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H09 Spearman's Rank Correlation

EDEXCEL 1ST0 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Spearman's rank correlation coefficient is given by the formula

rs = 1 - 6Σd² / (n(n²-1))

where n is the number of items ranked and d is the difference between a pair of ranks.

For a particular set of rankings, n = 5 and Σd² = 10.

(a)State what the letter n represents in the formula.(1)

(b)Substitute n = 5 and Σd² = 10 into the formula for rs.(1)

(c)Calculate the value of rs.(1)

(d)Interpret this value of rs in terms of the strength and direction of agreement between the two rankings.(1)

(Total for Question 1 is 4 marks)

At a village show, two judges rank 5 chutneys from 1 (best) to 5 (worst). Their rankings are shown in the table, with some of the differences d already filled in.

Chutney	Judge 1 rank	Judge 2 rank	d
Bramble	1	2	-1
Fig	2	1	1
Onion	3	3
Plum	4	5
Apple	5	4	1

(a)Complete the table by finding the missing values of d for Onion and Plum.(2)

(b)Calculate Σd², the sum of the squared differences.(1)

(c)Use the formula rs = 1 - 6Σd² / (n(n²-1)) to calculate rs.(2)

(d)Interpret the value rs = 0.8, referring to the strength and direction of agreement between the two judges.(1)

(Total for Question 2 is 6 marks)

Two food critics, Priya and Tom, independently score 5 new recipes out of 100. Neither critic sees the other's scores.

Recipe	Priya's score	Tom's score
Lentil dahl	82	78
Fish pie	75	80
Mushroom risotto	90	88
Bean chilli	68	65
Prawn curry	95	91

(a)Rank Priya's scores and Tom's scores separately, giving rank 1 to the highest score in each column.(2)

(b)Find the difference d between the two ranks for each recipe, and calculate Σd².(2)

(c)Calculate rs for this data.(2)

(d)Priya and Tom's editor claims: 'The two critics broadly agree on which recipes are best.' Use your value of rs to comment on this claim.(1)

(Total for Question 3 is 7 marks)

Two examiners, Examiner A and Examiner B, independently mark 6 students' extended writing pieces out of 40.

Student	Examiner A mark	Examiner B mark
Aisha	28	25
Ben	35	33
Carys	19	22
Dev	40	38
Ellis	24	20
Farah	31	30

(a)Rank the 6 students according to Examiner A's marks, and separately according to Examiner B's marks (rank 1 = highest mark).(2)

(b)Calculate the value of d for each student, and hence find Σd².(2)

(c)Calculate rs, giving your answer correct to 3 significant figures.(3)

(d)State, with a reason, whether Examiner A and Examiner B are marking consistently with each other.(1)

(Total for Question 4 is 8 marks)

In a cookery competition, 6 dishes are given a presentation score out of 20 by a judge. Two dishes receive exactly the same presentation score. Each dish has also already been given a taste rank from 1 (best) to 6 (worst) by a separate taste panel.

Dish	Presentation score	Taste rank
Tart	18	2
Terrine	14	4
Souffle	14	3
Consomme	9	6
Roulade	20	1
Galette	11	5

(a)Explain why Terrine and Souffle should be given the same rank for presentation score, instead of one being ranked 3rd and the other 4th.(1)

(b)Complete the ranking of presentation scores for all 6 dishes, giving the tied dishes the mean of the ranks they would otherwise share.(2)

(c)Calculate d for each dish and find Σd².(2)

(d)Calculate rs, giving your answer correct to 3 significant figures.(2)

(e)Interpret this value of rs in the context of the presentation scores and taste ranks.(1)

(Total for Question 5 is 8 marks)

At a regional bake-off, three separate pairs of judges each rank the same set of entries using different criteria. The Spearman's rank correlation coefficient is given for each pair.

Judging pair	rs
Pastry vs Presentation	0.92
Presentation vs Taste	0.15
Pastry vs Taste	-0.68

(a)Write down which pair of criteria shows the weakest association.(1)

(b)Write down which pair of criteria shows the strongest association, regardless of whether it is positive or negative.(2)

(c)Explain what the coefficient rs = -0.68 for Pastry vs Taste tells the competition organisers.(2)

(Total for Question 6 is 5 marks)

A talent show has 6 acts. Two judges, Judge Reyes and Judge Okafor, independently score each act out of 30, without seeing each other's scores until both have finished.

