Mathematics 9709 · Probability & Statistics 2

Unbiased estimates of mean & variance

You see one sample and have to guess the whole population from it. The sample mean is already an honest guess for $\mu$ — but the sample variance comes out too small (a sample huddles tighter than the population it came from), so we inflate it by dividing by $n-1$ instead of $n$ . That single swap is the whole chapter's first idea.

Estimate μ with x̄, σ² with s²

“Unbiased” just means right on average: if you repeated the sampling forever, the estimates would centre on the true value. The mean passes this test as-is, so $\hat\mu=\bar x$ . The variance does not — you “spent” one piece of information working out $\bar x$ first, so you divide by $n-1$ (one fewer) to undo the shrinkage.

$\hat\mu=\bar x=\dfrac{\sum x}{n},\qquad s^2=\dfrac{1}{n-1}\Big(\sum x^2-\dfrac{(\sum x)^2}{n}\Big)$

Estimate μ with x̄ (use as-is); estimate σ² with s² = (1/(n−1))(Σx² − (Σx)²/n). The key point is dividing by n−1, not n.

Worked example

Start simple: variance from summarised data

A conservationist samples 30 larks' nests: $\sum m=66,\ \sum m^2=162$ . Estimate the variance of eggs per nest:

Use the $n-1$ formula straight off the summary: $s^2=\tfrac{1}{29}\big(162-\tfrac{66^2}{30}\big)$ .
$\tfrac{66^2}{30}=145.2$ , so $s^2=\tfrac{162-145.2}{29}=\tfrac{16.8}{29}=0.579$ .

Answer

s^2\approx0.579

If the question had said “these 30 nests are the population”, you would divide by 30 (giving 0.56) — but “estimate / sample” always means $n-1$ .

Grouped frequency: same formula, weighted by f

When the data come as a frequency table, every $x$ becomes $\sum fx$ , every $x^2$ becomes $\sum fx^2$ , and $n=\sum f$ . Nothing else changes.

Worked example

Build up: grouped frequency → mean and variance

The cub counts in $24$ tiger litters: $c=1,2,3,4$ with frequencies $f=2,7,12,3$ . Estimate the mean and variance.

$n=\sum f=24$ ; $\sum fc=2+14+36+12=64$ , so $\bar c=\tfrac{64}{24}=2.67$ .
$\sum fc^2=(2{\cdot}1)+(7{\cdot}4)+(12{\cdot}9)+(3{\cdot}16)=186$ .
$s^2=\tfrac{1}{23}\big(186-\tfrac{64^2}{24}\big)=\tfrac{1}{23}\big(186-170.67\big)=\tfrac{2}{3}$ .

Answer

\bar c\approx2.67,\quad s^2=\tfrac23\approx0.667

Examiner note

Always the

n-1

divisor for an estimate; read the calculator's

\sigma_{n-1}

, and for frequency data use

n=\sum f

, not the number of categories. (9709/62/M/J/23 Q3) (9709/62/O/N/22 Q1)

Common mistake

Dividing by

n

instead of

n-1

(the classic mark-loser); forgetting the

-\tfrac{(\sum x)^2}{n}

term; for frequency data dividing by the number of categories instead of

n=\sum f

; and reading the calculator's

\sigma_n

rather than

\sigma_{n-1}

PreviousSample means of discrete distributions NextHypothesis test for a mean

Unbiased estimates of mean & variance

Estimate μ with x̄, σ² with s²

$\hat\mu=\bar x=\dfrac{\sum x}{n},\qquad s^2=\dfrac{1}{n-1}\Big(\sum x^2-\dfrac{(\sum x)^2}{n}\Big)$

Estimate μ with x̄ (use as-is); estimate σ² with s² = (1/(n−1))(Σx² − (Σx)²/n). The key point is dividing by n−1, not n.

Worked example

Start simple: variance from summarised data

A conservationist samples 30 larks' nests: $\sum m=66,\ \sum m^2=162$ . Estimate the variance of eggs per nest:

Use the $n-1$ formula straight off the summary: $s^2=\tfrac{1}{29}\big(162-\tfrac{66^2}{30}\big)$ .
$\tfrac{66^2}{30}=145.2$ , so $s^2=\tfrac{162-145.2}{29}=\tfrac{16.8}{29}=0.579$ .

Answer

s^2\approx0.579

If the question had said “these 30 nests are the population”, you would divide by 30 (giving 0.56) — but “estimate / sample” always means $n-1$ .

Grouped frequency: same formula, weighted by f

When the data come as a frequency table, every $x$ becomes $\sum fx$ , every $x^2$ becomes $\sum fx^2$ , and $n=\sum f$ . Nothing else changes.

Worked example

Build up: grouped frequency → mean and variance

The cub counts in $24$ tiger litters: $c=1,2,3,4$ with frequencies $f=2,7,12,3$ . Estimate the mean and variance.

$n=\sum f=24$ ; $\sum fc=2+14+36+12=64$ , so $\bar c=\tfrac{64}{24}=2.67$ .
$\sum fc^2=(2{\cdot}1)+(7{\cdot}4)+(12{\cdot}9)+(3{\cdot}16)=186$ .
$s^2=\tfrac{1}{23}\big(186-\tfrac{64^2}{24}\big)=\tfrac{1}{23}\big(186-170.67\big)=\tfrac{2}{3}$ .

Answer

\bar c\approx2.67,\quad s^2=\tfrac23\approx0.667

Examiner note

Always the

n-1

divisor for an estimate; read the calculator's

\sigma_{n-1}

, and for frequency data use

n=\sum f

, not the number of categories. (9709/62/M/J/23 Q3) (9709/62/O/N/22 Q1)

Common mistake

Dividing by

n

instead of

n-1

(the classic mark-loser); forgetting the

-\tfrac{(\sum x)^2}{n}

term; for frequency data dividing by the number of categories instead of

n=\sum f

; and reading the calculator's

\sigma_n

rather than

\sigma_{n-1}