SH2–SH4 - Confidence intervals with estimated variance | AQA A-Level Further Mathematics

Construct confidence intervals for a normal population mean when its variance must be estimated. You will distinguish the large-sample normal approximation from a small-sample t interval, choose degrees of freedom and interpret changes in interval width.

Large-sample normal intervals

If a sample quadruples in size, does the uncertainty in its mean become four times smaller? The square root in the standard error gives a different answer. First, decide how to handle an unknown population standard deviation.

When $\sigma$ is unknown, use the unbiased sample variance estimate $s^2=\sum(x_i-\bar x)^2/(n-1)$ , and estimated standard error $s/\sqrt n$ . Do not relabel $s$ as a known population standard deviation. For independent observations from a normal population, the exact standardised mean has a $t_{n-1}$ distribution. With a large sample, $s$ is more stable and the t distribution is close to standard normal, so a z interval is a useful approximation.

Large-sample interval with estimated variance

For a large independent random sample from a normal population with unknown positive variance, the approximate $100(1-\alpha)\%$ confidence interval for $\mu$ is

\bar x\pm z_{1-\alpha/2}\frac{s}{\sqrt n}

Here $s^2$ uses divisor $n-1$ , and $\Phi(z_{1-\alpha/2})=1-\alpha/2$ .

This approximate interval is not printed separately in the formula booklet. The unbiased variance estimator and normal and t sampling pivots are supplied.

The normal-population assumption remains part of the setting here. Large $n$ justifies using a normal critical value instead of a nearby t critical value; it does not turn an estimated standard deviation into a known one. The exact t interval remains valid for a normal population at large $n$ too. There is no mathematical jump at $n=30$ , and no sample-size rule can repair dependent observations or biased sampling. Applying a normal approximation to means from a non-normal population would need an additional adequacy argument; a large number alone is not that argument.

Worked example 1: using an estimated standard error

Question. A random sample of 100 independent normal scores gives $\bar x=70$ and sample standard deviation $s=10$ , calculated with divisor $n-1$ . Construct an approximate 95% confidence interval for the population mean.

Plan. Use the large-sample normal critical value with the estimated standard error.

\widehat{\operatorname{SE}}=\frac{10}{\sqrt{100}}=1,\qquad70\pm1.960(1)

Answer. The approximate 95% interval is $(68.04,71.96)$ .

Check. The midpoint is 70 and the half-width is 1.960 estimated standard errors. An exact normal-population t interval would be slightly wider, since $t_{99}(0.975)\approx1.984$ .

Keep the observed mean, sample standard deviation and confidence level fixed, but quadruple the sample size. This comparison isolates the square-root effect.

Worked example 2: quadrupling sample size

Question. A sample of 400 independent normal scores instead gives the same $\bar x=70$ and $s=10$ . Construct the approximate 95% interval.

Plan. Recalculate the standard error, retaining the same z value.

\widehat{\operatorname{SE}}=10/\sqrt{400}=0.5,\qquad70\pm1.960(0.5)=70\pm0.98

Answer. The interval is $(69.02,70.98)$ .

Check. Width falls from 3.92 to 1.96, exactly half. Algebraically $1/\sqrt{4n}=1/(2\sqrt n)$ , not $1/(4\sqrt n)$ .

Watch out

Wrong: use $s$ itself as the standard error, or call the normal interval exact just because the sample is large. Correct: use $s/\sqrt n$ and identify the z interval as an approximation when variance is estimated.

Small-sample t intervals

For a small independent random sample from a normal population with unknown variance, allow explicitly for the uncertainty in $s$ by using the t distribution. A small sample does not make the mean “not normal”: a normal population gives a normal sample mean. It is estimating the denominator that changes the standardised statistic to t.

Starting from the supplied pivot,

\mathrm{P}\left(-c\le\frac{\bar X-\mu}{S/\sqrt n}\le c\right)=1-\alpha

and rearranging gives the t interval. The positive critical value has cumulative probability $1-\alpha/2$ , because the interval is symmetric.

Confidence interval using t

For $n\ge2$ independent random observations from a normal population with unknown positive variance, and a positive observed sample standard deviation $s$ , an exact $100(1-\alpha)\%$ confidence interval for $\mu$ is

\bar x\pm t_{n-1}(1-\alpha/2)\frac{s}{\sqrt n}

Here $t_\nu(p)$ means the $p$ -quantile of the t distribution with $\nu$ degrees of freedom.

The pivot and t percentage points are in the formula booklet. This rearranged interval formula is not printed separately.

Worked example 3: a 95% small-sample interval

Question. A random sample of 10 independent normal observations has $\bar x=50$ and $s=6$ . Construct a 95% interval for the population mean.

Plan. Use 9 degrees of freedom and cumulative probability 0.975.

\begin{aligned} c&=t_9(0.975)=2.262\\ \widehat{\operatorname{SE}}&=6/\sqrt{10}=1.89737\ldots\\ \text{interval}&=50\pm2.262\frac6{\sqrt{10}} \end{aligned}

Answer. Using the booklet's rounded critical value gives approximately $(45.71,54.29)$ .

Check. The half-width is $4.29184\ldots$ ; dividing it by the estimated standard error returns 2.262. Using 1.960 would give a narrower interval that omits the finite-sample t adjustment.

Keep the same sample but reduce confidence to 90%. Only the tail allocation and critical value change.

Worked example 4: a 90% interval from the same sample

Question. Construct a 90% interval from the same $n=10$ , $\bar x=50$ , $s=6$ .

Plan. Put 5% in each tail and use $t_9(0.95)=1.833$ .

50\pm1.833\frac6{\sqrt{10}}=50\pm3.47787\ldots

Answer. The 90% interval is approximately $(46.52,53.48)$ .

Check. It is inside the 95% interval, with the same midpoint. Its half-width ratio is $1.833/2.262<1$ , as expected when confidence decreases.

Watch out

Wrong: use the t-table 0.90 column for a symmetric 90% interval. Correct: cumulative probability $(1+0.90)/2=0.95$ . Use $n-1$ , not $n$ , as the row.

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SH2–SH4 - Confidence intervals with estimated variance

Large-sample normal intervals

Small-sample t intervals

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