Further Maths Estimation and confidence intervals Practice Questions

Free Further Maths Estimation and confidence intervals practice questions with full step-by-step worked solutions. Covers unbiased-estimators, sample-mean, sample-variance, summary-statistics. Practise exam-style problems and check your method.

unbiased-estimatorssample-meansample-variancesummary-statisticsbiasdivisor-n
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A random sample of n=12n = 12 apples is taken and the mass, xx grams, of each is recorded. The sample summary statistics are ∑x=148.8\sum x = 148.8 and ∑x2=1869.9\sum x^{2} = 1869.9. Find the unbiased estimate of the population mean μ\mu, giving your answer to 44 decimal places.
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Worked solution

  1. Compute the unbiased estimate of the mean

    xˉ=∑xn=148.812=12.4\bar{x}=\frac{\sum x}{n}=\frac{148.8}{12}=12.4

    Since E(Xˉ)=μE(\bar{X})=\mu, the sample mean is the unbiased estimate of μ\mu.

  2. Justify that this is the unbiased estimate

    E(Xˉ)=μ  ⇒  μ^=xˉ=12.4000E(\bar{X})=\mu\;\Rightarrow\;\hat{\mu}=\bar{x}=12.4000

    The estimator Xˉ\bar{X} has expectation μ\mu, so its observed value is the unbiased estimate of μ\mu.

  3. State the required estimate

    xˉ=12.4000\bar{x}=12.4000

    This is the value asked for, correct to 44 decimal places, measured in grams.

Answer
xˉ=12.4000\bar{x}=12.4000
Question 2
2 markseasy
A random sample X1,X2,…,XnX_{1},X_{2},\ldots,X_{n} of size n=9n = 9 is taken from a population of steel pins whose length has unknown mean μ\mu and unknown variance σ2\sigma^{2}. Which of the following statements concerning unbiased estimators is correct?
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Worked solution

  1. State the definition of an unbiased estimator

    E(θ^)=θE(\hat{\theta})=\theta

    Unbiasedness is a property of the ESTIMATOR across all samples, not of the estimate from one sample.

  2. Take the expectation of the sample mean

    E(Xˉ)=19∑i=19E(Xi)=9μ9=μE(\bar{X})=\frac{1}{9}\sum_{i=1}^{9}E(X_{i})=\frac{9\mu}{9}=\mu

    So Xˉ\bar{X} is unbiased for μ\mu for every nn, whatever the population.

  3. Compare the two candidate variance estimators

    E(18∑i=19(Xi−Xˉ)2)=σ2,E(19∑i=19(Xi−Xˉ)2)=89σ2E\left(\frac{1}{8}\sum_{i=1}^{9}(X_{i}-\bar{X})^{2}\right)=\sigma^{2},\qquad E\left(\frac{1}{9}\sum_{i=1}^{9}(X_{i}-\bar{X})^{2}\right)=\frac{8}{9}\sigma^{2}

    The divisor n−1n-1 hits σ2\sigma^{2} exactly; the divisor nn falls short by the factor n−1n\frac{n-1}{n}, so it is biased low.

  4. Select the statement that follows from these expectations

    E(Xˉ)=μ,E(S2)=σ2,\qquadE(Sxx9)=89σ2E(\bar{X})=\mu,\qquad E(S^{2})=\sigma^{2},\qquadE\left(\frac{S_{xx}}{9}\right)=\frac{8}{9}\sigma^{2}

    These three expectations settle which estimators are unbiased and which is not.

Answer
E(Xˉ)=μ, E(S2)=σ2E(\bar{X})=\mu,\ E(S^{2})=\sigma^{2}
Question 3
4 marksintermediate
A random sample X1,X2,…,XnX_{1},X_{2},\ldots,X_{n} of size n=20n = 20 is taken from a population of batteries whose lifetime has unknown mean μ\mu and unknown variance σ2\sigma^{2}. Which of the following statements concerning unbiased estimators is correct?
Show worked solution

Worked solution

  1. State the definition of an unbiased estimator

    E(θ^)=θE(\hat{\theta})=\theta

    Unbiasedness is a property of the ESTIMATOR across all samples, not of the estimate from one sample.

