Edexcel International A Level Maths · S1 · Statistics 1
Correlation and regression
28 free, high-quality past-paper-style questions on Correlation and regression for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 240 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
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A fish farm measures the length x cm and the mass y g of 10 trout. The results are summarised by Σx = 308, Σy = 2980, Σx² = 9668, Σy² = 955750, Σxy = 95270.
(a)
Find Sxx, Syy and Sxy.
(3)
(b)
Calculate the product moment correlation coefficient between the length and the mass of these trout.
(2)
(c)
Interpret your value of r in context.
(1)
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Answer
Mark
Notes
Part (a)
Sxx = 9668 − 308²⁄10 = 181.6
B1
Syy = 955750 − 2980²⁄10 = 67710
B1
Sxy = 95270 − 308 × 2980⁄10 = 3486
B1
Part (b)
r = 3486⁄√(181.6 × 67710), ft their values
M1
= 0.994
A1
Part (c)
strong positive (linear) correlation: longer trout tend to have greater mass
B1
Exam tip: An interpretation needs the direction, the strength and the two variables named in context — “strong positive” on its own is not enough.
A courier records the distance x km and the time y minutes for 12 deliveries made by bicycle. The distances range from 3 km to 25 km. Σx = 157, Σy = 372, Sxx = 550.917 and Sxy = 1033.
(a)
Find the equation of the regression line of y on x in the form y = a + bx, giving a and b to 3 significant figures.
(3)
(b)
Interpret the value of b in context.
(1)
(c)
Estimate the time for a delivery of 14 km.
(1)
(d)
Give a reason why the line should not be used to estimate the time for a 60 km delivery.
(1)
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Answer
Mark
Notes
Part (a)
b = 1033⁄550.917 = 1.875…
M1
a = 372⁄12 − 1.875… × 157⁄12
M1
y = 6.47 + 1.88x
A1
Part (b)
each extra kilometre adds about 1.88 minutes to the delivery time
B1
Part (c)
6.468 + 1.875 × 14 = 32.7 minutes, ft their line
B1
Part (d)
60 km is far outside the range of the data (3 to 25 km): extrapolation, so the estimate is unreliable
B1
Common error: Rounding b to 3 s.f. before calculating a. Keep the full value of b in the calculator; only the final coefficients are rounded.
An engineer records the wind speed x m s−1 and the power output y kW of a wind turbine at 20 different times. The wind speeds recorded range from 4 m s−1 to 12 m s−1. The product moment correlation coefficient is 0.962 and the regression line of y on x is y = −62.4 + 31.5x.
(a)
State, with a reason, which variable is the explanatory variable.
(1)
(b)
Interpret the value 31.5 in context.
(1)
(c)
Estimate the power output when the wind speed is 9 m s−1, and comment on the reliability of your estimate.
(2)
(d)
Use the line to find the power output predicted for a wind speed of 1 m s−1, and explain why this value should not be trusted.
(1)
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Answer
Mark
Notes
Part (a)
the wind speed, because the wind drives the turbine: the power output depends on the wind speed, not the reverse
B1
Part (b)
for each increase of 1 m s−1 in wind speed, the power output increases by about 31.5 kW
B1
Part (c)
−62.4 + 31.5 × 9 = 221.1 kW
B1
reliable: 9 is within the range of the data (interpolation) and r = 0.962 shows strong positive correlation
B1
Part (d)
−30.9 kW, which is impossible (negative power); 1 m s−1 is outside the data, so this is extrapolation
B1
Exam tip: When commenting on reliability, give two reasons: whether the value is inside the data range, and what r says about the strength of the correlation.
A forester measures the age x years and the trunk diameter y mm of 9 trees of the same species. The data are coded using u = (x − 40)⁄5 and v = (y − 120)⁄3, giving Suu = 52, Svv = 98 and Suv = 61.5.
(a)
Calculate the product moment correlation coefficient between u and v.
(2)
(b)
Write down the product moment correlation coefficient between x and y, giving a reason.