Act	Judge Reyes score	Judge Okafor score
Magic	24	17
Ballad	19	22
Street dance	27	19
Stand-up	14	15
Juggling	22	25
Choir	30	29

(a)(Plan) Give one reason why the two judges should score the acts independently, without seeing each other's scores first.(1)

(b)(Collect) Rank the acts according to Judge Reyes's scores, and separately according to Judge Okafor's scores (rank 1 = highest score).(2)

(c)(Process) Calculate Σd² and hence find rs, giving your answer correct to 3 significant figures.(3)

(d)(Interpret) Judge Reyes says: 'Okafor and I are judging this competition very similarly.' Use your value of rs to evaluate this statement.(2)

(Total for Question 7 is 8 marks)

Two wine critics score 6 wines out of 20. Critic Bianchi favours bold, fruity wines; Critic Laurent favours subtle, dry wines.

Wine	Critic Bianchi score	Critic Laurent score
Merlot	15	12
Chardonnay	11	14
Rioja	19	8
Sauvignon	9	18
Shiraz	13	16
Pinot	17	10

(a)Rank the wines according to Critic Bianchi's scores, and separately according to Critic Laurent's scores (rank 1 = highest score).(2)

(b)Calculate d for each wine and find Σd².(2)

(c)Calculate rs, giving your answer correct to 3 significant figures.(3)

(d)Interpret the value rs = -0.943 in the context of the two critics' scores.(2)

(Total for Question 8 is 9 marks)

A survey ranks 7 cafe branches (P1 to P7) for speed of service and, separately, for customer rating, both from 1 (best) to 7 (worst). The scatter graph shows each branch's speed-of-service rank plotted against its customer-rating rank.

(a)Using the scatter graph, write down the coordinates (speed rank, rating rank) of point P3 and of point P6.(2)

(b)Using the graph, complete the table below to find d and d² for each branch, and hence find Σd².

Branch	Speed rank	Rating rank
P1	1	2
P2	2	1
P3	3	4
P4	4	3
P5	5	5
P6	6	7
P7	7	6

(3)

(c)Calculate rs for this data, giving your answer correct to 3 significant figures.(3)

(d)Interpret this value in the context of the survey.(1)

(Total for Question 9 is 9 marks)

Spearman's rank correlation coefficient and the product moment correlation coefficient (PMCC) are both used to measure the strength of a relationship between two variables.

(a)A dataset contains one extreme outlier that does not fit the general pattern of the rest of the data. Give one reason why Spearman's rank correlation coefficient is less affected by this outlier than the PMCC.(1)

(b)A researcher only has data given as finishing positions in a race (1st, 2nd, 3rd, and so on), with no times recorded. Explain why Spearman's rank correlation coefficient can be calculated for this data but the PMCC cannot be calculated directly.(1)

(c)A scientist measures the exact height (in cm) of 10 plants and the exact number of hours of sunlight they receive each day, and the scatter graph shows a clear straight-line pattern with no outliers. State which coefficient, Spearman's rank correlation coefficient or the PMCC, would normally be calculated directly from this raw data, and give a reason.(2)

(Total for Question 10 is 4 marks)

For each statement about Spearman's rank correlation coefficient, write down whether it is True or False.

(i)Spearman's rank correlation coefficient, rs, can only take values between -1 and 1 inclusive.(1)

(ii)If two rankings share tied values, the ranks used in the formula rs = 1 - 6Σd²/(n(n²-1)) should be the mean of the positions the tied values would otherwise occupy.(1)

(iii)A value of rs = -0.9 shows a weaker association than a value of rs = 0.4.(1)

(iv)If n increases but Σd² stays exactly the same, the value of rs calculated from the formula must also stay exactly the same.(1)

(Total for Question 11 is 4 marks)

At a regional science fair, two moderators, Moderator Chen and Moderator Osei, independently score 8 projects out of 50. Ties occur in both moderators' scores.

