Further Statistics — Hypothesis testing: A-level Further Maths OCR revision notes

5.05

Unbiased estimates and the central limit theorem

Unbiased estimates and the distribution of the sample mean 5.05

Definitions

Unbiased estimator: one whose expectation equals the parameter it estimates.
Sample mean $\overset{x}{ˉ} = \frac{\sum x}{n}$ , an unbiased estimate of $μ$ .
Unbiased estimate of $σ^{2}$ : $s^{2} = \frac{1}{n - 1} (\sum x^{2} - n \overset{x}{ˉ}^{2})$ .

Key results

For any population with mean $μ$ and variance $σ^{2}$ : $E (\overset{ˉ}{X}) = μ$ and $Var (\overset{ˉ}{X}) = \frac{σ ^{2}}{n}$ .
If $X \sim N (μ, σ^{2})$ , then $\overset{ˉ}{X} \sim N (μ, \frac{σ ^{2}}{n})$ exactly.
Central limit theorem: for any population, $\overset{ˉ}{X} \approx N (μ, \frac{σ ^{2}}{n})$ when $n$ is large.

Method

Standardise with the standard error: $Z = \frac{X ˉ - μ}{σ / n}$ .
If $σ$ is unknown and $n$ is large, use $s$ in its place.

In practiceA sample of 10 gives $\sum x = 52$ and $\sum x^{2} = 290$ . Find unbiased estimates of $μ$ and $σ^{2}$ , and the standard error of the mean.

$\overset{x}{ˉ} = \frac{52}{10} = 5.2$
$s^{2} = \frac{1}{9} (290 - 10 \times 5. 2^{2}) = \frac{19.6}{9} = 2.18$ divide by $n - 1 = 9$ , not 10
$SE = \frac{s}{10} = \frac{1.476}{3.162} = 0.467$

Notes

Divide by $n - 1$ , not $n$ , for an unbiased variance estimate: deviations from $\overset{x}{ˉ}$ are on average smaller than deviations from $μ$ .
The theorem is about the distribution of $\overset{ˉ}{X}$ , not of $X$ . The population itself can stay as skewed as it likes.
Say when the CLT is being used: 'since $n$ is large, $\overset{ˉ}{X}$ is approximately normal by the central limit theorem'.
How large is large: $n \geq 30$ is a common working rule for the central limit theorem, but the more skewed the population, the larger $n$ needs to be. For a normal population no condition on $n$ is needed at all.

Worked examples

Worked example

A sample of 10 values has $\sum x = 52$ and $\sum x^{2} = 290$ .

Find unbiased estimates of $μ$ and $σ^{2}$ .

Show worked solution

$\overset{x}{ˉ} = 5.2$ .

s^{2} = \frac{290 - 10 ( 5.2 ) ^{2}}{9} = \frac{19.6}{9} = 2.18

Worked example

$X$ has mean 4 and variance 9 but an unknown distribution.

Find approximately the probability that the mean of a random sample of 50 exceeds 4.5.

Show worked solution

By the CLT, since $n$ is large, $\overset{ˉ}{X} \approx N (4, \frac{9}{50})$ .

P (\overset{ˉ}{X} > 4.5) = P (Z > \frac{0.5}{3/ 50}) = P (Z > 1.179) = 0.119

5.05

Hypothesis tests for a population mean

Testing a population mean 5.05

Key results

If $X \sim N (μ, σ^{2})$ , or $n$ is large (central limit theorem), then under $H_{0} : μ = μ_{0}$ the test statistic $Z = \frac{X ˉ - μ _{0}}{σ / n}$ is (approximately) $N (0, 1)$ .
One-tailed tests at 5% and 1% use 1.645 and 2.326; two-tailed tests use 1.960 and 2.576.
If $σ$ is unknown and $n$ is large, the unbiased estimate $s$ is used in its place.

Method

State $H_{0}$ and $H_{1}$ in terms of the population mean $μ$ , with the significance level and whether the test is one- or two-tailed.
Calculate $z$ (or the $p$ -value, or the critical region for $\overset{x}{ˉ}$ ) and compare.
Conclude in context, without claiming certainty: 'there is sufficient evidence at the 5% level that…'.

In practiceBags are claimed to have mean mass 500 g, with $σ = 8$ g. A sample of 40 has mean 497.2 g. Test at the 5% level whether the mean is less than 500 g.

$H_{0} : μ = 500, H_{1} : μ < 500$ one-tailed, 5%
$z = \frac{497.2 - 500}{8/ 40} = \frac{- 2.8}{1.265} = - 2.21$ $\overset{ˉ}{X}$ approximately normal by the CLT, $n = 40$
$- 2.21 < - 1.645$ in the critical region
Reject $H_{0}$ : there is sufficient evidence at the 5% level that the mean mass is less than 500 g.

