FS2 6.1 - Testing and estimating a normal variance | Edexcel A-Level Further Mathematics

A process can keep its average while becoming much less consistent. You will test a normal population variance with a chi-squared statistic, find critical regions and construct confidence intervals for variance, keeping standard deviations and squared units distinct.

A sampling distribution for spread

If a sample estimates a variance as $9\,\mathrm{mm^2}$ , why might a confidence interval extend much further above $9$ than below it?

Variance cannot be negative, and its sampling distribution is not symmetric. For the sample of $11$ measurements used later, a $95\%$ interval will be approximately $(4.39,27.7)\,\mathrm{mm^2}$ . Its unequal margins are part of the method, not an arithmetic accident.

You already know

The unbiased sample variance is $s^2=\sum(x_i-\bar x)^2/(n-1)$ . A confidence-interval rule has a repeated-sampling coverage rate; after observing the data, its endpoints are fixed. Neither unbiased estimation nor confidence makes the unknown parameter random.

Write $S^2$ for the random variance estimator and $s^2$ for its observed value. We require independent observations from the same normal population. A large sample alone does not turn the following exact normal-population result into an assumption-free procedure.

Sampling distribution of a normal sample variance

For a random sample $X_1,\ldots,X_n$ from $\mathrm{N}(\mu,\sigma^2)$ , with $n\ge2$ and $\sigma^2>0$ ,

S^2=\frac{\sum(X_i-\bar X)^2}{n-1}, \qquad \frac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}

Here $\chi^2_\nu$ denotes the chi-squared distribution with $\nu$ degrees of freedom. Both the estimator and this sampling distribution are supplied. In the formula booklet.

The $n-1$ degrees of freedom match estimating the population mean from the sample: the deviations from $\bar X$ sum to zero. Use this stated result; no density formula or proof is needed. We are not assuming that the population mean is known.

The booklet's chi-squared table gives a value $c$ for which $P(Q>c)$ equals the column heading. Thus, for $Q\sim\chi^2_{10}$ , the $0.050$ column gives $18.307$ , but the lower $5\%$ cutoff is $3.940$ , from column $0.950$ . The curve is positive and skewed; the lower cutoff is not the negative of the upper one.

Key idea

Under a proposed variance, unusually large sample spread gives a large chi-squared statistic; unusually small spread gives a small positive statistic.

Check the assumptions before treating an observed variance as evidence.

Testing for increased or reduced variation

State hypotheses about the population variance $\sigma^2$ , not the sample estimate $s^2$ . Under $H_0:\sigma^2=\sigma_0^2$ , substitute the null variance in the denominator.

A one-sample variance test statistic

For a random sample of size $n$ from a normal population, under $H_0:\sigma^2=\sigma_0^2$ ,

Q=\frac{(n-1)S^2}{\sigma_0^2}\sim\chi^2_{n-1}

For $H_1:\sigma^2>\sigma_0^2$ , reject in the upper tail. For $H_1:\sigma^2<\sigma_0^2$ , reject in the lower tail. This is the supplied sampling-distribution result evaluated under the null hypothesis. In the formula booklet.

A claim about a standard deviation must first be squared. A stated standard deviation of $2\,\mathrm{mm}$ means the null variance is $4\,\mathrm{mm^2}$ .

Worked example 1: has the variance increased?

Question. A normal production process previously had standard deviation $2\,\mathrm{mm}$ . A random sample of $11$ independent measurements gives unbiased sample variance $9\,\mathrm{mm^2}$ . Test at $5\%$ for increased population variance.

Plan. Form the null-scaled statistic and use an upper-tail cutoff with $10$ degrees of freedom.

\begin{aligned} H_0&:\sigma^2=4,\qquad H_1:\sigma^2>4\\ q&=\frac{10(9)}{2^2}=22.5 &&\text{square the null standard deviation}\\ c&=18.307 &&\chi^2_{10},\text{ upper probability }0.050\\ 22.5&>18.307 &&\text{in the upper critical region} \end{aligned}

Why this step? Increased variance produces larger values of the sample spread and therefore of $Q$ .

