I have some confusion regarding confidence interval.

I was going through this link http://www.stat.yale.edu/Courses/1997-98/101/confint.htm

It has something like this:

Suppose a student measuring the boiling temperature of a certain liquid observes the readings (in degrees Celsius) 102.5, 101.7, 103.1, 100.9, 100.5, and 102.2 on 6 different samples of the liquid. He calculates the sample mean to be 101.82. If he knows that the standard deviation for this procedure is 1.2 degrees, what is the confidence interval for the population mean at a 95% confidence level?

In other words, the student wishes to estimate the true mean boiling temperature of the liquid using the results of his measurements. If the measurements follow a normal distribution, then the sample mean will have the distribution $\mathcal{N}(\mu_i,\sigma/\sqrt{n})$. Since the sample size is 6, the standard deviation of the sample mean is equal to $1.2/\sqrt{6}=0.49$.

My question is when it says, "then the sample mean will have the distribution", why is it saying the distribution of the sample mean? It should have said 'the distribution of the readings of boiling points', shouldn't it? Is it that we take sets of samples and from each set of samples we get a sample mean and this set of sample means follow normal distribution? I am confused regarding this.

Also how come it is $\mathcal{N}(\mu_i,\sigma/\sqrt{n})$. I mean since the mean is an unbiased estimator we have taken it as $\mu_i$ and since sample standard deviation is a biased estimator we have taken $\sigma/\sqrt{n}$. Still it should have been $\sigma/\sqrt{(n-1)}$, shouldn't it?

  • 1
    $\begingroup$ What kind of confusion would you like? $\endgroup$
    – Gschneider
    Commented Jun 5, 2012 at 21:41
  • $\begingroup$ I mean why it is saying distribution of sample mean. It should have been distribution of the reading isn't it. When we plot the normal curve, in the x axis is it the sample mean or the readings? $\endgroup$
    – user31820
    Commented Jun 5, 2012 at 21:43
  • 1
    $\begingroup$ Since gung's edit negates my previous comment, a good place to start would be: en.wikipedia.org/wiki/Sampling_distribution. $\endgroup$
    – Gschneider
    Commented Jun 5, 2012 at 21:49
  • 2
    $\begingroup$ There is an interesting linguistic ambiguity coming to light here: in almost any other context, "the sample mean" would be taken in the sense of "the mean of this sample," whereas the text intends that it be understood in the sense of "a random variable determined by the process of obtaining a sample and computing its mean." (However, the quotation was taken from a more extensive text which uses various locutions in an effort to avoid this ambiguity, such as the phrase "standard deviation for this procedure.") $\endgroup$
    – whuber
    Commented Jun 5, 2012 at 21:57

3 Answers 3


I think what would help here is a good, solid overview of distributions and standard deviations / standard errors. I take it you know what a distribution is, but note that there are typically 3 different entities that we work with, all of which have distributions that are relevant.

First there is the population distribution, which is understood to be infinite. Usually, we think of there being an infinite number of units which could be observed; in your case, I think it's being conceived as one true value of the boiling point, but the measurement system has random error associated with it. That is, I don't think they have sampling error in mind, but either way, there are infinite potential values of the boiling point, and these values are distributed in some way. (I gather it's being assumed that the distribution is normal, or normal-ish--which is good enough for our purposes.)

The second is your sample distribution. You have the observations for your sample, you can describe this completely.

But how can you connect what you know about your sample to what you want to know about the population? This is where the third distribution comes in. It is the sampling distribution of the sample mean. Specifically, it is the distribution that you would have, if you followed the same procedure as here (i.e., taking 6 samples in the same way, measuring the boiling point in the same way, calculating the mean in the same way, etc.), over and over again infinitely. It is the sampling distribution that helps you draw inferences about the population (more precisely, the true boiling point of the liquid). This issue is the crux of your confusion. When they write, "then the sample mean will have the distribution", the problem is that it is not clearly worded. Your sample mean is a single realized value; it does not have a distribution, except in a trivial sense. Instead, they are referring to the sampling distribution of the sample mean in the sense I just described.

An important issue concerns the shape of the sampling distribution. The Central Limit Theorem (also discussed here) is often invoked to assert that the (hypothetical, infinite set of such) sample means would be distributed normally. Typically, I would feel much more comfortable saying that if the $N$ were considerably greater than 6(!), but the situation described is clearly meant to be understood as the distribution of the data are either normal or very close to, and therefore we may be OK here. The CLT doesn't only tell us that the sampling distribution will be normally distributed, it also tells us (quite usefully) that the standard deviation of the sampling distribution (called the 'standard error') is $\sigma/\sqrt{N}$. You can then use this standard error to calculate your 95% confidence interval. The CLT is really very helpful.

Finally, you are right that the population formula for the variance:
$$ Var(X)=\frac{\sum(x_i-\bar{x})^2}{N}, $$
when applied to a sample, provides a biased estimate of the true population variance. This is because you first used the very same sample data to calculate the mean, $\bar{x}$, and you need to account for this fact. This is done by dividing by $N-1$ instead of $N$. After having done this, however, you have appropriately accounted for that source of bias--you're done with that. When it comes to estimating the variability in the sampling distribution of the mean, you don't have to worry about it any more. Thus, when the CLT tells us that the standard error will be $\sigma/\sqrt{N}$, it is talking about a different thing, and is assuming that we have already taken care of the issue and that the $\sigma$ we are using is correct.


The confidence interval is for the population mean and it is based on the sample mean. So you want the distribution of the sample mean. The sample standard deviation does not enter into the distribution of the sample mean. Since the independent observations are N(μ,σ) the average of n of them is N(μ,σ/√n).


Distribution of sample mean which is $\bar X$=$\Sigma_{i=1}^{n} X_i\over{n}$ comes from


If $X \sim N(\mu_X, \sigma^2_X)$, $Y \sim N(\mu_Y, \sigma^2_Y)$

$X+Y \sim N(\mu_X+\mu_Y, \sigma^2_X+\sigma^2_Y)$

So, $\bar X$=$\Sigma_{i=1}^{n}$$ X_i\over{n}$ when $X_i \sim N(\mu_X, \sigma^2_X)$ and independent with each other

and $ X_i\over{n}$ $\sim N(\mu_X/n,$$ \sigma^2_X \over{n^2}$)


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