Questions on estimators, standard error and confidence intervals

Question 1

Define an unbiased estimator, and name one unbiased estimator of a population variance.

Accepted 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.

Question 2

A sample of 36 from a normal population with σ = 12 has mean 48. Find a 95% confidence interval for the population mean.

Accepted 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.

Question 3

Explain what the standard error of the mean measures, and state the effect on it of multiplying the sample size by 9.

Accepted Answer

It is the standard deviation of the sample mean itself: a measure of how far the mean of one sample typically falls from the population mean, over repeated samples of that size. B1. Multiplying n by 9 divides σ/√n by √9 = 3, so the standard error becomes a third of what it was. B1.
It is not the spread of the readings, which is σ and does not change with the sample size at all. The two are the same only when n = 1.

Question 4

A random sample of 25 observations is taken from a population with standard deviation 8. Find the standard error of the sample mean, and find the sample size that would reduce that standard error to 0.8.

Accepted Answer

Standard error = σ/√n = 8/√25 = 8/5 = 1.6. B1. For a standard error of 0.8, solve 8/√n = 0.8, so √n = 10 and n = 100. M1 for the equation, A1 for 100. Halving the standard error took four times the readings, because n sits under the square root: dividing it by k costs a sample k² times as large.

Question 5

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).

Accepted 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; 1.96 is the 95% multiplier and gives a 95% interval, not a 99% one.

Question 6

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.

Accepted Answer

The total width is 2 × 1.96 × 12/√n, and the requirement is that this is at most 4. The multiplier applies to the half-width, so using the total width changes the required n 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. n = 138 would give an interval slightly wider than the stated maximum. 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.

Question 7

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.

Accepted 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.

Question 8

A random sample of 9 observations gives Σx = 180 and Σx² = 3720. Calculate unbiased estimates of the population mean and the population variance.

Accepted Answer

The sample mean is 180/9 = 20, and it is unbiased for μ exactly as it stands. B1. For the variance, build S_xx from the two totals first: S_xx = Σx² − (Σx)²/n = 3720 − 180²/9 = 3720 − 3600 = 120. M1 for the formula, A1 for 120. The unbiased estimate divides by n − 1, so s² = 120/8 = 15. A1. Dividing by n gives 13.3, an estimate that undershoots on average because the squared gaps are measured from the sample's own mean, which sits closer to its data than μ does. And s² being unbiased for σ² does not make √15 unbiased for σ, because the square root does not preserve unbiasedness.

Question 9

A random sample of 12 observations gives Σx = 342 and Σx² = 9846. Calculate the estimated standard error of the sample mean, and state what that standard error measures.

Accepted Answer

The population standard deviation is unknown, so estimate it from the sample. S_xx = Σx² − (Σx)²/n = 9846 − 342²/12 = 9846 − 9747 = 99. M1 for the formula.
s² = 99/11 = 9, so s = 3. A1.
Estimated standard error = s/√n = 3/√12 = 0.866. A1.
It measures how far the mean of a sample of 12 typically falls from the population mean, over repeated samples of that size, and it is what the width of any confidence interval for μ is built from. B1 for saying what it measures. It is not the spread of the twelve readings, which is s = 3, more than three times as large. Dividing S_xx by 12 rather than 11 gives s² = 8.25, an estimate that undershoots on average.

Question 10

A random sample of 100 light bulbs has mean lifetime 1180 hours with s = 90 hours. Find a 95% confidence interval for the mean lifetime of all bulbs of this type, and state the result that permits a normal multiplier here.

Accepted Answer

The population standard deviation is unknown, so use s in its place: standard error = 90/√100 = 9. B1.
The interval is 1180 ± 1.96 × 9 = 1180 ± 17.64, that is (1162, 1198) hours. M1 for the form, A1 for the interval.
The sample of 100 is large, so the Central Limit Theorem makes the sample mean approximately normal whatever the shape of the population of lifetimes, and with a sample this size s is close enough to σ for 1.96 to remain usable. B1 for the Central Limit Theorem named.
The interval is therefore approximate, and saying so is part of the answer. With a sample of 10 rather than 100, neither justification would hold and a t multiplier on 9 degrees of freedom would be needed instead.

