4.2 - Confidence intervals for one mean | Cambridge A-Level Further Mathematics

A sample estimate is more useful when accompanied by a measure of its uncertainty. You will use a t-distribution to construct a confidence interval for the mean of a normal population with unknown variance, then interpret the confidence level through repeated sampling.

A two-sided t multiplier

What must a 95% confidence interval do in repeated samples: contain 95% of the observations, or capture the fixed population mean about 95% of the time? It is the second idea, and it begins with a central probability for the t statistic.

For an independent normal sample of size $n$ and unknown variance,

\frac{\bar X-\mu}{S/\sqrt n}\sim t_{n-1}.

To leave central probability $0.95$ , put probability $0.025$ in each tail. The positive endpoint $c$ then satisfies $P(T\le c)=0.975$ . This is the cumulative column heading used in the provided t table.

Confidence level and multiplier

For a two-sided confidence level $100(1-\alpha)\%$ , choose

c=t_{n-1,\,1-\alpha/2}.

Then $P(-c\le T\le c)=1-\alpha$ . The t table is in the formula booklet; selecting the row and cumulative column is part of the method.

Common cumulative columns are $0.95$ for 90% confidence, $0.975$ for 95%, and $0.995$ for 99%. The degrees of freedom are $n-1$ , not $n$ .

Worked example 1: a 95% multiplier

Question. A random sample of 8 observations comes from a normal population with unknown variance. Find the multiplier for a two-sided 95% confidence interval. The relevant table entries are $t_{7,0.95}=1.895$ and $t_{7,0.975}=2.365$ .

Plan. Find the degrees of freedom, halve the excluded probability and select the cumulative column.

There are $8-1=7$ degrees of freedom. The excluded probability is $0.05$ , with $0.025$ in each tail. Therefore the positive cumulative probability is $1-0.025=0.975$ .

Answer. Use $c=2.365$ .

Check. By symmetry, $P(T<-2.365)=0.025$ and $P(T>2.365)=0.025$ . The central probability is $1-0.05=0.95$ .

Keep the sample size but increase only the requested confidence level. A larger central probability needs a larger multiplier.

Worked example 2: a 99% multiplier

Question. For the same sample size, choose the multiplier for a two-sided 99% confidence interval. Use $t_{7,0.99}=2.998$ and $t_{7,0.995}=3.499$ .

Plan. Divide the total excluded probability $0.01$ equally between two tails.

The degrees of freedom remain $7$ . Each tail has probability $0.005$ , so the positive endpoint uses cumulative probability $0.995$ .

Answer. Use $c=3.499$ .

Check. The tail probabilities add to $0.01$ , leaving $0.99$ centrally. Also $3.499>2.365$ , as required for the higher confidence level.

The column headed $0.99$ would leave $0.01$ in each tail, giving a central probability of $0.98$ , not $0.99$ .

A small-sample interval

A confidence interval rearranges the central t probability so that the unknown mean lies between two quantities calculated from the sample.

With $S>0$ , start from

-c\le\frac{\bar X-\mu}{S/\sqrt n}\le c.

Multiplying by the positive standard error and collecting $\mu$ gives

\bar X-c\frac S{\sqrt n}\le\mu\le \bar X+c\frac S{\sqrt n}.

Before sampling, both endpoints are random. After observing the data, they become the numerical interval you report.

t confidence interval for one mean

For an independent random sample from a normal population with unknown variance, a two-sided $100(1-\alpha)\%$ confidence interval for $\mu$ is

\left(\bar x-t_{n-1,\,1-\alpha/2}\frac{s}{\sqrt n},\quad \bar x+t_{n-1,\,1-\alpha/2}\frac{s}{\sqrt n}\right),

where $s^2$ is the unbiased variance estimate. This interval formula is not printed in the formula booklet; the variance formula and t tables are supplied.

The quantity $cs/\sqrt n$ is the margin of error, or half-width. Keep extra digits in it and round the final interval sensibly. Give both endpoints, rather than leaving the result only as a centre plus or minus a margin.

Worked example 3: a 95% interval

Question. Ten independent observations from a normal population give $\bar x=38\,\mathrm{mg}$ and $s^2=36\,\mathrm{mg}^2$ . Find a 95% confidence interval for the population mean, using $t_{9,0.975}=2.262$ .

Plan. Take the square root of the variance estimate, then multiply the estimated standard error by the t value.

s=6,\qquad \frac{s}{\sqrt n}=\frac6{\sqrt{10}}.

\text{margin}=2.262\frac6{\sqrt{10}}=4.291843\ldots

The endpoints are $38-4.291843\ldots$ and $38+4.291843\ldots$ .

Answer. A 95% confidence interval is $(33.7,42.3)\,\mathrm{mg}$ , to 3 significant figures.

Check. The unrounded endpoints have midpoint $38$ . Standardising either endpoint's distance from the sample mean by $6/\sqrt{10}$ gives $2.262$ , confirming the margin calculation.

Change only the confidence level to 90%. The centre and standard error remain unchanged, so the smaller multiplier must give a narrower interval.

Worked example 4: reducing the confidence level

Question. Find a 90% confidence interval for the same population mean from the same sample, using $t_{9,0.95}=1.833$ .

Plan. Replace only the t multiplier in the margin.

\text{margin}=1.833\frac6{\sqrt{10}}=3.477872\ldots

(38-3.477872\ldots,\ 38+3.477872\ldots) =(34.522127\ldots,41.477872\ldots).

Answer. A 90% confidence interval is $(34.5,41.5)\,\mathrm{mg}$ , to 3 significant figures.

Check. It has the same midpoint and sits inside the 95% interval. Its half-width ratio is $1.833/2.262$ , exactly the ratio of the chosen multipliers.

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4.2 - Confidence intervals for one mean

A two-sided t multiplier

A small-sample interval

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