4.2 - Confidence intervals for differences of means | Cambridge A-Level Further Mathematics

A confidence interval estimates the size and direction of a population mean difference, rather than only deciding whether there is evidence of a difference. You will choose a pooled t interval, a paired-difference interval or a normal interval from the sampling design and variance information.

Independent means with pooled variance

If a confidence interval for $\mu_X-\mu_Y$ is entirely positive, which population has the larger mean? Writing the order of the difference makes that interpretation immediate.

For independent random samples from normal populations with a common unknown variance, estimate the common variance by pooling within-sample deviations. The standard error is the same one used in the two-sample t-test, with $n_X+n_Y-2$ degrees of freedom.

Pooled t interval for a mean difference

A two-sided $100(1-\alpha)\%$ confidence interval for $\mu_X-\mu_Y$ is

(\bar x-\bar y)\pm t_{n_X+n_Y-2,\,1-\alpha/2} \sqrt{s_p^2\left(\frac1{n_X}+\frac1{n_Y}\right)},

where

s_p^2=\frac{(n_X-1)s_X^2+(n_Y-1)s_Y^2}{n_X+n_Y-2}.

Report the two endpoints. The pooled variance formula and t table are provided in the formula booklet; the confidence-interval formula is not printed there.

This interval assumes normal populations, equal population variances and independent samples. Similar-looking sample variances alone do not establish all three conditions.

Worked example 1: an interval containing zero

Question. Independent normal populations have a common unknown variance. Samples give $n_X=6$ , $\bar x=14$ , $s_X^2=4$ , and $n_Y=8$ , $\bar y=11$ , $s_Y^2=9$ . Find and interpret a 95% confidence interval for $\mu_X-\mu_Y$ , using $t_{12,0.975}=2.179$ .

Plan. Pool the variance, estimate the difference's standard error and place a two-sided margin about the observed difference.

\begin{aligned} s_p^2&=\frac{5(4)+7(9)}{12}=\frac{83}{12} \\ \widehat{\operatorname{SE}}&=\sqrt{\frac{83}{12}\left(\frac16+\frac18\right)} \\ \text{margin}&=2.179\sqrt{\frac{83}{12}\left(\frac16+\frac18\right)}=3.094917\ldots \end{aligned}

The observed difference is $14-11=3$ , giving endpoints $-0.094917\ldots$ and $6.094917\ldots$ .

Answer. The interval is $(-0.0949,6.09)$ , to 3 significant figures. It contains zero and includes both signs, so it does not establish a direction for the population mean difference at this confidence level.

Check. The associated equality-test statistic is $3/\widehat{\operatorname{SE}}=2.112\ldots$ , less than $2.179$ . Its two-sided 5% non-rejection agrees with zero lying inside the 95% interval.

Change only the first sample mean from $14$ to $15$ , retaining both sample spreads and sizes. The interval shifts by one without changing width.

Worked example 2: a positive interval

Question. With all other information as in Worked example 1, let $\bar x=15$ . Find and interpret the 95% interval for $\mu_X-\mu_Y$ .

Plan. Keep the standard error and multiplier, but centre the interval at the new difference.

The difference is now $15-11=4$ , while the margin remains $3.094917\ldots$ . Thus

(4-3.094917\ldots,\ 4+3.094917\ldots) =(0.905082\ldots,7.094917\ldots).

Answer. The interval is $(0.905,7.09)$ , to 3 significant figures. All values are positive, providing evidence at the corresponding two-sided 5% level that $\mu_X>\mu_Y$ .

Check. Both endpoints are exactly one above their previous unrounded values. Reversing the difference to $\mu_Y-\mu_X$ would give the negative interval with endpoints in increasing order.

An interval that includes zero does not prove equal means. It indicates that zero is among the values compatible with the sample at this confidence level under the model.

Paired differences

For paired observations, the parameter being estimated is the population mean of the within-pair differences.

If $D=X-Y$ , then $\mu_D=\mu_X-\mu_Y$ , but its estimated variance must come from the actual differences. Using the two separate sample variances as if the lists were independent loses the within-pair connection. Under independent normally distributed differences, apply the one-sample t confidence interval to the $n$ differences.

Paired-difference interval

For $n$ independent pairs whose differences are normally distributed with unknown variance, a two-sided $100(1-\alpha)\%$ interval for $\mu_D$ is

\bar d\pm t_{n-1,\,1-\alpha/2}\frac{s_d}{\sqrt n}.

Here $s_d^2$ is the unbiased variance estimate of the differences. This interval formula is not printed separately in the formula booklet; the variance formula and t table are supplied.

State the subtraction order before interpreting a positive or negative interval. For a time measurement, “after minus before” below zero means a reduction.

Worked example 3: an after-minus-before interval

Question. Nine independently selected participants have normal differences $d=\text{after}-\text{before}$ , in seconds. Their summaries are $\bar d=-2$ , $s_d^2=9$ . Find a 95% interval for the population mean difference, using $t_{8,0.975}=2.306$ , and interpret its sign.

Plan. Treat the nine differences as one sample, with eight degrees of freedom.

\frac{s_d}{\sqrt n}=\frac3{\sqrt9}=1, \qquad \bar d\pm2.306(1)=-2\pm2.306.

Answer. The interval is $(-4.31,0.306)\,\mathrm{s}$ , to 3 significant figures. It includes zero and both signs, so the sample does not determine a population mean decrease at the corresponding two-sided 5% level.

Check. The observed paired statistic for no change is $-2/1=-2$ , whose magnitude is below $2.306$ . This agrees with zero lying inside the interval.

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4.2 - Confidence intervals for differences of means

Independent means with pooled variance

Paired differences

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