Project	Moderator Chen score	Moderator Osei score
P1	45	39
P2	38	44
P3	45	33
P4	30	28
P5	27	30
P6	41	47
P7	33	28
P8	38	41

(a)Chen's scores for P1 and P3 are tied, and so are the scores for P2 and P8. Complete the ranking of Chen's scores for all 8 projects, using mean ranks for the tied pairs.(2)

(b)Osei's scores for P4 and P7 are tied. Complete the ranking of Osei's scores for all 8 projects.(2)

(c)Calculate d for each project and find Σd².(2)

(d)Calculate rs, giving your answer correct to 3 significant figures.(3)

(e)Interpret this value of rs in the context of the two moderators' scoring.(1)

(Total for Question 12 is 10 marks)

For a set of n = 6 pairs of rankings, Spearman's rank correlation coefficient is rs = 0.8.

(a)Show that Σd² = 7 for this data.(3)

(b)A different pair of rankings, also with n = 6, gives Σd² = 70, the largest value Σd² can take for n = 6. State the value of rs, and explain what this tells you about the two rankings.(2)

(Total for Question 13 is 5 marks)

A student calculates Spearman's rank correlation coefficient for 5 pairs of rankings, using Σd² = 8. Their working is shown below.

rs = 1 - (6 x 8) / 5² = 1 - 48/25 = 1 - 1.92 = -0.92

(a)Identify the mistake in the student's working.(2)

(b)Work out the correct value of rs.(2)

(c)Explain how the student's mistake affected their conclusion about the two rankings.(1)

(Total for Question 14 is 5 marks)

A gymnastics club has 7 gymnasts. Coach Ito ranks the gymnasts by technical difficulty score (out of 60), and Coach Novak ranks them by artistic presentation score (out of 60), in order to help choose a single squad captain. Coach Ito's technical scores include a tie.

Gymnast	Technical score (Ito)	Artistic score (Novak)
G1	52	48
G2	47	53
G3	52	44
G4	38	36
G5	44	50
G6	49	39
G7	41	46

(a)Complete the ranking of the gymnasts' technical scores, using a mean rank for the tied pair.(2)

(b)Rank the gymnasts' artistic scores (no ties occur here), then calculate Σd².(3)

(c)Calculate rs for this data.(3)

(d)The club wants to use these two coaches' rankings alone to choose a squad captain. Evaluate whether this is a sound approach, using your value of rs and the size of the sample to justify your answer.(4)

(Total for Question 15 is 12 marks)

Mark scheme · H09 Spearman's Rank Correlation

Question 1

(a) B1 the number of items/pairs being ranked oe
(a) Answer: The number of items (pairs of ranks) being compared.
(b) M1 rs = 1 - (6 x 10) / (5 x (5²-1)), oe
(b) Answer: rs = 1 - (6 x 10) / (5 x 24)
(c) A1 0.5 cao
(c) Answer: 0.5
(d) B1 a weak to moderate positive association oe
(d) Answer: A weak to moderate positive association between the two rankings.

Question 2

(a) B1 Onion d = 0
(a) B1 Plum d = -1
(a) Answer: Onion: d = 0; Plum: d = -1
(b) B1 4 cao
(b) Answer: 4
(c) M1 rs = 1 - (6 x 4) / (5 x 24), oe
(c) A1 0.8 cao
(c) Answer: 0.8
(d) B1 a strong positive agreement between the two judges' rankings oe
(d) Answer: A strong positive agreement - the two judges rank the chutneys in a very similar order.

Question 3

(a) B1 Priya's ranks 3, 4, 2, 5, 1 cao (Lentil dahl to Prawn curry)
(a) B1 Tom's ranks 4, 3, 2, 5, 1 cao (Lentil dahl to Prawn curry)
(a) Answer: Priya: 3, 4, 2, 5, 1. Tom: 4, 3, 2, 5, 1 (Lentil dahl, Fish pie, Mushroom risotto, Bean chilli, Prawn curry).
(b) M1 d values -1, 1, 0, 0, 0 oe
(b) A1 Σd² = 2 cao
(b) Answer: Σd² = 2
(c) M1 rs = 1 - (6 x 2) / (5 x 24), oe
(c) A1 0.9 cao
(c) Answer: 0.9
(d) B1 rs = 0.9 is close to 1, supporting the claim that the two critics broadly agree oe
(d) Answer: The claim is supported - rs = 0.9 is close to the maximum of 1, showing the critics broadly agree on the ranking of the recipes.

