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Estimators, standard error and confidence intervals questions
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.
6 original questions · 22 marks · the estimators, standard error and confidence intervals notes · Further Statistics 2
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Define an unbiased estimator, and name one unbiased estimator of a population variance.
Worked answer
An estimator is unbiased if its expected value equals the parameter it estimates, so that it is right on average over repeated samples. The sample variance with divisor n − 1, s² = Σ(x − sample mean)²/(n − 1), is unbiased for σ²; dividing by n instead gives an estimate that is too small on average. B1 for the definition, B1 for naming s² with divisor n − 1.A sample of 36 from a normal population with σ = 12 has mean 48. Find a 95% confidence interval for the population mean.
Worked answer
Standard error = 12/√36 = 2. The interval is 48 ± 1.96 × 2 = 48 ± 3.92, that is (44.08, 51.92). B1 for the standard error, M1 for 48 ± 1.96 × 2, A1 for the interval.A random sample of 36 observations from a normal population with σ = 12 has mean 48. Find a 99% confidence interval for the population mean, and compare it with the 95% interval (44.08, 51.92).
Worked answer
Standard error = 12/√36 = 2, unchanged, since the confidence level affects only the multiplier.
Using z = 2.5758: 48 ± 2.5758 × 2 = 48 ± 5.15, that is (42.85, 53.15).
It is wider than the 95% interval by about 1.23 units at each end. More confidence is bought with less precision, and only a larger sample improves both at once. B1 for z = 2.5758, M1 for 48 ± z × 2, A1 for the interval, B1 for the comparison. Quoting 2.58 rather than 2.5758 is acceptable, but 1.96 for a 99% interval loses every accuracy mark that follows.A normal population has σ = 12. Find the smallest sample size for which a 95% confidence interval for the mean has total width at most 4.
Worked answer
The total width is 2 × 1.96 × 12/√n, and the requirement is that this is at most 4. Working with the total width rather than the half-width is where most answers go wrong, and it changes the answer by a factor of four.
Rearranging: √n ≥ 2 × 1.96 × 12/4 = 11.76, so n ≥ 138.3.
Since n must be a whole number and the inequality runs upwards, round up to n = 139. Rounding down to 138 gives an interval slightly too wide and loses the final mark. M1 for an inequality in the total width, A1 for √n ≥ 11.76, dM1 for reaching n ≥ 138.3, A1 for 139.
Halving the width again would need four times as many readings, since n appears under a square root.A student says that a 95% confidence interval has a 95% probability of containing the population mean. Explain what is wrong with this, and give the correct interpretation.
Worked answer
The population mean is a fixed number rather than a random one, so once the interval has been calculated it either contains μ or it does not, and no probability remains. What varies from sample to sample is the interval, not μ.
The correct statement is that the method produces intervals of which 95% would contain μ in repeated sampling. The confidence attaches to the procedure, not to the one interval in front of you. B1 for μ being a fixed number, B1 for the interval being what varies, B1 for the repeated-sampling statement.A random sample of 4 observations has mean X̄, and an independent random sample of 6 observations has mean Ȳ. Both come from a population with mean μ and variance σ². The estimator T = λX̄ + (1 − λ)Ȳ is proposed, where λ is a constant. Show that T is unbiased for μ for every λ, find the value of λ that minimises Var(T), and state that minimum variance.
Worked answer
Unbiasedness. E(X̄) = E(Ȳ) = μ, so E(T) = λμ + (1 − λ)μ = μ for every λ. The coefficients were chosen to add to 1, which is exactly what unbiasedness requires, so no value of λ can be ruled out on those grounds. Unbiasedness alone never settles which estimator to use.
Variance. The two sample means are independent, with Var(X̄) = σ²/4 and Var(Ȳ) = σ²/6. Each coefficient squares:
Var(T) = λ²σ²/4 + (1 − λ)²σ²/6.
Minimising. Differentiate with respect to λ and set the result to zero: λσ²/2 − (1 − λ)σ²/3 = 0. The σ² cancels, which is the sign that the answer cannot depend on it.
So 3λ = 2(1 − λ), giving 5λ = 2 and λ = 0.4. The second derivative is positive, so this is a minimum; say so, because it is a mark.
Then Var(T) = 0.16σ²/4 + 0.36σ²/6 = 0.04σ² + 0.06σ² = 0.1σ², that is σ²/10.
B1 for E(T) = μ for every λ, M1 A1 for Var(T) = λ²σ²/4 + (1 − λ)²σ²/6, M1 for differentiating and equating to zero, A1 for λ = 0.4, A1 for 0.1σ².
Note what that means. σ²/10 is the variance of the mean of all ten observations, so the best weighting is the one that treats every original reading equally. Weighting the two sample means equally, at λ = 0.5, gives 5σ²/48 ≈ 0.104σ², which is worse.
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