Question 11

A 95% confidence interval for the mean of a normal population with σ = 15 is calculated from a random sample of n observations and comes out as (68.1, 77.9). Find the sample mean and the value of n.

Accepted Answer

The interval is symmetric about the sample mean, so the mean is the midpoint: (68.1 + 77.9)/2 = 73. B1.
The half-width is (77.9 − 68.1)/2 = 4.9, and that half-width is 1.96 × standard error, so the standard error is 4.9/1.96 = 2.5. M1 for reading the half-width as 1.96 times the standard error, A1 for 2.5.
Then 15/√n = 2.5 gives √n = 6 and n = 36. A1.
Use the half-width, not the total width of 9.8: taking 9.8 as 1.96 × standard error gives n = 9, a quarter of the true sample size.

Question 12

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.

Accepted 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 explicitly. 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. 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.

Question 13

From a sample of three observations, one student estimates the population mean with the mean of all three values and another uses the first value alone. Explain why both estimators are unbiased, and why the first is still the better choice.

Accepted Answer

Each observation has expectation μ, so E(X₁) = μ, and E(sample mean) = (μ + μ + μ)/3 = μ: both rules are right on average over repeated samples, and unbiasedness cannot tell them apart. B1 for both expectations. What separates them is spread: the single value has variance σ², the mean of three has variance σ²/3, so the mean lands near its target more often: it is the more efficient estimator, with a standard error smaller by a factor of √3. B1 for the variance comparison. Unbiasedness on its own is a weak recommendation; among unbiased estimators the one with the smaller variance is preferred, which is the same reason the standard error σ/√n shrinks as the sample grows. B1 for efficiency as the deciding property.

Question 14

A random sample X₁, X₂ of two observations is taken from a population with mean μ and variance σ². Show that V = ½(X₁ − X₂)² is an unbiased estimator of σ². The estimator W = ½[(X₁ − X̄)² + (X₂ − X̄)²], where X̄ is the sample mean, is proposed instead. Show that E(W) = ½σ², and state what has to change to repair it.

Accepted Answer

The first estimator. Write D = X₁ − X₂. The two observations are independent, so E(D) = μ − μ = 0 and Var(D) = σ² + σ² = 2σ². M1 for the mean and variance of the difference. Now use E(D²) = Var(D) + [E(D)]² = 2σ² + 0 = 2σ². A1. So E(V) = ½E(D²) = σ², and V is unbiased. A1. The second estimator. With two observations X̄ = (X₁ + X₂)/2, so X₁ − X̄ = (X₁ − X₂)/2 and X₂ − X̄ = −(X₁ − X₂)/2. M1 for both deviations written in terms of D. Each squares to D²/4, so their sum is D²/2 and W = ½ × D²/2 = D²/4. A1. Hence E(W) = 2σ²/4 = ½σ², so W is biased and undershoots by half. A1. The repair is the divisor. W averages the two squared deviations by dividing by n = 2; dividing instead by n − 1 = 1 gives D²/2, which is V, and V is unbiased. That is the general rule appearing in its smallest case: the deviations are measured from the sample's own mean, which sits closer to the data than μ does, so a divisor of n is too generous by exactly one degree of freedom.

Question 15

A population has σ = 20. A 95% confidence interval for its mean is wanted with half-width at most 3, but only 100 observations can be taken. Find the half-width the 100 observations actually give, find the confidence level at which a half-width of 3 would be achieved from those 100 observations, and state what has been given up.

Accepted Answer

Standard error = 20/√100 = 2. B1.
At 95% the half-width is 1.96 × 2 = 3.92, which exceeds the 3 asked for. A1.
For a half-width of exactly 3, the multiplier must satisfy 2z = 3, so z = 1.5. M1.
Since Φ(1.5) = 0.9332, the two tails carry 2(1 − 0.9332) = 0.1336 between them, so the confidence level is 1 − 0.1336 = 86.6%. A1.
What has been given up is confidence, not precision: with the sample size fixed, the only way to narrow the interval is to accept a method that captures μ 86.6 times in 100 rather than 95. B1.
Reaching a half-width of 3 while keeping 95% would need n ≥ (1.96 × 20/3)² = 170.7, that is 171 observations, which is what the budget of 100 rules out.