Hard Further Maths Spearman’s rank correlation Questions

Challenging, exam-style Further Maths Spearman’s rank correlation questions with worked solutions. Stretch yourself on the hardest spearman, ranking, rank-correlation, raw-data problems.

spearmanrankingrank-correlationraw-datatied-rankspmcc-of-ranks
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
The table shows the mass, in kg, and the price, in pounds, of each of 88 items sold by a shop. One item is a very heavy and very expensive machine. ItemABCDEFGHMass (kg)1112131415161890Price (£)3331302826252295\begin{array}{l|cccccccc} \text{Item} & A & B & C & D & E & F & G & H \\ \hline \text{Mass (kg)} & 11 & 12 & 13 & 14 & 15 & 16 & 18 & 90 \\ \text{Price (\pounds)} & 33 & 31 & 30 & 28 & 26 & 25 & 22 & 95 \end{array} Rank 11 is given to the largest value in each row. Which of the following gives the correct values of Spearman's rank correlation coefficient rsr_{s} and of the product moment correlation coefficient rr of the raw data, together with the correct reason for preferring the rank coefficient here?
Show worked solution

Worked solution

  1. Rank both variables

    ItemABCDEFGHRank of x87654321Rank of y23456781\begin{array}{l|cccccccc} \text{Item} & A & B & C & D & E & F & G & H \\ \hline \text{Rank of x} & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\ \text{Rank of y} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 1 \end{array}

    Rank 11 is given to the largest value of each variable.

  2. Find Spearman's coefficient from the ranks

    rs=−0.333r_{s}=-0.333

    This depends only on the order of the data, not on the values themselves.

  3. Find the product moment correlation coefficient of the raw data

    r=SxySxxSyy=0.974r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=0.974

    This uses the actual values, so it is sensitive to the extreme observation.

  4. Compare the two coefficients

    rs=−0.333,r=0.974r_{s}=-0.333,\qquad r=0.974

    They measure different things: rr measures LINEAR association and rsr_{s} measures MONOTONIC association.

  5. Quote Spearman's rank correlation coefficient

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

    This is the formula in the booklet: dd is the difference between the two ranks of a single item and nn is the number of items.

  6. Note what dd means

    d=rank of x−rank of yd=\text{rank of }x-\text{rank of }y

    The sign of dd never matters, because only d2d^{2} is used.

  7. Check the ranks against a known total

    ∑ranks=n(n+1)2\sum\text{ranks}=\frac{n\left(n+1\right)}{2}

    Both sets of ranks must add up to this, even when tied ranks have been averaged.

  8. Check that the differences sum to zero

    ∑d=0\sum d=0

    The two sets of ranks have the same total, so the differences must cancel; this also forces ∑d2\sum d^{2} to be even.

  9. Note the range of the coefficient

    −1≤rs≤1-1\le r_{s}\le 1

    A value outside this range can only be an arithmetic error.

  10. Recall what rsr_{s} actually is

    rs=PMCC of the ranksr_{s}=\text{PMCC of the ranks}

    Spearman's coefficient is the product moment correlation coefficient applied to the ranks instead of to the raw data.

  11. Note when the closed formula may be used

    no tied ranks⇒1−6∑d2n(n2−1)=PMCC of the ranks\text{no tied ranks}\Rightarrow 1-\frac{6\sum d^{2}}{n\left(n^{2}-1\right)}=\text{PMCC of the ranks}

    The closed formula is derived on the assumption that the ranks are 1,2,…,n1,2,\ldots,n in some order; once ranks are averaged that is no longer true.

  12. Recall the rule for tied values

    tied values share the mean of the ranks they would occupy\text{tied values share the mean of the ranks they would occupy}

    Two values tied for 33rd and 44th place each receive the rank 3.53.5.

  13. Recall the product moment correlation coefficient

    r=SxySxxSyyr=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}

    Applied to the ranks, this is exactly Spearman's coefficient.

  14. Note what perfect agreement means

    ∑d2=0  ⟺  rs=1\sum d^{2}=0\iff r_{s}=1

    The two rankings are then identical, item by item.

  15. Decide which coefficient is appropriate

    use rs\text{use }r_{s}

    A rank method is preferred here because one extreme observation would otherwise dominate the product moment coefficient, while ranking reduces it to a single ordinary rank.

