The problem it solves
You have one sample. You compute its mean and get 14.5. The next question is the only question that matters, and it is not "what is the mean?" — it is "how far from the truth might 14.5 be?"
Nothing in the sample answers that on its own. The answer comes from a thought experiment: if you could draw the sample again and again, what would the collection of sample means look like? That collection — its shape, its centre, its spread — is the sampling distribution of the mean, and every confidence interval and every test statistic in the rest of CS1 is read off one of these distributions.
The idea to hold on to is simple and easy to lose: a statistic is a random variable. It has a distribution, a mean and a variance of its own, and those are different from the distribution, mean and variance of the data.
The two results for the sample mean
For a random sample of size n from any distribution with mean μ and variance σ²:
In words: the sample mean is centred on the right answer, and its spread shrinks as the sample grows — but as 1/√n, not as 1/n, because the variance is what falls by the factor of n.
Both come straight out of the algebra of expectation and variance. The mean of a sum of n things each with mean μ is nμ, and dividing by n gives μ. The variance of a sum of n independent things each with variance σ² is nσ², and dividing by n divides the variance by n², leaving σ²/n.
The independence is doing real work in the second one and none in the first. The sample mean is unbiased whether or not the observations are independent; its variance formula is not. That is why sampling scheme matters, and why a sample of 1,000 policies from one city is not a sample of 1,000 policies.
The standard deviation of the sample mean, σ/√n, is called the standard error. It is one of the few pieces of vocabulary in the paper worth memorising precisely: the standard deviation of the data describes the spread of individual observations, the standard error describes the spread of an estimate.
Adding normality: the three results that unlock everything
If the sample comes from a normal distribution, three further things are true, and together they are the foundation of the whole inference half of the paper.
One. The sample mean is exactly normal, not approximately: X̄ ~ N(μ, σ²/n). A sum of independent normals is normal, which the generating-function argument gives you in one line.
Two. The scaled sample variance has a chi-square distribution:
In words: the sum of squared deviations, measured in units of the true variance, follows a chi-square distribution with one degree of freedom fewer than the sample size. The degree of freedom goes missing because the deviations are taken from the sample mean rather than from μ; one piece of information has already been spent locating the centre.
Three. X̄ and S² are independent. This is special to the normal distribution and it is what makes the next step legal.
Put two and three together. You would like to standardise the sample mean, but σ is unknown, so you use S instead. What you get is a normal divided by the square root of an independent chi-square over its degrees of freedom — which is the definition of a t distribution:
So the t distribution is not a patch bolted on to handle small samples. It is the exact distribution of that ratio, and it has heavier tails than the normal for a concrete reason: the denominator is random too, and sometimes it comes out small.
Two facts about the chi-square worth carrying
A chi-square with k degrees of freedom has mean k and variance 2k. For k = 7 that is a mean of 7 and a standard deviation of √14 = 3.7416574 — a spread more than half the size of the mean. A chi-square is a badly skewed distribution at small degrees of freedom, and that single fact is why a confidence interval for a variance is nothing like symmetric. The worked example makes that uncomfortably clear.
Why the sample variance divides by n minus 1
Deviations are taken from X̄, and X̄ is pulled towards whatever the sample happens to contain. So the sample's deviations are, on average, slightly too small — the data is closer to its own mean than to the true mean. Dividing the sum of squared deviations by n would inherit that shortfall and produce an estimate that is on average too low by the factor (n − 1)/n. Dividing by n − 1 corrects it exactly:
This is usually asserted. It can be verified, completely, on a population small enough to enumerate, and part (a) of the worked example does exactly that: 16 samples, every one of them written down, and the average of the 16 values of S² equal to the population variance to the last digit.
Where it is used after this topic
Every interval and every test. A confidence interval for a mean is the t result above, rearranged. A test for a variance is the chi-square result, rearranged. Nothing new is needed; the sampling distributions are the inference.
Regression. The t statistic on a slope is the same construction with a different standard error, and the F statistic is a ratio of two independent chi-squares over their degrees of freedom.