Notes

Say why the distribution of $\overset{ˉ}{X}$ is normal: either the population is normal, or $n$ is large enough for the central limit theorem.
A two-tailed test at the 5% level rejects $μ_{0}$ exactly when $μ_{0}$ lies outside the 95% confidence interval.

Worked example

The mass of cereal in a box has standard deviation 8 g.

A random sample of 64 boxes has mean 52.1 g.

Test at the 5% level whether the mean mass exceeds 50 g.

Show worked solution

$H_{0} : μ = 50$ , $H_{1} : μ > 50$ .

Since $n = 64$ is large,

\overset{ˉ}{X} \approx N (50, \frac{8 ^{2}}{64}) = N (50, 1)

under $H_{0}$ by the central limit theorem.

z = \frac{52.1 - 50}{1} = 2.1 > 1.645

(equivalently:

p = 0.018 < 0.05

Reject $H_{0}$ : there is evidence at the 5% level that the mean mass exceeds 50 g.

5.05

Confidence intervals

Confidence intervals for a mean 5.05

Definitions

A 95% confidence interval: an interval constructed by a method which, over repeated samples, contains the true parameter 95% of the time.

Key results

Normal population with known $σ$ , or large $n$ : $\overset{x}{ˉ} \pm z \frac{σ}{n}$ , with $σ$ replaced by $s$ when it is unknown and $n$ is large.
$z = 1.645$ , $1.960$ and $2.576$ for 90%, 95% and 99% intervals.
The width is $2 z σ / n$ : four times the sample size halves the width.

Method

Compute $\overset{x}{ˉ}$ and the standard error, then $\overset{x}{ˉ} \pm z \times SE$ .
To judge a claim $μ = μ_{0}$ : if $μ_{0}$ lies outside a 95% interval, a two-tailed test at the 5% level would reject it.

In practiceFind a 95% confidence interval for $μ$ , given $n = 50$ , $\overset{x}{ˉ} = 12.4$ and $s = 2.1$ .

$SE = \frac{2.1}{50} = 0.297$ $s$ in place of $σ$ : $n$ is large
$12.4 \pm 1.96 \times 0.297 = 12.4 \pm 0.582$
$(11.82, 12.98)$ to 2 decimal places

Notes

Never say 'there is a 95% probability that $μ$ lies in this interval'. $μ$ is fixed; the probability belongs to the method that produced the interval.
Higher confidence gives a wider interval. More data gives a narrower one.

Worked example

A random sample of 80 bags has mean mass 32.4 g and standard deviation 5.1 g.

Find a 95% confidence interval for the population mean, and comment on a claim that $μ = 33$ g.

Show worked solution

$n$ is large, so use $s$ : SE:

= \frac{5.1}{80} = 0.570

The interval is:

32.4 \pm 1.96 (0.570) = 32.4 \pm 1.12

that is:

(31.28, 33.52)

33 lies inside the interval, so the data do not contradict the claim at the 5% level.

5.06

χ² goodness-of-fit tests

The χ² statistic and its distribution 5.06

Definitions

Test statistic: $X^{2} = \sum \frac{( O _{i} - E _{i} ) ^{2}}{E _{i}}$ , where $O_{i}$ are observed and $E_{i}$ expected frequencies under $H_{0}$ .
Degrees of freedom $ν$ : the number of cells minus the number of independent constraints on the expected frequencies.

Key results

If every $E_{i} \geq 5$ , $X^{2}$ is approximately $χ_{ν}^{2}$ under $H_{0}$ .
Large $X^{2}$ means observed and expected frequencies disagree, so the critical region is always the upper tail.
$χ_{ν}^{2}$ has mean $ν$ and is skewed to the right, less so as $ν$ grows.

Method

Find each $E_{i}$ , combine adjacent cells if any $E_{i} < 5$ , then count the cells that remain.
Calculate $X^{2}$ , find $ν$ , and compare with the critical value from tables or the calculator.
Conclude in context: 'there is (or is not) sufficient evidence that…'.

In practiceObserved frequencies 18, 30, 52 are compared with expected 25, 25, 50 from a fully specified model. Test at the 5% level.

$X^{2} = \frac{( 18 - 25 ) ^{2}}{25} + \frac{( 30 - 25 ) ^{2}}{25} + \frac{( 52 - 50 ) ^{2}}{50}$ every $E_{i} \geq 5$
$= 1.96 + 1 + 0.08 = 3.04$
$ν = 3 - 1 = 2, χ_{2}^{2} (5%) = 5.991$ one constraint: the totals agree
$3.04 < 5.991$ do not reject $H_{0}$ : the data are consistent with the model

Notes

Use frequencies, never proportions or percentages: $X^{2}$ scales with the sample size.
Not rejecting $H_{0}$ does not prove the model is right. It only shows the data are consistent with it.

