this situation by saying that the random variable $\bar{X}_N$ has a 95% probability of falling in the fixed interval $I=[\mu-1.96\frac{\sigma}{\sqrt{N}},\mu+1.96\frac{\sigma}{\sqrt{N}}]$. Now I won't call $I$ a confidence interval (it's not random), but still it's true that $\bar{X}_N$ has a 95% probability of being in $I$. I considered a normal population with known variance, in order to have a simpler example, but the concept is valid in general. Do you agree with me?

ps I know I could verify this with a simulation in R, but I don't have R here right now.

EDIT: as noted in the comments, I know that $I$ cannot be used to perform inference in practice, since $\mu$ is not known, and if it was, we wouldn't need to do any inference at all, since then the population pdf would be perfectly determined. It was just an intellectual curiosity, and anyway it can be seen as a mathematical model of the distribution of a statistic (in this case, the sample mean for a normal population). In other words, even if $\mu$ and $\sigma$ are unknown, we know that $\bar{X}_N$ is normally distributed $\forall N$, and we know that the event $A=\{\bar{X}_N \in I\}$ has $p=0.95$.

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    $\begingroup$ That seems unexceptionable: what's the nature of the disagreement? (The only odd thing is talking of an "alternative interpretation" of a confidence interval, when, as you say, $I$ is not a confidence interval.) $\endgroup$ – Scortchi - Reinstate Monica Jan 2 '16 at 13:36
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    $\begingroup$ This "concept" is perfectly valid, but it's also useless, because any application of it would require knowing both $\mu$ and $\sigma$--and at that point we don't even need any data to give us full information about the distribution! $\endgroup$ – whuber Jan 2 '16 at 13:47
  • $\begingroup$ @Scortchi, thanks for the confirmation! After I showed your comments, the disagreement has been resolved. Turns out it was partly due to habit (the situation is usually described in terms of multiple intervals which may or may not cointain $\mu$, so the person I was discussing with didn't immediately see the equivalence) and partly due to improper language on my part. As you note, since I know that $I$ is not a confidence interval, I shouldn't really talk of an "alternative interpretation" of confidence intervals. $\endgroup$ – DeltaIV Jan 2 '16 at 15:06
  • $\begingroup$ @whuber, sure, I expected such a comment and I also thought of writing a disclaimer, but then I stopped because I didn't want to write a long post for such a simple question. Anyway, since the comment actually arised, I will update the question with the disclaimer I had in mind. $\endgroup$ – DeltaIV Jan 2 '16 at 15:09
  • $\begingroup$ What is the question, then? $\endgroup$ – whuber Jan 2 '16 at 15:28

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