Confidence intervals

Statistics · First-year university statistics · about 10 min

A sample estimate is rarely identical to the population value. A confidence interval adds a margin of error to an estimate, showing a range produced by a repeatable method. The interval width depends on variability, sample size and confidence level. Exam questions often require both the arithmetic and a careful interpretation of what the confidence percentage means.

The key idea

For a population mean when $\sigma$ is known, a confidence interval is

\bar{x}\pm z^*\frac{\sigma}{\sqrt{n}}.

Here $\bar{x}$ is the sample mean, $z^*$ the normal critical value, $\sigma$ the population standard deviation, and $n$ the sample size.

A horizontal number line with a central filled dot at x_bar. Draw a solid horizontal segment from a left endpoint L to a right endpoint U with short vertical end caps. Use two thin double-headed dimension arrows above, one from L to x_bar and the other from x_bar to U, and label both E. The two distances are exactly equal. No numeric endpoints. — A two-sided confidence interval extends a margin of error on both sides of the sample mean.

Try it yourself

Shade the middle of the normal curve to see where the critical values for an interval come from.

Understand standard error

The sample mean changes from sample to sample. Its standard error, when observations are independent, is $SE=\sigma/\sqrt{n}$ . Increasing the sample size reduces the uncertainty in the mean, but the improvement follows a square-root rule: multiplying $n$ by four halves $SE$ . A normal-based method is appropriate when the parent population is normal, or when a representative sample is sufficiently large for a normal approximation. The interval concerns the population mean, not the spread of individual observations. A wide population distribution can still yield a precise estimate of its mean if the sample is large.

Choose a critical value

The confidence level fixes the amount of probability between two critical values of a reference distribution. Common two-sided normal critical values are $z^*\approx1.645$ for $90\%$ , $1.960$ for $95\%$ , and $2.576$ for $99\%$ . Calculate the margin $E=z^*SE$ , then state the lower and upper endpoints, keeping units and sensible precision. If $\sigma$ is unknown and the assumptions support a one-sample t-interval, replace $\sigma$ by the sample standard deviation $s$ and use a t critical value with $n-1$ degrees of freedom. Do not silently treat $s$ as known $\sigma$ .

Interpret repeated coverage

A $95\%$ confidence procedure captures the fixed population mean in about $95\%$ of intervals produced by repeated independent samples under the model assumptions. For one interval already calculated, the population mean does not move randomly between its endpoints. Therefore, the conventional frequentist interpretation is about long-run performance of the method, not a $95\%$ probability assigned to the fixed mean. A higher confidence level produces a wider interval for the same sample. A smaller interval indicates more precision, but neither narrowness nor nominal coverage can remove bias from poor sampling.

Draw 20 thin horizontal confidence-interval segments at twenty equally spaced vertical levels, each with a filled central sample-mean dot and two short end caps. The endpoints vary modestly. A vertical dashed line marks the single fixed true mean mu across the full stack. Construct exactly 19 segments crossing the vertical dashed line and one segment wholly to its right. No axes, percentages, numeric values or extra text. — Most repeated-sample intervals cross the true mean, but occasional intervals miss it.

Worked example: Estimating the average length of rods

A random independent sample of $36$ rods has mean length $500\,\mathrm{mm}$ . The population standard deviation is known to be $12\,\mathrm{mm}$ . Assuming a normal sampling distribution for the mean, construct a two-sided $95\%$ confidence interval for the population mean length.

Show the solution

Identify the inputs: $\bar{x}=500\,\mathrm{mm}$ , $\sigma=12\,\mathrm{mm}$ , $n=36$ , and $z^*=1.96$ .
Standard error:

SE=\frac{12\,\mathrm{mm}}{\sqrt{36}}=2\,\mathrm{mm}.

Margin of error:

E=1.96(2\,\mathrm{mm})=3.92\,\mathrm{mm}.

Endpoints:

500\,\mathrm{mm}\pm3.92\,\mathrm{mm}=(496.08\,\mathrm{mm},\ 503.92\,\mathrm{mm}).

Answer: The $95\%$ confidence interval is $(496.08\,\mathrm{mm},\ 503.92\,\mathrm{mm})$ . This procedure has $95\%$ long-run coverage if its assumptions hold.

Common mistakes

Using $\sigma$ rather than $\sigma/\sqrt{n}$ for the uncertainty in a sample mean.

Calculate the standard error before multiplying by the critical value.

Treating $95\%$ confidence as a probability that a fixed mean is inside this specific interval.

Interpret confidence as repeated-sampling coverage of the interval procedure.

Using $1.96$ automatically when the population standard deviation is unknown.

Check whether the problem calls for an appropriate t-interval.

Reporting an interval without measurement units.

Attach the original variable's units to both endpoints and the margin of error.

Check your understanding

1. For a $90\%$ two-sided z-interval, $\sigma=10\,\mathrm{mm}$ and $n=100$ . Using $z^*=1.645$ , what is the margin of error?

2. A sample mean is $100\,\mathrm{mm}$ , with known $\sigma=6\,\mathrm{mm}$ and $n=9$ . Assume the population is normally distributed. Using $z^*=1.96$ , which $95\%$ confidence interval is correct?

3. Which statement correctly interprets a $95\%$ confidence procedure?

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