Goodness-of-fit tests 5.06

Key results

$ν = (number of cells after combining) - 1 - (number of parameters estimated from the data)$ .
Models tested include the discrete uniform, binomial, Poisson and geometric distributions, and distributions given by a table of probabilities.

Method

State $H_{0}$ : the model fits, with any given parameter values. State $H_{1}$ : it does not.
If a parameter is not given, estimate it from the data (the sample mean for a Poisson $λ$ ) and subtract an extra degree of freedom.
Expected frequency = total × model probability. The last cell is usually 'this value or more', so the probabilities sum to 1.

In practice100 counts are grouped as $0, 1, 2, \geq 3$ with observed frequencies 25, 40, 20, 15 and sample mean 1.2. Test a Poisson model at the 5% level.

$Po (1.2) : p_{0} = 0.301, p_{1} = 0.361, p_{2} = 0.217, p_{\geq 3} = 0.121$ $λ$ estimated by the sample mean
$E_{i} = 100 p_{i} = 30.12, 36.14, 21.69, 12.05$ all at least 5, so no merging
$X^{2} = \frac{( 25 - 30.12 ) ^{2}}{30.12} + \frac{( 40 - 36.14 ) ^{2}}{36.14} + \frac{( 20 - 21.69 ) ^{2}}{21.69} + \frac{( 15 - 12.05 ) ^{2}}{12.05} = 2.13$
$ν = 4 - 1 - 1 = 2, 2.13 < 5.991$ one degree lost for the total, one for estimating $λ$
Do not reject $H_{0}$ : a Poisson model is consistent with the data.

Notes

Combining cells is done before counting $ν$ , and it reduces $ν$ .
Check the subtraction for estimated parameters. It is the most common error in this topic.

Worked example

Goals in 100 matches: 0 goals 15 times, 1 goal 30, 2 goals 28, 3 goals 15, 4 or more 12.

The sample mean is 1.8.

Test at the 5% level whether a Poisson model fits.

Show worked solution

$H_{0}$ : goals follow a Poisson distribution.

Using Po(1.8), the probabilities are 0.1653, 0.2975, 0.2678, 0.1607 and 0.1087 (by subtraction), so:

E = 16.53, 29.75, 26.78, 16.07, 10.87

all at least 5.

$X^{2} = 0.388$ .

One parameter was estimated, so $ν = 5 - 1 - 1 = 3$ , with critical value 7.815.

0.388 < 7.815

do not reject $H_{0}$ .

The Poisson model is consistent with the data.

5.06

χ² tests for association

Contingency tables: tests for association 5.06

Definitions

Contingency table: frequencies classified by two factors, with $r$ rows and $c$ columns.

Key results

Under $H_{0}$ (no association), $E = \frac{row total \times column total}{grand total}$ .
$ν = (r - 1) (c - 1)$ , counted after any rows or columns are merged.

Method

State $H_{0}$ : there is no association between the two factors. State $H_{1}$ : there is an association.
Tabulate the contributions $\frac{( O - E ) ^{2}}{E}$ ; after a significant result, the largest contributions show where the association lies, and whether $O$ is above or below $E$ there.

In practiceIn a $3 \times 4$ table, a cell has row total 60, column total 45 and observed frequency 20; the grand total is 200.

$E = \frac{60 \times 45}{200} = 13.5$ row total × column total ÷ grand total
$\frac{( O - E ) ^{2}}{E} = \frac{( 20 - 13.5 ) ^{2}}{13.5} = 3.13$ this cell's contribution to $X^{2}$
$ν = (3 - 1) (4 - 1) = 6, χ_{6}^{2} (5%) = 12.59$ before any merging
If the total $X^{2}$ exceeds 12.59 there is evidence of association, and a large contribution with $O > E$ , like this one, shows where it lies.

Notes

Association is not causation: a significant result says the factors are related in this population, not why.
Merging must make sense in context: combine neighbouring age bands, not unrelated categories.

Worked example

120 students are classified by whether they play an instrument and by grade.

Yes: A 25, B 15, C 10.

No: A 15, B 25, C 30.

Test at the 1% level for association.

Show worked solution

Column totals are 40 each and row totals 50 and 70, so $E = 16.67$ in the 'Yes' row and $23.33$ in the 'No' row.

$X^{2} = 12.0$ with:

ν = (2 - 1) (3 - 1) = 2

The 1% critical value is 9.210 and:

12.0 > 9.210

so reject $H_{0}$ : there is evidence of association.

The largest contributions show that players gain more A grades and fewer C grades than expected.