Lesson 2 · 25 min

Confidence intervals and Student’s t

Only four measurements — what does the mean tell you about the true value? Student’s t gives an honest confidence interval, and shows why extra replicates soon bring diminishing returns.

In this lesson

Calculate a confidence interval from the mean, the standard deviation and Student’s t
Interpret a “95 % confidence interval” correctly
Explain why small samples need t rather than 1.96
Estimate how many replicates are needed for a given interval width

A tablet is labelled 250 mg of active ingredient, and four analyses average 247.5 mg. Is the tablet short? It depends on how far that mean might be from the truth: if the four results agree within 0.2 mg the shortfall is probably real; if they differ by 5 mg a mean 2.5 mg low could easily be luck. A confidence interval turns “how far it might be” into a number: a range around the mean that contains the true value with a stated level of confidence.

Predict first

A hundred laboratories each make 4 measurements and each compute a 95 % confidence interval. About how many of the intervals contain the true value?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

All 100 About 95 About 50 Impossible to say; it is luck

Student’s t distribution

If the true σ were known, the 95 % interval for the mean would be $\bar{x} \pm 1.96\,\sigma/\sqrt{n}$ . But in the lab we only have the sample standard deviation s, and with few measurements s is itself unreliable — sometimes too small, which makes the interval too narrow. In 1908 William Gosset, a chemist at the Guinness brewery, published the fix under the pen name “Student”: replace the normal distribution with the t distribution. It is lower and wider, with heavier tails — the more so the fewer the degrees of freedom ν = n − 1 — and turns back into the normal distribution as n grows.

\mu = \bar{x} \pm \frac{t\,s}{\sqrt{n}},\qquad \nu = n - 1

Two-sided t for 95 % confidence: 12.706 for ν = 1, 2.776 for ν = 4, 2.262 for ν = 9 and 1.960 as ν → ∞.

$\mu$: the true value (population mean)
$t$: Student’s t, set by the confidence level and the degrees of freedom
$s/\sqrt{n}$: the standard error of the mean
$\nu$: degrees of freedom: n results minus the one used up by the mean

Half-width of the 95 % interval in units of s (t/√n): the first few replicates pay most
n	t (ν = n − 1)	Half-width / s
2	12.706	8.98
3	4.303	2.48
4	3.182	1.59
5	2.776	1.24
10	2.262	0.72
20	2.093	0.47

Source: Computed from the t distribution (checked against the t table in Harris, Quantitative Chemical Analysis, ch. 4).

Going from 2 to 3 measurements shrinks the interval more than threefold; going from 10 to 20 shrinks it by only another third. That is why routine analyses usually run 3–5 replicates. You can also plan backwards: expecting s = 0.40 and wanting a 95 % half-width of 0.25 or less, solve $t\,s/\sqrt{n} \le 0.25$ — it takes 13 measurements.

Does each interval x̄ ± t·s/√n catch the true value?

Confidence intervals: 50 repeated experiments

Replicates per experiment n · 3

Wrongly use z (normal) instead of t

t distribution (ν = 2)
Standard normal

Critical value t: 4.303
Critical value z: 1.960

Half-width = 1.960 × s/√n; for n = 3, t is 120 % larger than z.

38 of the 50 intervals contain the true value

In the long run (4000 experiments): 82.1 %

Here z = 1.96 is used on purpose, with only 3 measurements per experiment: count the misses among the 50 intervals and read the long-run coverage below — well under 95 %. Untick the box to use t = 4.303 and the coverage returns to about 95 %; then raise n to 30 and the difference between t and z all but vanishes.

Worked example

Carbohydrate in a glycoprotein

Five determinations of carbohydrate in a glycoprotein give 12.6 %, 11.9 %, 13.0 %, 12.7 % and 12.5 % by mass. Find the 95 % and 90 % confidence intervals.

Concept check

With s unchanged, going from 4 to 16 replicates shrinks the half-width of the 95 % confidence interval by about what factor?

1/2 About 1/3 1/4 1/16

Try the calculation

Five determinations of nitrate in a water sample give $\bar{x} = 10.08$ mg/L and $s = 0.15$ mg/L. What is the half-width of the 95 % confidence interval in mg/L?

Key takeaways

Confidence interval μ = x̄ ± t·s/√n with ν = n − 1; the 95 % t value falls from 12.706 (ν = 1) to 1.960 (ν → ∞).
“95 %” is the method’s success rate: in the long run 95 % of intervals contain the true value.
Small samples need t; using 1.96 makes the interval too narrow.
The first few replicates pay most; 3–5 replicates is the usual compromise in routine analysis.

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Data and references

Student [W. S. Gosset], “The probable error of a mean”, Biometrika 6, 1–25 (1908).
D. C. Harris, Quantitative Chemical Analysis, 9th ed., W. H. Freeman (2016), chapter 4 (confidence intervals and Student’s t).
JCGM 100:2008, Evaluation of measurement data — Guide to the expression of uncertainty in measurement (GUM), BIPM.