Question 4

(a) B1 Examiner A ranks: Aisha 4, Ben 2, Carys 6, Dev 1, Ellis 5, Farah 3
(a) B1 Examiner B ranks: Aisha 4, Ben 2, Carys 5, Dev 1, Ellis 6, Farah 3
(a) Answer: A: 4, 2, 6, 1, 5, 3. B: 4, 2, 5, 1, 6, 3 (Aisha, Ben, Carys, Dev, Ellis, Farah).
(b) M1 d values 0, 0, 1, 0, -1, 0 oe
(b) A1 Σd² = 2 cao
(b) Answer: Σd² = 2
(c) M1 n(n²-1) = 6 x 35 = 210 oe
(c) M1 rs = 1 - (6 x 2) / 210, oe
(c) A1 awrt 0.943
(c) Answer: rs = 0.943 (3sf)
(d) B1 Yes - rs is very close to 1, so the two examiners are ranking the students in almost the same order oe
(d) Answer: Yes - rs = 0.943 is very close to 1, so the two examiners' rankings agree closely.

Question 5

(a) B1 they have exactly the same presentation score (14), so there is no fair way to place one above the other oe
(a) Answer: Terrine and Souffle both scored 14, so they are tied - neither can fairly be placed above the other.
(b) M1 Terrine and Souffle would occupy positions 3 and 4, so both are given rank 3.5 oe
(b) A1 full set of presentation ranks: Tart 2, Terrine 3.5, Souffle 3.5, Consomme 6, Roulade 1, Galette 5
(b) Answer: Presentation ranks: Roulade 1, Tart 2, Terrine 3.5, Souffle 3.5, Galette 5, Consomme 6.
(c) M1 d values 0, -0.5, 0.5, 0, 0, 0 oe
(c) A1 Σd² = 0.5 cao
(c) Answer: Σd² = 0.5
(d) M1 rs = 1 - (6 x 0.5) / (6 x 35), oe
(d) A1 awrt 0.986
(d) Answer: rs = 0.986 (3sf)
(e) B1 a very strong positive association - dishes that look good tend to also taste good (and vice versa) oe
(e) Answer: A very strong positive association: dishes with a high presentation score also tend to have a good taste rank.

Question 6

(a) B1 Presentation vs Taste cao
(a) Answer: Presentation vs Taste
(b) M1 compares the size of each coefficient ignoring sign (0.92, 0.15, 0.68)
(b) A1 Pastry vs Presentation cao
(b) Answer: Pastry vs Presentation
(c) B1 identifies a fairly strong negative association: entries that rank highly for pastry tend to rank poorly for taste, and vice versa
(c) B1 a sensible conclusion for the organisers, e.g. good pastry technique does not reliably predict good taste, so both criteria need to be judged separately oe
(c) Answer: Entries that score well for pastry technique tend to score less well for taste, and vice versa; the organisers should judge and score both criteria independently rather than relying on one to predict the other.

Question 7

(a) B1 so that neither judge is influenced by the other's opinion, keeping the two sets of scores independent oe
(a) Answer: So that neither judge's score is influenced by seeing the other's opinion first.
(b) B1 Reyes ranks: Magic 3, Ballad 5, Street dance 2, Stand-up 6, Juggling 4, Choir 1
(b) B1 Okafor ranks: Magic 5, Ballad 3, Street dance 4, Stand-up 6, Juggling 2, Choir 1
(b) Answer: Reyes: 3, 5, 2, 6, 4, 1. Okafor: 5, 3, 4, 6, 2, 1 (Magic, Ballad, Street dance, Stand-up, Juggling, Choir).
(c) M1 d values -2, 2, -2, 0, 2, 0 oe
(c) A1 Σd² = 16 cao
(c) A1 rs = awrt 0.543
(c) Answer: Σd² = 16, rs = 0.543 (3sf)
(d) B1 rs = 0.543 is positive, so there is some genuine agreement between the two judges oe
(d) B1 however 0.543 is only a moderate value (well below 1), so the statement overstates how closely they agree oe
(d) Answer: Reyes's statement is only partly justified - rs = 0.543 shows a positive but moderate association, so the judges agree to some extent but not as closely as 'very similarly' suggests.

