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Minitab blog claims that:

"As you increase the sample size, the sampling error decreases and the intervals become narrower. If you could increase the sample size to equal the population, there would be no sampling error. In this case, the confidence interval would have a width of zero and be equal to the true population parameter."

That makes sense intuitively, but I don't see it mathematically. Lets say I had a population size of 100 with a known mean and population standard deviation. If I were to sample the entire population, my confidence interval would not be 0:

$$Mean_{sample} (+ OR-) z*\sigma= CI$$

Mean sample would become population mean and z*standard devation would become some nonzero number (depending on the confidence level chosen for z). Why doesn't the $z*\sigma$ term go to zero for a noninfinite sized population?

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  • $\begingroup$ What will happen with the standard deviation when the population size increases? How does this affect your formula for CI? $\endgroup$
    – imranfat
    Commented Oct 6, 2016 at 15:48

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The notion of confidence interval is somewhat intuitive, but that may be keeping you from understanding what it means in more depth.

Say I have multiple samples $x_i$ from a population, and I wish to estimate the population mean $\mu$. A CI of, say, 95\% represents an interval of possible values of $\mu$ such that given my samples, the "probability" that the $\mu$ lies in that interval is 95%.

We immediately see that there can be more than one such interval, since I could trade probability past the upper end for probability at the lower end of the interval, thus shifting the interval. Let's skirt that issue by demanding a symmetric interval about my sample mean.

But the "probability" is not well defined from the information I just presented!

In order to assign a probability, I have to make some assumptions about the population. The usual assumption is that the population variance is equal to the unbiased estimator of variance obtained from our sample. But we still have things backward: We can't honestly talk about the probability of the population mean being in some range, without any assumption about the a priori (before I saw my samples) probabilities of the mean being various values.

So we apply the usual sleight-of-mind logic employed by the frequentist point of view. We ask:

Given that the population variance is our unbiased sample variance estimate, What are the highest and lowest values of the population mean $\mu$ such that the chance of or sample being as far away from $\mu$ as it is, is lower than 100%-95% = 5%.

Now let's go back to your problem. Since the population is finite, as you draw more samples (without replacement) you actually do learn something about the population. If you had drawn all the objects but one, if you take your sample unbiased variance as the population variance, your 95% confidence interval for the value of that remaining one object would be roughly $2\sigma$ but your estimate of the population mean will have a variance of $\sigma/N$. This is quite a bit smaller than would be the case for an infinite population or a small sampling of a large population.

Now when you draw that last sample, you know everything about the distribution. In particular, you know the mean exactly. Therefore any interval that includes the actual mean is a 100% CI. If you then say that the real CI is the tightest such interval, then it has width zero.

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  • $\begingroup$ No, you don't know everything about the distribution. You still only have a finite number of samples from it. $\endgroup$
    – TonyK
    Commented Oct 6, 2016 at 16:10
  • $\begingroup$ Hmm, so there is no way to show the interval going to zero mathematically, when we "draw that last sample"? $\endgroup$
    – Nova
    Commented Oct 6, 2016 at 16:15
  • $\begingroup$ @TonyK if you have sampled all the population without replacement, you in fact do know everything about the distribution, including the mean, sigma, and all higher moments. $\endgroup$
    – Mark Fischler
    Commented Oct 6, 2016 at 20:55
  • $\begingroup$ The CI is given by $\left[\bar{x}-z\frac{\sigma}{\sqrt{n}}, \bar{x}+z\frac{\sigma}{\sqrt{n}}\right]$ so its length is $2z\frac{\sigma}{\sqrt{n}}$ so as $n\to\infty$ the length goes to zero and you end up with the degenerate CI $\left[\bar{x}, \bar{x}\right]$ which is the true population mean $\mu$. $\endgroup$
    – Filip
    Commented Nov 27, 2020 at 23:03
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The standard deviation will go to 0, since you do not have uncertainty.

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I deleted my previous answer because, according to @Mark Fischler, it was not relevant. Here's another explanation:

To calculate the confidence interval of the mean, we need the standard deviation of the sample mean, which is $\sigma/\sqrt{n}$.

$$CI = (\mu-z*\sigma/\sqrt{n}, \mu+z*\sigma/\sqrt{n}).$$

If standard deviation is not known, we can use standard error of the mean instead.

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    $\begingroup$ But the population standard deviation is know (we sampled the entire population)...and I'm not sure how you are answering my question. $\endgroup$
    – Nova
    Commented Oct 6, 2016 at 16:12
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The confidence interval for the mean will go to zero, because the mean is just the sum of the population, divided by its size. If you can sample the whole population, then you know this exactly.

But if you are estimating the mean $\mu$ of the underlying distribution, then (as you say) its confidence interval will not go to zero. These are two different things.

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  • $\begingroup$ Well a confidence interval always uses the sample mean. The point is why would confidence interval go to 0 when the sample mean is equivalent to the population mean (because entire population is sampled). $\endgroup$
    – Nova
    Commented Oct 6, 2016 at 16:10

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