Sampling distributions

In depth

Why σ ÷ √n

When n values are drawn independently, the variance of their total is n times the variance of one value, $n\sigma^2$. The mean is the total divided by n, so its variance is $n\sigma^2 \div n^2 = \sigma^2 \div n$, and its standard deviation $\sigma \div \sqrt{n}$. In words: in a sample, high and low values partly cancel, and the more values there are the more they cancel, but only with the square root of n. This is the law of large numbers made precise: the sample mean settles towards μ as n grows, and σ ÷ √n says how fast.

Notice what the formula leaves out: the size of the population. A random sample of 1,000 customers is as precise for a retailer with a million customers as for one with 50,000, provided the sample is a small part of each. Precision depends on the sample size and on how variable the population is, not on what fraction of it you see.

Sampling without replacement: the finite population correction

Real samples are usually drawn without replacement: an order can’t be picked twice. When the sample is a large share of the population, each draw changes what is left, and the sample mean varies less than $\sigma \div \sqrt{n}$ says. The finite population correction $\sqrt{(N - n) \div (N - 1)}$ shrinks the standard error to match. OpenStax’s business text uses it when more than 5% of a known population is sampled.

Thirty of the 231 orders is 13% of the population: the correction is $\sqrt{201 \div 230} = 0.935$ and the standard error falls from $18.38 to $17.18. Ana’s 40 bags came from thousands that Kilnworth fills, so for them the correction is almost exactly 1 and can be ignored. If the whole population is sampled, $n = N$, the correction is 0: a census has no sampling error.

How large must n be?

OpenStax’s business text says the theorem needs a sample that is “large enough”, which it puts at 30 observations or more. That rule of thumb works for populations that are not badly skewed. For skewed populations a stricter guide is Cochran’s rule, as Kott and Levine (2024) state it: a 95% interval of the usual kind around a sample mean is reasonable when n is greater than 25 times the square of the population’s skewness.

The rule needs the population’s skewness, which Excel’s SKEW.P calculates (SKEW adjusts for a sample). For the 231 orders as the population, SKEW.P is 13.57, so the rule asks for $25 \times 13.57^2 \approx 4,607$ orders. Without the office order SKEW.P is 2.28 and the rule asks for more than $25 \times 2.28^2 \approx 130$. Both are rules of thumb, not theorems; they say the same thing as the figure: the more skewed the population, and the more its mean depends on a few extreme values, the larger the sample it takes before x̄ behaves normally.

Estimating the standard error from one sample

The formula needs σ, the population’s standard deviation, which is rarely known. In practice you put the sample’s s in its place: $s \div \sqrt{n}$, the line the ToolPak labels Standard Error. Ana’s 40 bags give $2.66 \div \sqrt{40} = 0.42$ g: her sample mean of 1,002.75 g is probably within about 1 g of the machine’s true mean. Because s is itself an estimate that varies from sample to sample, a z built from it is a little more variable than one built from σ; the next lesson’s t distribution allows for that.

Two habits follow. Report a sample mean with its standard error, so the reader can see how much it could move. And remember that the standard error estimates chance variation only if the sample was drawn at random: from 40 bags chosen because they looked underfilled, s ÷ √n would be a precise-looking answer to the wrong question.

When is a sample proportion close to normal?

A proportion’s sampling distribution is a binomial count divided by n, so the previous lesson’s condition carries over: it is close to normal when the sample is expected to contain enough of each outcome. With p = 0.45 and n = 50, a sample should hold about $50 \times 0.45 = 22.5$ subscriptions and 27.5 one-off orders, plenty of each. With n = 10 it would hold about 4.5 and 5.5: then $\hat{p}$ can take only 11 values, 0, 0.1, …, 1, and its histogram is a set of separate bars rather than a curve. The Explore widget of sample proportions shows both.

Using the normal model for a sample mean

When the sampling distribution is close to normal, standardise x̄ as you would one value, with the standard error in place of σ: $z = (\bar{x} - \mu) \div (\sigma \div \sqrt{n})$. NORM.DIST then gives probabilities for x̄, with the standard error as its standard_dev argument. The challenging example asks how often a sample of 30 orders averages more than $50, and finds that the normal model gets it badly wrong for this population.

The same works for a proportion, with $z = (\hat{p} - p) \div \sqrt{p(1 - p) \div n}$. How often would a random sample of 50 orders, drawn with replacement, contain 30% subscriptions or fewer, if 45.0% of all orders are subscriptions? $z = (0.30 - 0.4502) \div 0.0704 = -2.13$, and =NORM.S.DIST(-2.13,TRUE) gives about 1.7%. The exact binomial, =BINOM.DIST(15,50,104/231,TRUE), gives 2.2%: about once in 46 samples. The normal model is a little low this far into the tail, a reminder that it is an approximation.

How samples are drawn

The formulas above assume a simple random sample (B01): every set of n members of the frame equally likely. In Excel, put =RAND() beside each row (it returns a number from 0 up to 1, and a new one each time the sheet recalculates), copy the column and paste it back as values so it stops changing, sort by it, and take the first n rows. Three other designs are common.

Stratification pays when the strata differ. Lantern’s regions do: 75% of East orders were subscriptions against 33% of North orders. A proportional stratified sample of 40 takes 40 × 58 ÷ 231 = 10.04 North orders, and so on; rounding by largest remainders gives 10 North, 8 South, 9 East, 9 West, 4 Central. The share of subscriptions it finds can’t swing because chance put too many East orders in the sample.

Biases that no sample size cures

Every formula here measures sampling error, the chance part. None measures bias, and a larger sample doesn’t reduce it. If the sampling frame leaves people out (a list of web customers that misses app-only buyers), if the people who answer differ from those who don’t (non-response bias), or if the measurement is off (a scale that reads 2 g light), every sample is wrong in the same direction, however large.

B01’s September survey is the example to remember: 240 customers were invited and 79 replied, 32.9%. The next lesson puts a margin of error on the 61 of 79 who would recommend Lantern, but that margin covers only the luck of which customers replied, not the possibility that happy customers were keener to reply. Survey design and response rates are M04 Market Research’s subject; this lesson’s point is that the standard error is silent about them.