Answer. Reject $H_0$ . There is sufficient evidence at $5\%$ that the population variance of the process measurements has increased.

Check. The upper-tail p-value is about $0.0128$ , below $0.05$ , agreeing with the critical-value decision.

Worked example 2 changes only the pre-specified question to whether the population variance has decreased. This is a different investigation, not a change of direction chosen after seeing the sample.

Worked example 2: a lower-tail question

Question. Test for decreased population variance at $5\%$ , retaining the same sample and normal-population assumption.

Plan. Keep the statistic but select the lower tail.

\begin{aligned} H_0&:\sigma^2=4,\qquad H_1:\sigma^2<4\\ q&=\frac{10(9)}4=22.5\\ c&=3.940 &&P(Q>3.940)=0.950\\ 22.5&>3.940 &&\text{outside the lower critical region} \end{aligned}

Why this step? A reduction requires unusually small spread. A large statistic cannot support that directional claim.

Answer. Do not reject $H_0$ . There is insufficient evidence at $5\%$ of reduced population variance.

Check. The lower-tail p-value is about $0.987$ ; this observation is far from the lower rejection tail.

If asked for the critical region in $S^2$ , rearrange the statistic inequality. For example, the upper $5\%$ test above rejects when

\frac{10S^2}{4}>18.307 \quad\Longleftrightarrow\quad S^2>7.3228\,\mathrm{mm^2}

This is a rule for future sample variances under a fixed null value, rather than a confidence interval for $\sigma^2$ .

Watch out

Wrong: divide by the null standard deviation $2$ , or use column $0.050$ for a lower $5\%$ tail. Right: divide by $2^2=4$ ; the lower cutoff uses upper probability $0.950$ .

Change only the observed sample variance from Worked example 2.

Detecting a change in either direction

A two-tailed variance test must allow both unusually small and unusually large spread. Split the significance level equally between the tails, using their own positive chi-squared cutoffs.

At $5\%$ with $10$ degrees of freedom, the lower and upper cutoffs are $3.247$ and $20.483$ . The table headings are $0.975$ and $0.025$ , because they describe the area to the right of each value.

Worked example 3: a two-tailed test from summaries

Question. A random sample of $11$ independent measurements from a normal process gives

\sum x=110,\qquad\sum x^2=1190

where measurements are in millimetres. Test at $5\%$ whether the population variance differs from $4\,\mathrm{mm^2}$ .

Plan. Recover the unbiased sample variance, then compare the null-scaled statistic with both $2.5\%$ tail cutoffs.

\begin{aligned} H_0&:\sigma^2=4,\qquad H_1:\sigma^2\ne4\\ s^2&=\frac{1190-110^2/11}{10}=9 &&\text{divide the centred total by }n-1\\ q&=\frac{10(9)}4=22.5\\ \text{critical region: }&Q<3.247\text{ or }Q>20.483 &&2.5\%\text{ in each tail}\\ 22.5&>20.483 \end{aligned}

Why this step? The alternative permits either direction, so the upper $5\%$ cutoff alone would assign the wrong total significance level.

Answer. Reject $H_0$ . There is sufficient evidence at $5\%$ that the population variance of the process measurements differs from $4\,\mathrm{mm^2}$ .

Check. The same critical region in sample-variance units is $S^2<1.2988$ or $S^2>8.1932$ , in $\mathrm{mm^2}$ . The observed $9$ reaches the upper region.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Core Pure Mathematics

Further Pure Mathematics 1

Further Pure Mathematics 2

Further Statistics 1

Further Statistics 2

Further Mechanics 1

Further Mechanics 2

Decision Mathematics 1

Decision Mathematics 2

FS2 6.1 - Testing and estimating a normal variance

A sampling distribution for spread

Testing for increased or reduced variation

Detecting a change in either direction

Continue this lesson.

Learn faster with Nova