The bootstrap. When the sampling distribution of a statistic cannot be derived, you can simulate it by resampling. That idea only makes sense once you see that the object you want has been a distribution all along.
Where marks are lost
Confusing the standard deviation with the standard error. In the worked example the data has a standard deviation of 3.8913824 days and the mean has a standard error of 1.3758114 days. Both are correct; they answer different questions, and an interval built with the wrong one is wrong by a factor of √n.
Using σ²/n when the sampling is not independent. Nothing in the formula warns you.
Writing n degrees of freedom for the sample variance. It is n − 1, and the loss is not a convention — it is the price of having estimated the centre from the same data.
Assuming a symmetric interval for a variance. The chi-square distribution is skewed, so the two tail points are not reflections of each other and the resulting interval is not centred on s². Students who build it by taking "estimate plus or minus something" get an answer that can include negative variances.
Forgetting that results two and three need normality. The mean and variance of X̄ hold for any distribution. The chi-square result, the independence and the t distribution are all normal-theory results.
Questions students ask
Is the sampling distribution something I could ever see?
Directly, no: you have one sample, not a thousand. With a computer, yes — draw 10,000 samples from a known distribution, compute the mean of each, and histogram them. The histogram is the sampling distribution, and this is exactly what the practical half of the paper asks you to do. Part (a) of the worked example is the same exercise done by hand on a population small enough that "all possible samples" is only sixteen.
Why does anyone care about the distribution of S²? I only ever want the mean.
Three reasons. It is what justifies the n minus 1 divisor. It is what the t distribution is built out of, so you cannot have intervals for a mean without it. And variability is sometimes the quantity of interest in its own right — a reserve is set on the spread of outcomes, not on their average.
If the data is not normal, is any of this usable?
The mean and variance of X̄ are unaffected. The distributional results are not, but the central limit theorem rescues the mean: for a reasonable sample size X̄ is approximately normal whatever the data looks like, so t intervals for a mean are robust. The chi-square result for S² is not robust at all — it is sensitive to non-normality even in large samples, which is worth knowing before you quote an interval for a variance.
A worked example
Two parts. The first proves the n minus 1 divisor by brute force, on a population small enough to write down completely. The second builds an interval for a variance from a real-sized sample and checks it against the test it inverts. Both data sets are invented for this example.
Part (a): all sixteen samples
Take a population with just four equally likely values: 2, 4, 6, 8. Its mean is (2+4+6+8)/4 = 5, and its variance — dividing by 4, because this is the whole population and nothing is being estimated — is
Now take samples of size 2 with replacement. There are 4 × 4 = 16 of them, and for each one the sample mean is (x1+x2)/2 and the sample variance, with n − 1 = 1 in the denominator, is (x1−x2)²/2.
| Second draw → | 2 | 4 | 6 | 8 |
|---|---|---|---|---|
| 2 | 2, 0 | 3, 2 | 4, 8 | 5, 18 |
| 4 | 3, 2 | 4, 0 | 5, 2 | 6, 8 |
| 6 | 4, 8 | 5, 2 | 6, 0 | 7, 2 |
| 8 | 5, 18 | 6, 8 | 7, 2 | 8, 0 |
Each cell holds the sample mean and then the sample variance. Now average down the whole table.
The sample means. They add to 80, so their average is 80/16 = 5 — the population mean exactly. Their variance works out at 2.50, and σ²/n = 5/2 = 2.50. The formula for the standard error is confirmed, not assumed.
The sample variances. They add to 80 as well, so their average is 80/16 = 5, which is σ² exactly. Dividing by n − 1 is unbiased, and here that is a fact about sixteen numbers rather than a claim.
And the version that divides by n. Every entry halves, so the average is 2.50 — which is σ²(n−1)/n = 5 × ½. It understates the population variance by half. With n = 2 the error is as large as it ever gets, but it is the same error at every sample size, shrinking as (n−1)/n.
A second route to the same conclusion, in one line. The sixteen values of (x1−x2)² add to 160, so E[(X1−X2)²] = 160/16 = 10. But for independent draws that expectation is 2σ², giving σ² = 5; and since S² = (X1−X2)²/2, its expectation is 10/2 = 5 too. The enumeration and the algebra meet.