Answer
use rs\text{use }r_{s}
Question 2
9 markschallenging
The table shows the length, xx cm, of the side of a cube and its volume, yy cm3^{3}, for 77 cubes. CubeABCDEFGx2345678y82764125216343512\begin{array}{l|ccccccc} \text{Cube} & A & B & C & D & E & F & G \\ \hline \text{x} & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \text{y} & 8 & 27 & 64 & 125 & 216 & 343 & 512 \end{array} Rank 11 is given to the largest value in each row. Which of the following gives the correct values of Spearman's rank correlation coefficient rsr_{s} and of the product moment correlation coefficient rr of the raw data, together with the correct reason for preferring the rank coefficient here?
Show worked solution

Worked solution

  1. Rank both variables

    CubeABCDEFGRank of x7654321Rank of y7654321\begin{array}{l|ccccccc} \text{Cube} & A & B & C & D & E & F & G \\ \hline \text{Rank of x} & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\ \text{Rank of y} & 7 & 6 & 5 & 4 & 3 & 2 & 1 \end{array}

    Rank 11 is given to the largest value of each variable.

  2. Find Spearman's coefficient from the ranks

    rs=1.000r_{s}=1.000

    This depends only on the order of the data, not on the values themselves.

  3. Find the product moment correlation coefficient of the raw data

    r=SxySxxSyy=0.953r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=0.953

    This uses the actual values, so it is sensitive to the curvature of the relationship.

  4. Compare the two coefficients

    rs=1.000,r=0.953r_{s}=1.000,\qquad r=0.953

    They measure different things: rr measures LINEAR association and rsr_{s} measures MONOTONIC association.

  5. Quote Spearman's rank correlation coefficient

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

    This is the formula in the booklet: dd is the difference between the two ranks of a single item and nn is the number of items.

  6. Note what dd means

    d=rank of x−rank of yd=\text{rank of }x-\text{rank of }y

    The sign of dd never matters, because only d2d^{2} is used.

  7. Check the ranks against a known total

    ∑ranks=n(n+1)2\sum\text{ranks}=\frac{n\left(n+1\right)}{2}

    Both sets of ranks must add up to this, even when tied ranks have been averaged.

  8. Check that the differences sum to zero

    ∑d=0\sum d=0

    The two sets of ranks have the same total, so the differences must cancel; this also forces ∑d2\sum d^{2} to be even.

  9. Note the range of the coefficient

    −1≤rs≤1-1\le r_{s}\le 1

    A value outside this range can only be an arithmetic error.

  10. Recall what rsr_{s} actually is

    rs=PMCC of the ranksr_{s}=\text{PMCC of the ranks}

    Spearman's coefficient is the product moment correlation coefficient applied to the ranks instead of to the raw data.

  11. Note when the closed formula may be used

    no tied ranks⇒1−6∑d2n(n2−1)=PMCC of the ranks\text{no tied ranks}\Rightarrow 1-\frac{6\sum d^{2}}{n\left(n^{2}-1\right)}=\text{PMCC of the ranks}

    The closed formula is derived on the assumption that the ranks are 1,2,…,n1,2,\ldots,n in some order; once ranks are averaged that is no longer true.

  12. Recall the rule for tied values

    tied values share the mean of the ranks they would occupy\text{tied values share the mean of the ranks they would occupy}

    Two values tied for 33rd and 44th place each receive the rank 3.53.5.

  13. Recall the product moment correlation coefficient

    r=SxySxxSyyr=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}

    Applied to the ranks, this is exactly Spearman's coefficient.

  14. Note what perfect agreement means

    ∑d2=0  ⟺  rs=1\sum d^{2}=0\iff r_{s}=1

    The two rankings are then identical, item by item.

  15. Note what perfect disagreement means

    ∑d2=n(n2−1)3  ⟺  rs=−1\sum d^{2}=\frac{n\left(n^{2}-1\right)}{3}\iff r_{s}=-1

    This is the largest value ∑d2\sum d^{2} can take, reached when one ranking is the exact reverse of the other.

  16. Decide which coefficient is appropriate

    use rs\text{use }r_{s}

    A rank method is preferred here because the relationship is monotonic but not linear, so the ranks capture it perfectly even though the points are not collinear.

