Unbiased estimates of mean and variance | CAIE 9709 Probability & Statistics 2 Notes

Chapters

Probability & Statistics 2

Mathematics9709

Know which sample statistic estimates which parameter

Definition

1 mark

What is meant by unbiased estimator?

Model answer: An estimator T of a parameter θ is unbiased when $E(T)=\theta$ .

Sample statistics estimate parameters

Random sample

12, 15, 11, 14, …

calculate ↓

x̄ varies

s² varies

estimate ↓

μ fixed

σ² fixed

Unbiased means correct on average over repeated random samples.

Statistics vary from sample to sample; population parameters are fixed but unknown.

The sample mean $\bar x$ estimates the population mean $\mu$ .
The corrected sample variance $s^2$ estimates the population variance $\sigma^2$ .
Unbiased does not mean every estimate is correct. It means the estimates are centred on the true parameter over repeated samples.

Use n minus one for the variance estimate

\bar x=\frac{\sum x}{n}

Symbols

$\bar x$ = sample mean (same as x)
$n$ = sample size (none)

s^2=\frac{1}{n-1}\left(\sum x^2-\frac{(\sum x)^2}{n}\right)

Symbols

$s^2$ = unbiased estimate of population variance (squared units)
$\sum x^2$ = sum of squared data values (squared units)

The sample mean has already been estimated from the same data. This removes one degree of freedom, so the variance estimate divides by $n-1$ .

Examiner note

Calculator route: one-variable statistics usually reports both population standard deviation

\sigma_x

and sample standard deviation

s_x

. For an unbiased estimate of population variance, square

s_x

, not

\sigma_x

. Write the formula when method marks are required.

Substitute summary data, then interpret it

Worked example

Current-paper random sample and variance estimate

A sample of 30 students has $\sum x=610$ and $\sum x^2=12405$ .

\bar x=\frac{610}{30}=20.3\ \text{(3 s.f.)}

s^2=\frac{1}{29}\left(12405-\frac{610^2}{30}\right)

Answer

\hat\mu=20.3

\hat\sigma^2=0.0575

The estimated standard deviation is about $\sqrt{0.0575}=0.240$ hours. A small spread is not impossible just because the mean is near 20; the mean gives the centre, while the standard deviation measures spread. (9709/62/M/J/24 Q2)

Your turnIndependent unbiased-estimate check [3 marks]

A random sample of 40 values has $\sum x=510$ and $\sum x^2=6668.25$ . Estimate the population mean and variance.

Show worked answer

\bar x=510/40=12.75

\frac{510^2}{40}=6502.5

s^2=\frac{6668.25-6502.5}{39}

Answer

\hat\mu=12.75

\hat\sigma^2=4.25

Common mistake

Do not divide by n when estimating population variance. Also keep

(\sum x)^2

separate from

\sum x^2

; they are different quantities.

Know which sample statistic estimates which parameter

Definition

1 mark

What is meant by unbiased estimator?

Model answer: An estimator T of a parameter θ is unbiased when $E(T)=\theta$ .

Sample statistics estimate parameters

Random sample

12, 15, 11, 14, …

calculate ↓

x̄ varies

s² varies

estimate ↓

μ fixed

σ² fixed

Unbiased means correct on average over repeated random samples.

Statistics vary from sample to sample; population parameters are fixed but unknown.

The sample mean $\bar x$ estimates the population mean $\mu$ .
The corrected sample variance $s^2$ estimates the population variance $\sigma^2$ .
Unbiased does not mean every estimate is correct. It means the estimates are centred on the true parameter over repeated samples.

Use n minus one for the variance estimate

\bar x=\frac{\sum x}{n}

Symbols

$\bar x$ = sample mean (same as x)
$n$ = sample size (none)

s^2=\frac{1}{n-1}\left(\sum x^2-\frac{(\sum x)^2}{n}\right)

Symbols

$s^2$ = unbiased estimate of population variance (squared units)
$\sum x^2$ = sum of squared data values (squared units)

The sample mean has already been estimated from the same data. This removes one degree of freedom, so the variance estimate divides by $n-1$ .

Examiner note

Calculator route: one-variable statistics usually reports both population standard deviation

\sigma_x

and sample standard deviation

s_x

. For an unbiased estimate of population variance, square

s_x

, not

\sigma_x

. Write the formula when method marks are required.

Substitute summary data, then interpret it

Worked example

Current-paper random sample and variance estimate

A sample of 30 students has $\sum x=610$ and $\sum x^2=12405$ .

\bar x=\frac{610}{30}=20.3\ \text{(3 s.f.)}

s^2=\frac{1}{29}\left(12405-\frac{610^2}{30}\right)

Answer

\hat\mu=20.3

\hat\sigma^2=0.0575

Your turnIndependent unbiased-estimate check [3 marks]

A random sample of 40 values has $\sum x=510$ and $\sum x^2=6668.25$ . Estimate the population mean and variance.

Show worked answer

\bar x=510/40=12.75

\frac{510^2}{40}=6502.5

s^2=\frac{6668.25-6502.5}{39}

Answer

\hat\mu=12.75

\hat\sigma^2=4.25

Common mistake

Do not divide by n when estimating population variance. Also keep

(\sum x)^2

separate from

\sum x^2

; they are different quantities.