  2. Take the expectation of the sample mean

    E(Xˉ)=120∑i=120E(Xi)=20μ20=μE(\bar{X})=\frac{1}{20}\sum_{i=1}^{20}E(X_{i})=\frac{20\mu}{20}=\mu

    So Xˉ\bar{X} is unbiased for μ\mu for every nn, whatever the population.

  3. Compare the two candidate variance estimators

    E(119∑i=120(Xi−Xˉ)2)=σ2,E(120∑i=120(Xi−Xˉ)2)=1920σ2E\left(\frac{1}{19}\sum_{i=1}^{20}(X_{i}-\bar{X})^{2}\right)=\sigma^{2},\qquad E\left(\frac{1}{20}\sum_{i=1}^{20}(X_{i}-\bar{X})^{2}\right)=\frac{19}{20}\sigma^{2}

    The divisor n−1n-1 hits σ2\sigma^{2} exactly; the divisor nn falls short by the factor n−1n\frac{n-1}{n}, so it is biased low.

  4. Explain where the missing degree of freedom goes

    ∑i=120(Xi−Xˉ)=0\sum_{i=1}^{20}(X_{i}-\bar{X})=0

    The deviations are forced to sum to zero, so only n−1n-1 of them are free.

  5. Warn that unbiasedness does not survive a square root

    E(S2)=σ2butE(S)<σE(S^{2})=\sigma^{2}\quad\text{but}\quad E(S)<\sigma

    The square root is concave, so the square root of an unbiased variance estimator is biased low for σ\sigma.

  6. Note that unbiasedness says nothing about variance

    E(X1)=μ too, yet Var⁡(X1)=σ2>σ220=Var⁡(Xˉ)E(X_{1})=\mu\text{ too, yet }\operatorname{Var}(X_{1})=\sigma^{2}>\frac{\sigma^{2}}{20}=\operatorname{Var}(\bar{X})

    The first observation alone is also unbiased for μ\mu; Xˉ\bar{X} is preferred because it is far more precise, not because it is unbiased.

  7. Select the statement that follows from these expectations

    E(Xˉ)=μ,E(S2)=σ2,\qquadE(Sxx20)=1920σ2E(\bar{X})=\mu,\qquad E(S^{2})=\sigma^{2},\qquadE\left(\frac{S_{xx}}{20}\right)=\frac{19}{20}\sigma^{2}

    These three expectations settle which estimators are unbiased and which is not.

Answer
E(Xˉ)=μ, E(S2)=σ2E(\bar{X})=\mu,\ E(S^{2})=\sigma^{2}
Question 4
6 markshard
Two independent random samples, of sizes n1=10n_{1} = 10 and n2=14n_{2} = 14, are taken in order to construct a 99%99\% confidence interval for the difference μ1−μ2\mu_{1}-\mu_{2} between two mean reaction times. The two population variances are unknown but may be assumed equal, so a pooled estimate is used. Which of the following is the correct critical value to use, given to 44 decimal places?
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Worked solution

  1. Decide which distribution supplies the percentage point

    σ12=σ22 unknown  ⇒  t-distribution\sigma_{1}^{2}=\sigma_{2}^{2}\text{ unknown}\;\Rightarrow\;t\text{-distribution}

    An unknown variance always forces a tt percentage point, never a zz.

  2. Count the degrees of freedom

    ν=n1+n2−2=10+14−2=22\nu=n_{1}+n_{2}-2=10+14-2=22

    Each of the two samples spends one degree of freedom on its own mean.

  3. Split the tail area for a two-sided interval

    1−0.992=0.005 in each tail\frac{1-0.99}{2}=0.005\text{ in each tail}

    A 99%99\% interval leaves 0.0050.005 in each tail, not 0.010.01 in one tail.