Answer
use rs\text{use }r_{s}
Question 3
9 markschallenging
The table shows the monthly price, in pounds, of each of 1010 mobile phone contracts and the satisfaction score given to it by customers. ContractABCDEFGHIJPrice (£)100959085807570656055Satisfaction score72846088568052687664\begin{array}{l|cccccccccc} \text{Contract} & A & B & C & D & E & F & G & H & I & J \\ \hline \text{Price (\pounds)} & 100 & 95 & 90 & 85 & 80 & 75 & 70 & 65 & 60 & 55 \\ \text{Satisfaction score} & 72 & 84 & 60 & 88 & 56 & 80 & 52 & 68 & 76 & 64 \end{array} Rank 11 is given to the largest value in each row. A two-tailed test of H0H_{0}: ρs=0\rho_{s} = 0 against H1H_{1}: ρs≠0\rho_{s} \neq 0 is carried out at the 10%10\% significance level, and the critical value of rsr_{s} for n=10n = 10 is 0.56360.5636. Which of the following is the correct conclusion?
Show worked solution

Worked solution

  1. State the ranking convention

    rank 1=largest value in each row\text{rank }1=\text{largest value in each row}

    Both variables must be ranked in the same direction; ranking one of them the other way round would simply reverse the sign of rsr_{s}.

  2. Rank the values of price

    ContractABCDEFGHIJPrice (£)100959085807570656055Rank12345678910\begin{array}{l|cccccccccc} \text{Contract} & A & B & C & D & E & F & G & H & I & J \\ \hline \text{Price (\pounds)} & 100 & 95 & 90 & 85 & 80 & 75 & 70 & 65 & 60 & 55 \\ \text{Rank} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \end{array}

    The 1010 values are placed in order and given the ranks 11 to 1010.

  3. Rank the values of satisfaction score

    ContractABCDEFGHIJSatisfaction score72846088568052687664Rank52819310647\begin{array}{l|cccccccccc} \text{Contract} & A & B & C & D & E & F & G & H & I & J \\ \hline \text{Satisfaction score} & 72 & 84 & 60 & 88 & 56 & 80 & 52 & 68 & 76 & 64 \\ \text{Rank} & 5 & 2 & 8 & 1 & 9 & 3 & 10 & 6 & 4 & 7 \end{array}

    The second variable is ranked in exactly the same direction as the first.

  4. Find the difference dd between the two ranks of each item

    ContractABCDEFGHIJd−40−53−43−3253\begin{array}{l|cccccccccc} \text{Contract} & A & B & C & D & E & F & G & H & I & J \\ \hline \text{d} & -4 & 0 & -5 & 3 & -4 & 3 & -3 & 2 & 5 & 3 \end{array}

    These differences add up to 00, which is a useful check on the ranking.

  5. Square each difference

    \begin{array}{l|cccccccccc} \text{Contract} & A & B & C & D & E & F & G & H & I & J \\ \hline \text{d^{2}} & 16 & 0 & 25 & 9 & 16 & 9 & 9 & 4 & 25 & 9 \end{array}

    Squaring removes the signs, so only the size of each disagreement counts.

  6. Add the squared differences

    ∑d2=122\sum d^{2}=122

    This single number measures the total disagreement between the two rankings.

  7. Apply Spearman's formula

    rs=1−6×122990=43165=0.261r_{s}=1-\frac{6\times 122}{990}=\frac{43}{165}=0.261

    There are no tied ranks, so the closed formula gives rsr_{s} exactly.

  8. State the hypotheses

    H0:ρs=0,H1:ρs≠0H_{0}:\rho_{s}=0,\qquad H_{1}:\rho_{s}\neq0

    ρs\rho_{s} is the population rank correlation coefficient; H0H_{0} says there is no association between the price of a contract and the satisfaction score.

  9. Write down the critical value for this test

    n=10, two-tailed at 10% ⇒ critical value=0.5636n=10,\ \text{two-tailed at }10\%\ \Rightarrow\ \text{critical value}=0.5636

    The critical value is read from the table of critical values of Spearman's coefficient using the sample size and the significance level.

  10. Compare the test statistic with the critical value

    ∣rs∣=0.261 < 0.5636\left|r_{s}\right|=0.261\ <\ 0.5636

    The value is less extreme than the critical value, so it does not lie in the critical region.

