Spearman’s rank correlation Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Spearman’s rank correlation questions. See exactly how to solve problems on spearman, rank-correlation, formula, rearrangement.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A set of n=8n = 8 bivariate observations is ranked, with no tied ranks, and dd denotes the difference between the two ranks of an observation. Given that ∑d2=30\sum d^{2} = 30, calculate Spearman's rank correlation coefficient rsr_{s}, giving your answer correct to 33 decimal places.

Worked solution

  1. Quote the formula for Spearman's rank correlation coefficient

    rs=1−6∑d2n(n2−1)r_{s}=1-\dfrac{6\sum d^{2}}{n\left(n^{2}-1\right)}

    This is the standard form, valid when there are no tied ranks.

  2. Substitute nn and ∑d2\sum d^{2}

    rs=1−6×308(82−1)r_{s}=1-\frac{6\times 30}{8\left(8^{2}-1\right)}

    Only the sample size and the total of the squared rank differences are needed.

  3. Evaluate the fraction

    6×30504=514\frac{6\times 30}{504}=\frac{5}{14}

    Keeping the fraction exact means no rounding error can creep in.

  4. State the value of rsr_{s}

    rs=1−514=914=0.643r_{s}=1-\frac{5}{14}=\frac{9}{14}=0.643

    This is Spearman's rank correlation coefficient for the data.

Answer
0.6430.643
Question 2
2 markseasy
A set of n=10n = 10 bivariate observations is ranked, with no tied ranks, and dd denotes the difference between the two ranks of an observation. Given that ∑d2=44\sum d^{2} = 44, calculate Spearman's rank correlation coefficient rsr_{s}, giving your answer correct to 33 decimal places.

Worked solution

  1. Quote the formula for Spearman's rank correlation coefficient

    rs=1−6∑d2n(n2−1)r_{s}=1-\dfrac{6\sum d^{2}}{n\left(n^{2}-1\right)}

    This is the standard form, valid when there are no tied ranks.

  2. Substitute nn and ∑d2\sum d^{2}

    rs=1−6×4410(102−1)r_{s}=1-\frac{6\times 44}{10\left(10^{2}-1\right)}

    Only the sample size and the total of the squared rank differences are needed.

  3. State the value of rsr_{s}

    rs=1−415=1115=0.733r_{s}=1-\frac{4}{15}=\frac{11}{15}=0.733

    This is Spearman's rank correlation coefficient for the data.

Answer
0.7330.733
Question 3
2 markseasy
A set of n=7n = 7 bivariate observations is ranked, with no tied ranks, and dd denotes the difference between the two ranks of an observation. Given that ∑d2=24\sum d^{2} = 24, calculate Spearman's rank correlation coefficient rsr_{s}, giving your answer correct to 33 decimal places.

Worked solution

  1. Quote the formula for Spearman's rank correlation coefficient

    rs=1−6∑d2n(n2−1)r_{s}=1-\dfrac{6\sum d^{2}}{n\left(n^{2}-1\right)}

    This is the standard form, valid when there are no tied ranks.

  2. Substitute nn and ∑d2\sum d^{2}

    rs=1−6×247(72−1)r_{s}=1-\frac{6\times 24}{7\left(7^{2}-1\right)}

    Only the sample size and the total of the squared rank differences are needed.

  3. Evaluate the fraction

    6×24336=37\frac{6\times 24}{336}=\frac{3}{7}

    Keeping the fraction exact means no rounding error can creep in.

  4. State the value of rsr_{s}

    rs=1−37=47=0.571r_{s}=1-\frac{3}{7}=\frac{4}{7}=0.571

    This is Spearman's rank correlation coefficient for the data.

Answer
0.5710.571
Question 4
2 markseasy
A set of n=9n = 9 bivariate observations is ranked, with no tied ranks, and dd denotes the difference between the two ranks of an observation. Given that ∑d2=96\sum d^{2} = 96, calculate Spearman's rank correlation coefficient rsr_{s}, giving your answer correct to 33 decimal places.

Worked solution

  1. Quote the formula for Spearman's rank correlation coefficient

    rs=1−6∑d2n(n2−1)r_{s}=1-\dfrac{6\sum d^{2}}{n\left(n^{2}-1\right)}

    This is the standard form, valid when there are no tied ranks.

  2. Substitute nn and ∑d2\sum d^{2}

    rs=1−6×969(92−1)r_{s}=1-\frac{6\times 96}{9\left(9^{2}-1\right)}

    Only the sample size and the total of the squared rank differences are needed.

  3. Evaluate the fraction

    6×96720=45\frac{6\times 96}{720}=\frac{4}{5}

    Keeping the fraction exact means no rounding error can creep in.

  4. State the value of rsr_{s}

    rs=1−45=15=0.200r_{s}=1-\frac{4}{5}=\frac{1}{5}=0.200

    This is Spearman's rank correlation coefficient for the data.

Answer
0.2000.200
Question 5
2 markseasy
A set of n=12n = 12 bivariate observations is ranked, with no tied ranks, and dd denotes the difference between the two ranks of an observation. Given that ∑d2=350\sum d^{2} = 350, calculate Spearman's rank correlation coefficient rsr_{s}, giving your answer correct to 33 decimal places.

Worked solution

  1. Quote the formula for Spearman's rank correlation coefficient

    rs=1−6∑d2n(n2−1)r_{s}=1-\dfrac{6\sum d^{2}}{n\left(n^{2}-1\right)}

    This is the standard form, valid when there are no tied ranks.

  2. Substitute nn and ∑d2\sum d^{2}

    rs=1−6×35012(122−1)r_{s}=1-\frac{6\times 350}{12\left(12^{2}-1\right)}

    Only the sample size and the total of the squared rank differences are needed.

  3. State the value of rsr_{s}

    rs=1−175143=−32143=−0.224r_{s}=1-\frac{175}{143}=- \frac{32}{143}=-0.224

    This is Spearman's rank correlation coefficient for the data.

Answer
−0.224-0.224

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