  4. Read the value from the tt-table

    t22=2.819t_{22}=2.819

    Row ν=22\nu=22, column 0.0050.005.

  5. Recall what it means for an estimator to be unbiased

    θ^ is unbiased for θ  ⟺  E(θ^)=θ\hat{\theta}\text{ is unbiased for }\theta\iff E(\hat{\theta})=\theta

    Unbiasedness is a statement about the mean of the estimator's sampling distribution, not about any one sample.

  6. Show that the sample mean is unbiased for μ\mu

    E(Xˉ)=1n∑i=1nE(Xi)=nμn=μE(\bar{X})=\frac{1}{n}\sum_{i=1}^{n}E(X_{i})=\frac{n\mu}{n}=\mu

    Expectation is linear, so the nn copies of μ\mu divide out exactly.

  7. Recall the unbiased estimator of the population variance

    S2=1n−1∑i=1n(Xi−Xˉ)2,E(S2)=σ2S^{2}=\frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2},\qquad E(S^{2})=\sigma^{2}

    The divisor n−1n-1 is what makes the expectation come out as σ2\sigma^{2} exactly.

  8. Recall why the divisor nn gives a biased estimator

    E(1n∑i=1n(Xi−Xˉ)2)=n−1nσ2<σ2E\left(\frac{1}{n}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}\right)=\frac{n-1}{n}\sigma^{2}<\sigma^{2}

    The deviations are taken about Xˉ\bar{X} rather than μ\mu, so they are too small on average; dividing by nn leaves the estimate biased low.

  9. Recall the computational form of the sum of squares

    Sxx=∑x2−(∑x)2nS_{xx}=\sum x^{2}-\frac{\left(\sum x\right)^{2}}{n}

    This is the identity that turns the summary statistics into the sum of squared deviations without listing the data.

  10. Recall the variance of the sample mean

    Var⁡(Xˉ)=σ2n\operatorname{Var}(\bar{X})=\frac{\sigma^{2}}{n}

    The observations are independent, so the variances add and the factor 1n\frac{1}{n} is squared.

  11. Select the correct percentage point

    2.81902.8190

    This is the value the confidence interval requires.

Answer
2.81902.8190
Question 5
9 markschallenging
A random sample X1,X2,…,XnX_{1},X_{2},\ldots,X_{n} of size n=25n = 25 is taken from a population of nylon threads whose breaking strain has unknown mean μ\mu and unknown variance σ2\sigma^{2}. Which of the following statements concerning unbiased estimators is correct?
Show worked solution

Worked solution

  1. State the definition of an unbiased estimator

    E(θ^)=θE(\hat{\theta})=\theta

    Unbiasedness is a property of the ESTIMATOR across all samples, not of the estimate from one sample.

  2. Take the expectation of the sample mean

    E(Xˉ)=125∑i=125E(Xi)=25μ25=μE(\bar{X})=\frac{1}{25}\sum_{i=1}^{25}E(X_{i})=\frac{25\mu}{25}=\mu

    So Xˉ\bar{X} is unbiased for μ\mu for every nn, whatever the population.

  3. Compare the two candidate variance estimators

    E(124∑i=125(Xi−Xˉ)2)=σ2,E(125∑i=125(Xi−Xˉ)2)=2425σ2E\left(\frac{1}{24}\sum_{i=1}^{25}(X_{i}-\bar{X})^{2}\right)=\sigma^{2},\qquad E\left(\frac{1}{25}\sum_{i=1}^{25}(X_{i}-\bar{X})^{2}\right)=\frac{24}{25}\sigma^{2}

    The divisor n−1n-1 hits σ2\sigma^{2} exactly; the divisor nn falls short by the factor n−1n\frac{n-1}{n}, so it is biased low.

  4. Explain where the missing degree of freedom goes

    ∑i=125(Xi−Xˉ)=0\sum_{i=1}^{25}(X_{i}-\bar{X})=0

    The deviations are forced to sum to zero, so only n−1n-1 of them are free.