Part (b): an interval for a variance
Eight claims are sampled and the number of days each took to settle is recorded: 12, 15, 9, 21, 14, 18, 11, 16. Invented data.
The total is 116, so x̄ = 116/8 = 14.5 days. The deviations are −2.5, 0.5, −5.5, 6.5, −0.5, 3.5, −3.5 and 1.5, and their squares add to 106.0.
Check that sum the other way, from the raw totals: Σx² = 1788 and n x̄² = 8(210.25) = 1682, and 1788 − 1682 = 106. Agreed.
standard error of the mean = 3.8913824/√8 = 1.3758114 days
Those last two numbers are the ones students interchange. The spread of settlement times is about 3.9 days; the uncertainty in the average settlement time is about 1.4 days.
Now the interval for σ². Start from the chi-square result and rearrange it so that σ² is in the middle:
→ (n−1)s²/χ²upper < σ² < (n−1)s²/χ²lower
Note that the upper critical value produces the lower limit. Dividing reverses the order, and that swap is where most of the marks on this calculation are lost.
With 7 degrees of freedom the two 2.5 per cent points are χ²upper = 16.0127643 and χ²lower = 1.6898692:
upper limit = 106/1.6898692 = 62.7267
Taking square roots, a 95 per cent interval for σ itself runs from 2.5729 to 7.9200 days. Four decimal places, not seven: the upper limit divides by a critical value of 1.69, so it inherits every bit of that figure's rounding and magnifies it.
The check: the interval is the test, inverted
Suppose someone claims σ² = 6.6197, the lower limit. Test it with the same chi-square statistic:
which is the upper 2.5 per cent point, 16.0127643, to the four decimals the limit carries — the test sits exactly on the boundary of rejection. Repeat at the upper limit, 106/62.7267 = 1.6899, and you land on the lower critical value, 1.6898692, in the same way. The interval is not a separate construction from the test; it is the set of hypothesised variances the test would not reject, and the arithmetic proves it.
Read the interval before you leave it
The estimate is s² = 15.14 and the interval runs from 6.62 to 62.73. The upper limit is 9.48 times the lower one, and s² sits only 15 per cent of the way along — nowhere near the middle.
Both features are the chi-square skewness showing up in an answer, and the honest conclusion is the one worth writing down: eight observations tell you almost nothing about a variance. The same eight observations give a perfectly serviceable interval for the mean, 14.5 ± 2.3646243 × 1.3758114, which is 11.2467229 to 17.7532771 days — a range of about six days around an average of fourteen and a half. Means are cheap to estimate; variances are expensive. Any exercise in reserving or reinsurance pricing that hinges on a variance from a short run of data deserves the scepticism this interval is handing you.
Where this fits in CS1
This page explains one idea from CS1 Actuarial Statistics. The full CS1 syllabus, with what each part is examined on, sits on the course page, and past-paper style mock exams with full solutions are where you find out whether the idea has stuck under time pressure.
Other CS1 topics explained
- Moment generating functions, and how to get moments out of them
- The central limit theorem, and when the normal approximation is safe
- Maximum likelihood estimation, step by step
- Confidence intervals: what the 95% actually means
- Hypothesis testing and p-values, without the folklore
- Simple linear regression: least squares, fitted by hand
- Generalised linear models: the three parts, and the link function
- Bayesian estimation, conjugate priors and the credibility formula inside them
- Conditional expectation, and the variance decomposition that comes with it
- The method of moments, and when maximum likelihood is worth the extra work
- The likelihood ratio test, and why the Wald and score tests exist too
- The chi-square goodness of fit test, and how to get its degrees of freedom right
- Contingency tables: testing whether two classifications are related
- Analysis of variance: comparing three or more means at once
- Multiple linear regression, and what controlling for another variable really does
Stuck on a term rather than the idea? The actuarial glossary defines the notation used above. If something here is wrong or unclear, tell us — these pages get corrected.