Question 8

(a) B1 Bianchi ranks: Merlot 3, Chardonnay 5, Rioja 1, Sauvignon 6, Shiraz 4, Pinot 2
(a) B1 Laurent ranks: Merlot 4, Chardonnay 3, Rioja 6, Sauvignon 1, Shiraz 2, Pinot 5
(a) Answer: Bianchi: 3, 5, 1, 6, 4, 2. Laurent: 4, 3, 6, 1, 2, 5 (Merlot, Chardonnay, Rioja, Sauvignon, Shiraz, Pinot).
(b) M1 d values -1, 2, -5, 5, 2, -3 oe
(b) A1 Σd² = 68 cao
(b) Answer: Σd² = 68
(c) M1 n(n²-1) = 6 x 35 = 210 oe
(c) M1 rs = 1 - (6 x 68) / 210, oe
(c) A1 awrt -0.943
(c) Answer: rs = -0.943 (3sf)
(d) B1 a strong negative association oe
(d) B1 context: wines Bianchi rates highly tend to be rated poorly by Laurent, and vice versa - the two critics strongly disagree oe
(d) Answer: A strong negative association - wines that Bianchi scores highly tend to be scored poorly by Laurent, and vice versa; the two critics strongly disagree.

Question 9

(a) B1 P3 = (3, 4)
(a) B1 P6 = (6, 7)
(a) Answer: P3 = (3, 4); P6 = (6, 7)
(b) M1 d values -1, 1, -1, 1, 0, -1, 1 oe
(b) A1 d² values 1, 1, 1, 1, 0, 1, 1
(b) A1 Σd² = 6 cao
(b) Answer: Σd² = 6
(c) M1 n(n²-1) = 7 x 48 = 336 oe
(c) M1 rs = 1 - (6 x 6) / 336, oe
(c) A1 awrt 0.893
(c) Answer: rs = 0.893 (3sf)
(d) B1 a strong positive association - branches with fast service also tend to receive good customer ratings oe
(d) Answer: A strong positive association - branches that are fast tend to also be rated highly by customers.

Question 10

(a) B1 Spearman's rank correlation is calculated using the ranks (order) of the data rather than the actual values, so an extreme value only occupies the highest or lowest rank rather than distorting the calculation with its size oe
(a) Answer: Because Spearman's rank uses only the order (ranks) of the data, not the actual values, so one extreme value cannot distort the coefficient as much as it could distort the PMCC.
(b) B1 the PMCC needs actual numerical data values to work with, but finishing positions are already ranks/ordinal data, so only Spearman's rank correlation coefficient can be applied directly oe
(b) Answer: The PMCC requires actual numerical measurements, but finishing positions are already just an order (ranks), so Spearman's rank correlation coefficient can be used directly, whereas the PMCC cannot.
(c) B1 the PMCC
(c) B1 reason: the data is already precise numerical data with a linear pattern, so using the PMCC directly makes full use of the actual values rather than converting to ranks and losing information oe
(c) Answer: The PMCC - the data is precise numerical measurements with a linear pattern, so using the actual values (rather than converting to ranks) makes full use of the information available.

Question 11

(i) B1 True cao
(i) Answer: True
(ii) B1 True cao
(ii) Answer: True
(iii) B1 False cao
(iii) Answer: False
(iv) B1 False cao
(iv) Answer: False