  5. Warn that unbiasedness does not survive a square root

    E(S2)=σ2butE(S)<σE(S^{2})=\sigma^{2}\quad\text{but}\quad E(S)<\sigma

    The square root is concave, so the square root of an unbiased variance estimator is biased low for σ\sigma.

  6. Note that unbiasedness says nothing about variance

    E(X1)=μ too, yet Var⁡(X1)=σ2>σ225=Var⁡(Xˉ)E(X_{1})=\mu\text{ too, yet }\operatorname{Var}(X_{1})=\sigma^{2}>\frac{\sigma^{2}}{25}=\operatorname{Var}(\bar{X})

    The first observation alone is also unbiased for μ\mu; Xˉ\bar{X} is preferred because it is far more precise, not because it is unbiased.

  7. Recall what it means for an estimator to be unbiased

    θ^ is unbiased for θ  ⟺  E(θ^)=θ\hat{\theta}\text{ is unbiased for }\theta\iff E(\hat{\theta})=\theta

    Unbiasedness is a statement about the mean of the estimator's sampling distribution, not about any one sample.

  8. Show that the sample mean is unbiased for μ\mu

    E(Xˉ)=1n∑i=1nE(Xi)=nμn=μE(\bar{X})=\frac{1}{n}\sum_{i=1}^{n}E(X_{i})=\frac{n\mu}{n}=\mu

    Expectation is linear, so the nn copies of μ\mu divide out exactly.

  9. Recall the unbiased estimator of the population variance

    S2=1n−1∑i=1n(Xi−Xˉ)2,E(S2)=σ2S^{2}=\frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2},\qquad E(S^{2})=\sigma^{2}

    The divisor n−1n-1 is what makes the expectation come out as σ2\sigma^{2} exactly.

  10. Recall why the divisor nn gives a biased estimator

    E(1n∑i=1n(Xi−Xˉ)2)=n−1nσ2<σ2E\left(\frac{1}{n}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}\right)=\frac{n-1}{n}\sigma^{2}<\sigma^{2}

    The deviations are taken about Xˉ\bar{X} rather than μ\mu, so they are too small on average; dividing by nn leaves the estimate biased low.

  11. Recall the computational form of the sum of squares

    Sxx=∑x2−(∑x)2nS_{xx}=\sum x^{2}-\frac{\left(\sum x\right)^{2}}{n}

    This is the identity that turns the summary statistics into the sum of squared deviations without listing the data.

  12. Recall the variance of the sample mean

    Var⁡(Xˉ)=σ2n\operatorname{Var}(\bar{X})=\frac{\sigma^{2}}{n}

    The observations are independent, so the variances add and the factor 1n\frac{1}{n} is squared.

  13. Recall the standard error of the mean

    se⁡(Xˉ)=σn\operatorname{se}(\bar{X})=\frac{\sigma}{\sqrt{n}}

    It is σ\sigma divided by n\sqrt{n}, not by nn.

  14. Recall the degrees of freedom of the tt-distribution used here

    ν=n−1\nu=n-1

    One degree of freedom is spent estimating μ\mu by xˉ\bar{x}.

  15. Recall why tt and not zz is used when σ\sigma is unknown

    Xˉ−μS/n∼tn−1,tν>z for every finite ν\frac{\bar{X}-\mu}{S/\sqrt{n}}\sim t_{n-1},\qquad t_{\nu}>z\text{ for every finite }\nu

    The tt-distribution has heavier tails, so it gives a wider (more honest) interval.

  16. Select the statement that follows from these expectations

    E(Xˉ)=μ,E(S2)=σ2,\qquadE(Sxx25)=2425σ2E(\bar{X})=\mu,\qquad E(S^{2})=\sigma^{2},\qquadE\left(\frac{S_{xx}}{25}\right)=\frac{24}{25}\sigma^{2}

    These three expectations settle which estimators are unbiased and which is not.

Answer
E(Xˉ)=μ, E(S2)=σ2E(\bar{X})=\mu,\ E(S^{2})=\sigma^{2}

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