# Interesting question on confidence intervals

Say I conducted a sample and did the relevant processing to get a 95% confidence interval of a parameter to fall between a [10,14].

My friend did another sample, and got a x% confidence interval of a parameter to fall between [11,13].

Is x = 95, <95 or >95?

Edit: Made a mistake here. The friend was supposed to use the same data in the qns so answer should be x< 95% :D

• Re the edit: the answer is still not determined: your friend might have used a more powerful procedure than you did.
– whuber
Oct 23, 2020 at 14:23
• @whuber, It can also be a less powerful procedure. Oct 23, 2020 at 15:10
• @Sextus That seems unlikely when that procedure shrank the interval so much, but it is possible in principle.
– whuber
Oct 23, 2020 at 15:30
• @whuber so you are suggesting that [10,14] wasn't truly a 95% confidence interval after all (creating boundaries more wide than necessary)? Oct 23, 2020 at 15:38
• Yes, I was assuming both methods are using the same model for the distribution/probability of the data. Oct 23, 2020 at 16:20

Demetri Pananos argued that the question is underdetermined based on the idea that the sample might be different. However, even when the same sample is used (as clarified on your edit) then we can still not know whether the confidence interval [11,13] is a lower or higher % confidence interval than the confidence interval [10,14].

### Example

Below is an example that is a bit far-fetched, and this answer is pedantic. But it does illustrate how confidence intervals should be interpreted.

Say we compute the mean of a sample from a normal distributed population that is parametized by the 4-th root $$\theta^{1/4}$$ of $$\theta$$. Let the sample distribution of the mean be:

$$\bar{x} \sim \mathcal{N(\mu_{\bar{x}} = \text{sign}(\theta)\text{abs}(\theta)^{1/4}\, ,\, \sigma_{\bar{x}} = 1})$$

this expression $$\text{sign}(\theta)\text{abs}(\theta)^{1/4}$$ is like $$\theta^{1/4}$$ but also has a solution for negative $$\theta$$.

The image below shows how we can create different 95% confidence intervals for this case (see for more about this: The basic logic of constructing a confidence interval). We sketch two cases.

• In one case, the solid lines, boundaries are chosen by the upper and lower 2.5% test intervals.
• In the other case, the broken line, the boundaries are chosen by the upper 5% for $$\theta<0$$ and the lower 5% for $$\theta>0$$

In both these cases, the confidence intervals will contain the true value 5% of the time, independent of the true parameter $$\theta$$.

If we would observe $$\bar{X} = 0$$, then we would have two different intervals approximately $$[-7.320,7.320]$$ and $$[-14.757,14.757]$$. That's a difference by a factor two, just like your case.

I agree that this answer is pedantic, and shows how the question is technically underdetermined. But in practice, you will not find the situation of the example above. In a 'normal' situation we the confidence intervals will not be constructed in these strange/extreme ways and when one confidence interval is smaller (for the same data) then it typically relates to a smaller % confidence. But, as this answer shows, it is not necessary.

• This is truly educational. Thank you. I don't think any of my Profs can make a lesson this interesting. Nov 5, 2020 at 13:04
• @JoelTan to be honest, I find the intuition that [11,13] represents a confidence interval with a smaller x% than [10,14] not so bad. In practice, this difference, a factor 2, is large and you would not encounter this unless the confidence level is different. So, my answer here is only to show the theoretical possibility that the confidence levels for these two different intervals can be the same. In practice, you will not get a situation with such a strong difference in intervals for the same confidence levels. (and that is why my teachers often did not like me) Nov 5, 2020 at 13:10

The latter interval is narrower, but that could be explained by smaller sample variance (say by chance) or by the use of a different quantile (i.e. different $$\alpha$$ and thus different confidence).