  11. Quote Spearman's rank correlation coefficient

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

    This is the formula in the booklet: dd is the difference between the two ranks of a single item and nn is the number of items.

  12. Note what dd means

    d=rank of x−rank of yd=\text{rank of }x-\text{rank of }y

    The sign of dd never matters, because only d2d^{2} is used.

  13. Check the ranks against a known total

    ∑ranks=n(n+1)2\sum\text{ranks}=\frac{n\left(n+1\right)}{2}

    Both sets of ranks must add up to this, even when tied ranks have been averaged.

  14. Check that the differences sum to zero

    ∑d=0\sum d=0

    The two sets of ranks have the same total, so the differences must cancel; this also forces ∑d2\sum d^{2} to be even.

  15. Note the range of the coefficient

    −1≤rs≤1-1\le r_{s}\le 1

    A value outside this range can only be an arithmetic error.

  16. Recall what rsr_{s} actually is

    rs=PMCC of the ranksr_{s}=\text{PMCC of the ranks}

    Spearman's coefficient is the product moment correlation coefficient applied to the ranks instead of to the raw data.

  17. State the conclusion in context

    Do not reject H0: insufficient evidence of association\text{Do not reject }H_{0}\text{: insufficient evidence of association}

    The conclusion must be given at the stated significance level and in the context of the price of a contract and the satisfaction score.

Answer
Do not reject H0: insufficient evidence of association\text{Do not reject }H_{0}\text{: insufficient evidence of association}
Question 4
9 markschallenging
The table shows the daily maximum temperature, in degrees Celsius, and the number of ice creams sold by a kiosk on each of 88 days. DayABCDEFGHTemperature1724192822311526Ice creams sold4884961326015036108\begin{array}{l|cccccccc} \text{Day} & A & B & C & D & E & F & G & H \\ \hline \text{Temperature} & 17 & 24 & 19 & 28 & 22 & 31 & 15 & 26 \\ \text{Ice creams sold} & 48 & 84 & 96 & 132 & 60 & 150 & 36 & 108 \end{array} Rank 11 is given to the largest value in each row. A two-tailed test of H0H_{0}: ρs=0\rho_{s} = 0 against H1H_{1}: ρs≠0\rho_{s} \neq 0 is carried out at the 10%10\% significance level, and the critical value of rsr_{s} for n=8n = 8 is 0.64290.6429. Which of the following is the correct conclusion?
Show worked solution

Worked solution

  1. State the ranking convention

    rank 1=largest value in each row\text{rank }1=\text{largest value in each row}

    Both variables must be ranked in the same direction; ranking one of them the other way round would simply reverse the sign of rsr_{s}.

  2. Rank the values of temperature

    DayABCDEFGHTemperature1724192822311526Rank74625183\begin{array}{l|cccccccc} \text{Day} & A & B & C & D & E & F & G & H \\ \hline \text{Temperature} & 17 & 24 & 19 & 28 & 22 & 31 & 15 & 26 \\ \text{Rank} & 7 & 4 & 6 & 2 & 5 & 1 & 8 & 3 \end{array}

    The 88 values are placed in order and given the ranks 11 to 88.

  3. Rank the values of ice creams sold

    DayABCDEFGHIce creams sold4884961326015036108Rank75426183\begin{array}{l|cccccccc} \text{Day} & A & B & C & D & E & F & G & H \\ \hline \text{Ice creams sold} & 48 & 84 & 96 & 132 & 60 & 150 & 36 & 108 \\ \text{Rank} & 7 & 5 & 4 & 2 & 6 & 1 & 8 & 3 \end{array}

    The second variable is ranked in exactly the same direction as the first.

  4. Find the difference dd between the two ranks of each item

    DayABCDEFGHd0−120−1000\begin{array}{l|cccccccc} \text{Day} & A & B & C & D & E & F & G & H \\ \hline \text{d} & 0 & -1 & 2 & 0 & -1 & 0 & 0 & 0 \end{array}

    These differences add up to 00, which is a useful check on the ranking.

  5. Square each difference

    \begin{array}{l|cccccccc} \text{Day} & A & B & C & D & E & F & G & H \\ \hline \text{d^{2}} & 0 & 1 & 4 & 0 & 1 & 0 & 0 & 0 \end{array}

    Squaring removes the signs, so only the size of each disagreement counts.

