Estimation and confidence intervals Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Estimation and confidence intervals questions. See exactly how to solve problems on unbiased-estimators, sample-mean, sample-variance, summary-statistics.

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.

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 of n=8n = 8 steel pins is taken and the length, xx millimetres, of each is recorded. The observed values are 4.1,5.3,4.8,6,5.2,4.6,5.9,5.14.1, 5.3, 4.8, 6, 5.2, 4.6, 5.9, 5.1. Find the unbiased estimate of the population mean μ\mu, giving your answer to 44 decimal places.

Worked solution

  1. Reduce the raw data to the two summary statistics

    ∑x=4.1+5.3+4.8+6+5.2+4.6+5.9+5.1=41,∑x2=4.12+5.32+4.82+62+5.22+4.62+5.92+5.12=212.96\sum x=4.1+5.3+4.8+6+5.2+4.6+5.9+5.1=41,\qquad\sum x^{2}=4.1^{2}+5.3^{2}+4.8^{2}+6^{2}+5.2^{2}+4.6^{2}+5.9^{2}+5.1^{2}=212.96

    Every unbiased estimate below is built from these two totals; note that ∑x2\sum x^{2} squares each value BEFORE adding.

  2. Compute the unbiased estimate of the mean

    xˉ=∑xn=418=5.125\bar{x}=\frac{\sum x}{n}=\frac{41}{8}=5.125

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

  3. Justify that this is the unbiased estimate

    E(Xˉ)=μ  ⇒  μ^=xˉ=5.1250E(\bar{X})=\mu\;\Rightarrow\;\hat{\mu}=\bar{x}=5.1250

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

  4. State the required estimate

    xˉ=5.1250\bar{x}=5.1250

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

Answer
xˉ=5.1250\bar{x}=5.1250
Question 3
2 markseasy
A random sample of n=10n = 10 bottles of cordial is taken and the volume, xx millilitres, of each is recorded. The sample summary statistics are ∑x=252\sum x = 252 and ∑x2=6367.4\sum x^{2} = 6367.4. Find the unbiased estimate of the population variance σ2\sigma^{2}, giving your answer to 44 decimal places.

Worked solution

  1. Compute the sum of squared deviations

    Sxx=∑x2−(∑x)2n=6367.4−252210=17S_{xx}=\sum x^{2}-\frac{\left(\sum x\right)^{2}}{n}=6367.4-\frac{252^{2}}{10}=17

    The computational form avoids having to find each deviation separately.

  2. Divide by n−1n-1 to obtain the unbiased estimate of the variance

    s2=Sxxn−1=179=179=1.8889s^{2}=\frac{S_{xx}}{n-1}=\frac{17}{9}=\frac{17}{9}=1.8889

    Only the divisor n−1n-1 gives E(S2)=σ2E(S^{2})=\sigma^{2}.

  3. Compute the unbiased estimate of the mean

    xˉ=∑xn=25210=25.2\bar{x}=\frac{\sum x}{n}=\frac{252}{10}=25.2

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

  4. State the required estimate

    s2=1.8889s^{2}=1.8889

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

Answer
s2=1.8889s^{2}=1.8889
Question 4
2 markseasy
A random sample of n=8n = 8 batteries is taken and the lifetime, xx hours, of each is recorded. The observed values are 45,52,48,50,47,53,49,5145, 52, 48, 50, 47, 53, 49, 51. Find the unbiased estimate of the population variance σ2\sigma^{2}, giving your answer to 44 decimal places.

Worked solution

  1. Reduce the raw data to the two summary statistics

    ∑x=45+52+48+50+47+53+49+51=395,∑x2=452+522+482+502+472+532+492+512=19553\sum x=45+52+48+50+47+53+49+51=395,\qquad\sum x^{2}=45^{2}+52^{2}+48^{2}+50^{2}+47^{2}+53^{2}+49^{2}+51^{2}=19553

    Every unbiased estimate below is built from these two totals; note that ∑x2\sum x^{2} squares each value BEFORE adding.

  2. Compute the sum of squared deviations

    Sxx=∑x2−(∑x)2n=19553−39528=49.875S_{xx}=\sum x^{2}-\frac{\left(\sum x\right)^{2}}{n}=19553-\frac{395^{2}}{8}=49.875

    The computational form avoids having to find each deviation separately.

  3. Divide by n−1n-1 to obtain the unbiased estimate of the variance

    s2=Sxxn−1=49.8757=7.125s^{2}=\frac{S_{xx}}{n-1}=\frac{49.875}{7}=7.125

    Only the divisor n−1n-1 gives E(S2)=σ2E(S^{2})=\sigma^{2}.

  4. State the required estimate

    s2=7.1250s^{2}=7.1250

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

Answer
s2=7.1250s^{2}=7.1250
Question 5
2 markseasy
A random sample of n=8n = 8 trial responses is taken and the reaction time, xx milliseconds, of each is recorded. The sample summary statistics are ∑x=98.4\sum x = 98.4 and ∑x2=1215.86\sum x^{2} = 1215.86. Find the value of ss, where s2s^{2} denotes the unbiased estimate of σ2\sigma^{2} obtained from this sample. Give your answer to 44 decimal places.

Worked solution

  1. Compute the sum of squared deviations

    Sxx=∑x2−(∑x)2n=1215.86−98.428=5.54S_{xx}=\sum x^{2}-\frac{\left(\sum x\right)^{2}}{n}=1215.86-\frac{98.4^{2}}{8}=5.54

    The computational form avoids having to find each deviation separately.

  2. Divide by n−1n-1 to obtain the unbiased estimate of the variance

    s2=Sxxn−1=5.547=277350=0.7914s^{2}=\frac{S_{xx}}{n-1}=\frac{5.54}{7}=\frac{277}{350}=0.7914

    Only the divisor n−1n-1 gives E(S2)=σ2E(S^{2})=\sigma^{2}.

  3. Take the square root of the unbiased variance estimate

    s=s2=0.7914=0.8896s=\sqrt{s^{2}}=\sqrt{0.7914}=0.8896

    This is the value that goes into a tt interval, even though SS is not itself unbiased for σ\sigma.

  4. State the required estimate

    s=0.8896s=0.8896

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

Answer
s=0.8896s=0.8896

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