MathsFurther Statistics 2 › Estimators, standard error and confidence intervals

Estimators, standard error and confidence intervals

A sample statistic is a random variable with a distribution of its own. Knowing that distribution turns a single estimate into an interval, and gives a realistic account of how much the estimate can be trusted.

Builds on The Central Limit Theorem and The normal distribution.

IN THIS TOPIC

  • Explain what an unbiased estimator is and check whether a given one qualifies.
  • Compute an unbiased estimate of a population variance from summary statistics.
  • Calculate a standard error and say what it measures.
  • Construct and interpret a confidence interval for a normal mean.

COMMON MISCONCEPTION

A 95% confidence interval has a 95% chance of containing the sample mean.

Estimators and bias

An estimator is a rule for turning a sample into a guess at a population parameter. Feed it a different sample and it gives a different answer, so it is itself a random variable. It is unbiased when its expected value equals the parameter it is aiming at. The sample mean is unbiased for the population mean. The sample variance needs the n − 1 divisor to be unbiased for σ²:

s2=1n-1Σ(xi-sample mean)2s^{2} = \frac{1}{n - 1} Σ (x_{i} - \text{sample mean})^{2}IN THE FORMULAE BOOKLET

That statement is in the booklet under Sampling distributions, which also records that the sample mean is unbiased for μ with variance σ²/n. The computational version you actually use on summary statistics is not printed in this form, so learn it.

s2=1n-1(Σx2-(Σx)2n)s^{2} = \frac{1}{n - 1}(Σ x^{2} - \frac{(Σ x)^{2}}{n})NOT IN THE BOOKLET — LEARN IT

The bracket is Sxx, and Sxx = Σx² − (Σx)²/n does appear in the booklet, so if the form deserts you, look up Sxx and divide by n − 1. Dividing by n instead would systematically undershoot, because the squared gaps are measured from the sample mean and the sample mean sits closer to its own data than μ does. One warning that catches people out: s² is unbiased for σ², but s is not unbiased for σ. Taking a square root does not preserve unbiasedness.

WORKED EXAMPLE

An unbiased estimate from summary statistics

A sample of 10 gives Σx = 240 and Σx² = 5850. Find unbiased estimates of the population mean and variance.

The sample mean is 240/10 = 24, and it is unbiased for μ as it stands.

Sxx = 5850 − 240²/10 = 5850 − 5760 = 90.

s² = 90/9 = 10. Dividing by 10 instead of 9 would have given 9, which is the bias the n − 1 divisor removes.

Choosing between unbiased estimators

Unbiasedness on its own is a weak recommendation. Among unbiased estimators, prefer the one with the smaller variance, because it lands near the target more often. That property is called efficiency. The standard error of the sample mean, σ/√n, is exactly its variance expressed as a standard deviation, and it shrinks like the square root of the sample size, so quadrupling the data halves the uncertainty.

The sampling distribution of the mean narrowing as the sample grows: standard error 10, 5 and 2n = 1: se 10n = 4: se 5n = 25: se 2population meanse = σ / √n: four times the data, half the spread
FIG. 1The sampling distribution of the mean narrowing as the sample grows, with standard errors of 10, 5 and 2.

WORKED EXAMPLE

Comparing two estimators

A population has mean μ. Two estimators are proposed from a sample of 3: the mean of all three values, and the first value alone. Compare them.

Both are unbiased, since E(sample mean) = μ and E(X1) = μ.

Their variances are σ²/3 and σ². The sample mean has a third of the variance, so its standard error is smaller by a factor of √3.

Both are unbiased; only one is efficient. Being unbiased does not by itself make an estimator worth using.

Turning an estimate into an interval

Suppose the population variance is known and either the population is normal or the sample is large enough for the Central Limit Theorem. Then the sample mean is normal with standard error σ/√n, so a fixed proportion of samples land within a fixed number of standard errors of μ. Reversing that statement gives a confidence interval:

sample mean±z×σn,z=1.96 at 95 per cent\text{sample mean} ± z \times \frac{σ}{\sqrt{n}}, \qquad z = 1.96 \text{ at 95 per cent}NOT IN THE BOOKLET — LEARN IT

No confidence interval appears anywhere in the booklet. It prints the standardised statistic and tabulates the percentage points, and you build the interval from those, so learn its shape. The interpretation needs care. The population mean is a fixed number, not a random one, so it either is or is not inside your interval and no probability attaches to that. What is random is the interval itself. Build one this way from each of many repeated samples and 95% of them would contain μ. And the quantity captured is μ, not the sample mean, which sits at the centre of its own interval every time.

A 95% confidence interval for a mean: the sample mean 104, plus and minus 1.96 standard errors100104108100.08107.92sample mean 10495% of such intervals contain the population mean
FIG. 2A 95% interval about a sample mean of 104, running from 100.08 to 107.92.

GUIDED PRACTICE

Building the interval

A sample of 25 from a normal population with σ = 10 has mean 104. Find 95% and 99% confidence intervals for μ, and comment.

Show the working

The standard error is 10/√25 = 2.

95%: 104 ± 1.96 × 2 = 104 ± 3.92, giving (100.08, 107.92).

99%: 104 ± 2.5758 × 2 = 104 ± 5.15, giving (98.85, 109.15).

More confidence costs width. Since 100 lies inside both, a claim that μ = 100 would survive a two-tailed test at either level.

ASSESSMENT FOCUS

  • Calculate the standard error on a line of its own. It earns a mark and it prevents slips later.
  • Use the n − 1 divisor whenever the question says 'unbiased estimate'. A calculator's population standard deviation key gives the wrong one.
  • Quote the z value you used, and use 2.5758 instead of 2.58 when the question asks for accuracy.
  • Interpret the interval as a statement about the procedure, never as the probability that μ lies inside.
  • Link the interval to a test: a value outside the interval is rejected by the matching two-tailed test.

CHECK YOURSELF

A sample of 100 from a population with σ = 20 has mean 55. Find a 95% confidence interval for μ.

Show a hint

Standard error first, then 1.96 of them either side.

Show the answer

Standard error 20/√100 = 2. The interval is 55 ± 1.96 × 2 = 55 ± 3.92, that is (51.08, 58.92).

An estimator is unbiased when its expected value is the parameter, and among unbiased ones you prefer the smaller variance.

The standard error σ/√n sets the width: a 95% interval is the sample mean ± 1.96 standard errors, and 95% of such intervals capture μ.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

6 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the estimators, standard error and confidence intervals questions page.

CHECK YOUR PROGRESS

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  • Explain what an unbiased estimator is and check whether a given one qualifies.
  • Compute an unbiased estimate of a population variance from summary statistics.
  • Calculate a standard error and say what it measures.
  • Construct and interpret a confidence interval for a normal mean.

Open the full revision checklist to see every objective in the course in one place.