  6. Add the squared differences

    ∑d2=6\sum d^{2}=6

    This single number measures the total disagreement between the two rankings.

  7. Apply Spearman's formula

    rs=1−6×6504=1314=0.929r_{s}=1-\frac{6\times 6}{504}=\frac{13}{14}=0.929

    There are no tied ranks, so the closed formula gives rsr_{s} exactly.

  8. State the hypotheses

    H0:ρs=0,H1:ρs≠0H_{0}:\rho_{s}=0,\qquad H_{1}:\rho_{s}\neq0

    ρs\rho_{s} is the population rank correlation coefficient; H0H_{0} says there is no association between the temperature and the number of ice creams sold.

  9. Write down the critical value for this test

    n=8, two-tailed at 10% ⇒ critical value=0.6429n=8,\ \text{two-tailed at }10\%\ \Rightarrow\ \text{critical value}=0.6429

    The critical value is read from the table of critical values of Spearman's coefficient using the sample size and the significance level.

  10. Compare the test statistic with the critical value

    ∣rs∣=0.929 > 0.6429\left|r_{s}\right|=0.929\ >\ 0.6429

    The value is more extreme than the critical value, so it lies in the critical region.

  11. Quote Spearman's rank correlation coefficient

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

    This is the formula in the booklet: dd is the difference between the two ranks of a single item and nn is the number of items.

  12. Note what dd means

    d=rank of x−rank of yd=\text{rank of }x-\text{rank of }y

    The sign of dd never matters, because only d2d^{2} is used.

  13. Check the ranks against a known total

    ∑ranks=n(n+1)2\sum\text{ranks}=\frac{n\left(n+1\right)}{2}

    Both sets of ranks must add up to this, even when tied ranks have been averaged.

  14. Check that the differences sum to zero

    ∑d=0\sum d=0

    The two sets of ranks have the same total, so the differences must cancel; this also forces ∑d2\sum d^{2} to be even.

  15. State the conclusion in context

    Reject H0: sufficient evidence of association\text{Reject }H_{0}\text{: sufficient evidence of association}

    The conclusion must be given at the stated significance level and in the context of the temperature and the number of ice creams sold.

Answer
Reject H0: sufficient evidence of association\text{Reject }H_{0}\text{: sufficient evidence of association}
Question 5
9 markschallenging
The table shows the number of complete years of service and the annual appraisal rating of each of 1010 employees. EmployeeABCDEFGHIJYears of service25583107519Appraisal rating4536487527\begin{array}{l|cccccccccc} \text{Employee} & A & B & C & D & E & F & G & H & I & J \\ \hline \text{Years of service} & 2 & 5 & 5 & 8 & 3 & 10 & 7 & 5 & 1 & 9 \\ \text{Appraisal rating} & 4 & 5 & 3 & 6 & 4 & 8 & 7 & 5 & 2 & 7 \end{array} Rank 11 is given to the largest value in each row, and tied values are given the mean of the ranks they would otherwise have occupied, so Spearman's coefficient must be found as the product moment correlation coefficient of the ranks. A one-tailed test of H0H_{0}: ρs=0\rho_{s} = 0 against H1H_{1}: ρs>0\rho_{s} > 0 is carried out at the 1%1\% significance level, and the critical value of rsr_{s} for n=10n = 10 is 0.74550.7455. Which of the following is the correct conclusion?
Show worked solution

Worked solution

  1. State the ranking convention

    rank 1=largest value in each row\text{rank }1=\text{largest value in each row}

    Both variables must be ranked in the same direction; ranking one of them the other way round would simply reverse the sign of rsr_{s}.

  2. Rank the values of years of service

    EmployeeABCDEFGHIJYears of service25583107519Rank96638146102\begin{array}{l|cccccccccc} \text{Employee} & A & B & C & D & E & F & G & H & I & J \\ \hline \text{Years of service} & 2 & 5 & 5 & 8 & 3 & 10 & 7 & 5 & 1 & 9 \\ \text{Rank} & 9 & 6 & 6 & 3 & 8 & 1 & 4 & 6 & 10 & 2 \end{array}

    The 1010 values are placed in order and given the ranks 11 to 1010.