Question 12

(a) M1 P1 and P3 both share rank 1.5 (mean of 1 and 2); P2 and P8 both share rank 4.5 (mean of 4 and 5)
(a) A1 full set of Chen ranks: P1 1.5, P2 4.5, P3 1.5, P4 7, P5 8, P6 3, P7 6, P8 4.5
(a) Answer: Chen ranks: P1 1.5, P2 4.5, P3 1.5, P4 7, P5 8, P6 3, P7 6, P8 4.5
(b) M1 P4 and P7 both share rank 7.5 (mean of 7 and 8)
(b) A1 full set of Osei ranks: P1 4, P2 2, P3 5, P4 7.5, P5 6, P6 1, P7 7.5, P8 3
(b) Answer: Osei ranks: P1 4, P2 2, P3 5, P4 7.5, P5 6, P6 1, P7 7.5, P8 3
(c) M1 d values -2.5, 2.5, -3.5, -0.5, 2, 2, -1.5, 1.5 oe
(c) A1 Σd² = 37.5 cao
(c) Answer: Σd² = 37.5
(d) M1 n(n²-1) = 8 x 63 = 504 oe
(d) M1 rs = 1 - (6 x 37.5) / 504, oe
(d) A1 awrt 0.554
(d) Answer: rs = 0.554 (3sf)
(e) B1 a moderate positive association - the two moderators show some agreement in how they rank the projects, but not especially strong agreement oe
(e) Answer: A moderate positive association - the moderators agree on the general ranking of the projects to some extent, but not very strongly.

Question 13

(a) M1 rearranges the formula to Σd² = n(n²-1)(1-rs) / 6, oe
(a) M1 substitutes n = 6 and rs = 0.8: Σd² = (6 x 35 x 0.2) / 6, oe
(a) A1 Σd² = 7 cao, correctly shown
(a) Answer: Σd² = 7
(b) B1 rs = -1 cao
(b) B1 this means the two rankings are exact reverses of one another (perfect disagreement) oe
(b) Answer: rs = -1; the two rankings are exact opposites of one another - whichever item is ranked highest by one ranking is ranked lowest by the other, and so on.

Question 14

(a) B1 the denominator used is n² instead of the correct n(n²-1)
(a) B1 states the correct denominator: n(n²-1) = 5 x 24 = 120, not n² = 25
(a) Answer: The student used n² (=25) instead of the correct denominator n(n²-1) (=120).
(b) M1 rs = 1 - (6 x 8) / 120, oe
(b) A1 0.6 cao
(b) Answer: 0.6
(c) B1 the student's incorrect value (-0.92) suggested a strong negative association/disagreement, but the correct value (0.6) actually shows a moderate positive association/agreement - the error leads to the opposite conclusion oe
(c) Answer: The student's error changed both the size and the sign of the result: -0.92 wrongly suggests the two rankings strongly disagree, when the correct value of 0.6 actually shows a moderate positive agreement between them.

Question 15

(a) M1 G1 and G3 share rank 1.5 (mean of 1 and 2)
(a) A1 full set of technical ranks: G1 1.5, G2 4, G3 1.5, G4 7, G5 5, G6 3, G7 6
(a) Answer: Technical ranks: G1 1.5, G2 4, G3 1.5, G4 7, G5 5, G6 3, G7 6
(b) B1 artistic ranks: G1 3, G2 1, G3 5, G4 7, G5 2, G6 6, G7 4
(b) M1 d values -1.5, 3, -3.5, 0, 3, -3, 2 oe
(b) A1 Σd² = 45.5 cao
(b) Answer: Artistic ranks: G1 3, G2 1, G3 5, G4 7, G5 2, G6 6, G7 4. Σd² = 45.5
(c) M1 n(n²-1) = 7 x 48 = 336 oe
(c) M1 rs = 1 - (6 x 45.5) / 336, oe
(c) A1 0.1875 or exact equivalent, e.g. 3/16
(c) Answer: rs = 0.1875
(d) B1 rs = 0.1875 is positive but small, showing only a weak association between the technical and artistic rankings
(d) B1 explains that this means the two coaches do not agree closely on the order of the gymnasts, so relying on either ranking alone risks an unfair/inconsistent decision
(d) B1 notes the sample is small (n=7), so a single tie and a few swapped ranks have a large effect on rs, making the result less reliable
(d) B1 reaches a sensible recommendation, e.g. the club should combine both criteria (or bring in further judges) rather than relying on either coach's ranking alone
(d) Answer: This is not a sound approach: rs = 0.1875 is only weakly positive, showing the two coaches' rankings do not closely agree, and with just 7 gymnasts the result is easily affected by a single tie or swapped rank. The club should combine both technical and artistic scores (or involve further judges) rather than relying on one ranking alone.

Spearman's Rank Correlation - Worksheets, Questions and Revision