  3. Rank the values of appraisal rating

    EmployeeABCDEFGHIJAppraisal rating4536487527Rank7.55.5947.512.55.5102.5\begin{array}{l|cccccccccc} \text{Employee} & A & B & C & D & E & F & G & H & I & J \\ \hline \text{Appraisal rating} & 4 & 5 & 3 & 6 & 4 & 8 & 7 & 5 & 2 & 7 \\ \text{Rank} & 7.5 & 5.5 & 9 & 4 & 7.5 & 1 & 2.5 & 5.5 & 10 & 2.5 \end{array}

    The second variable is ranked in exactly the same direction as the first.

  4. Note that the closed formula is not valid here

    tied ranks⇒rs=PMCC of the ranks\text{tied ranks}\Rightarrow r_{s}=\text{PMCC of the ranks}

    The expression 1−6∑d2n(n2−1)1-\frac{6\sum d^{2}}{n\left(n^{2}-1\right)} assumes the ranks are 1,2,…,n1,2,\ldots,n in some order, which averaging destroys; it becomes an approximation only.

  5. Find the mean rank of each variable

    aˉ=bˉ=n+12=112\bar{a}=\bar{b}=\frac{n+1}{2}=\frac{11}{2}

    Averaging tied ranks leaves the total of the ranks equal to n(n+1)2\frac{n\left(n+1\right)}{2}, so the mean rank is unchanged.

  6. Compute the summary statistics of the two sets of ranks

    ∑a2=383,∑b2=7672,∑ab=7512\sum a^{2}=383,\quad\sum b^{2}=\frac{767}{2},\quad\sum ab=\frac{751}{2}

    These three sums are all that is needed for SaaS_{aa}, SbbS_{bb} and SabS_{ab}.

  7. Find SaaS_{aa} and SbbS_{bb}

    Saa=∑a2−naˉ2=1612,Sbb=∑b2−nbˉ2=81S_{aa}=\sum a^{2}-n\bar{a}^{2}=\frac{161}{2},\qquad S_{bb}=\sum b^{2}-n\bar{b}^{2}=81

    The two are no longer equal: tying ranks reduces the spread of the variable that contains the ties.

  8. Find SabS_{ab}

    Sab=∑ab−naˉbˉ=73S_{ab}=\sum ab-n\bar{a}\bar{b}=73

    This measures the extent to which the two sets of ranks vary together.

  9. Apply the product moment correlation coefficient to the ranks

    rs=SabSaaSbb=731612×81r_{s}=\frac{S_{ab}}{\sqrt{S_{aa}S_{bb}}}=\frac{73}{\sqrt{\frac{161}{2}\times 81}}

    This is the definition of Spearman's coefficient, and it is what must be used whenever ranks are tied.

  10. Evaluate the coefficient

    rs=0.904r_{s}=0.904

    The value is rounded only at this final stage.

  11. State the hypotheses

    H0:ρs=0,H1:ρs>0H_{0}:\rho_{s}=0,\qquad H_{1}:\rho_{s}>0

    ρs\rho_{s} is the population rank correlation coefficient; H0H_{0} says there is no association between the length of service and the appraisal rating.

  12. Write down the critical value for this test

    n=10, one-tailed at 1% ⇒ critical value=0.7455n=10,\ \text{one-tailed at }1\%\ \Rightarrow\ \text{critical value}=0.7455

    The critical value is read from the table of critical values of Spearman's coefficient using the sample size and the significance level.

  13. Compare the test statistic with the critical value

    rs=0.904 > 0.7455r_{s}=0.904\ >\ 0.7455

    The value is more extreme than the critical value, so it lies in the critical region.

  14. Quote Spearman's rank correlation coefficient

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

    This is the formula in the booklet: dd is the difference between the two ranks of a single item and nn is the number of items.

  15. Note what dd means

    d=rank of x−rank of yd=\text{rank of }x-\text{rank of }y

    The sign of dd never matters, because only d2d^{2} is used.

  16. State the conclusion in context

    Reject H0: sufficient evidence of positive association\text{Reject }H_{0}\text{: sufficient evidence of positive association}

    The conclusion must be given at the stated significance level and in the context of the length of service and the appraisal rating.

Answer
Reject H0: sufficient evidence of positive association\text{Reject }H_{0}\text{: sufficient evidence of positive association}

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