(1)
(c)
Find the values of Sxx and Sxy.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
61.5⁄√(52 × 98)
M1
= 0.862
A1
Part (b)
0.862: subtracting a constant and dividing by a positive constant do not change the correlation coefficient
B1
Part (c)
Sxx = 5² × 52 = 1300
B1
Sxy = 5 × 3 × 61.5 = 922.5
B1
Common error: Multiplying Sxx by 5 instead of 5². Sxx is a sum of squared deviations, so scaling x by 5 scales Sxx by 25; Sxy scales by 5 × 3.
Figure 1 shows four scatter diagrams, A, B, C and D, each for a different set of paired data. The product moment correlation coefficients for diagrams A, B and C, in a different order, are −0.55, 0.98 and −0.02.
(a)
Match each of the diagrams A, B and C to its product moment correlation coefficient.
(2)
(b)
For diagram D the product moment correlation coefficient is exactly 0. A student concludes that there is no relationship between the two variables in diagram D. Comment on this conclusion.
(1)
(c)
State, with a reason, the diagram for which the regression line of y on x would give the most reliable estimates.
(2)
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Answer
Mark
Notes
Part (a)
A: 0.98
B1
B: −0.55 and C: −0.02, both correct
B1
Part (b)
wrong: the points lie exactly on a curve, so there is a strong (non-linear) relationship; r = 0 only means there is no linear correlation
B1
Part (c)
diagram A
B1
its points lie close to a straight line: strong linear correlation, r = 0.98 is close to 1
B1
Common error: Reading r = 0 as “no relationship”. The product moment correlation coefficient measures only how close the points are to a straight line; a perfect curve can give r = 0.
A cyclist records the gradient, x %, and her average speed, y km h−1, on 8 different hill climbs. Figure 1 shows the data and the regression line of y on x. The points P and Q lie on the regression line at gridline intersections.
(a)
State, with a reason, which variable is the explanatory variable.
(1)
(b)
Use Figure 1 to find the equation of the regression line in the form y = a + bx.
(3)
(c)
Interpret the value of b in context.
(1)
(d)
Estimate her average speed on a hill of gradient 6 %.
(1)
(e)
Explain why the line should not be used to estimate her average speed on a hill of gradient 18 %.
(1)
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Answer
Mark
Notes
Part (a)
the gradient of the hill: the steepness affects the speed, not the other way round
B1
Part (b)
P(3, 28) and Q(10, 14): b = (14 − 28) ÷ (10 − 3) = −2
M1
a = 28 − (−2)(3) = 34
M1
y = 34 − 2x
A1
Part (c)
for each extra 1 % of gradient, the average speed falls by about 2 km h−1
B1
Part (d)
34 − 2 × 6 = 22 km h−1, ft their line
B1
Part (e)
18 % is outside the range of the data (1 % to 11 %), so this is extrapolation; the line even gives −2 km h−1, which is impossible
B1
Exam tip: To read a line from a graph, use two points on the line that sit exactly on gridline intersections and are far apart. Do not use data points: they are not on the line.
A hotel records the number of guests, x, and the mass of laundry, y kg, on 8 days. Figure 1 shows the data together with the regression line of y on x, which has equation y = 6 + 1.5x.
(a)
Describe the correlation shown in Figure 1.
(1)
(b)
Explain why the number of guests is the explanatory variable.
(1)
(c)
Use Figure 1 to write down the range of the numbers of guests in the data. Hence estimate the mass of laundry on a day with 40 guests, and comment on the reliability of your estimate.
(3)
(d)
Point A shows one of the days. Find how much more laundry there was on that day than the line predicts.
(1)
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Answer
Mark
Notes
Part (a)
positive correlation (strong): days with more guests tend to produce more laundry
B1
Part (b)
the amount of laundry depends on how many guests there are, not the other way round
B1
Part (c)
from 20 to 64 guests
B1
6 + 1.5 × 40 = 66 kg
B1
reliable: 40 lies inside the range of the data (interpolation) and the points lie close to the line
B1
Part (d)
A is (52, 90); line gives 6 + 1.5 × 52 = 84, so 6 kg more
B1
Exam tip: “Comment on reliability” needs a reason: say whether the value is inside the range of the data (interpolation) and how close the points lie to the line.
A biologist measures the length, x mm, and the mass, y g, of 12 shells: 6 from species P and 6 from species Q. Figure 1 shows the data. For all 12 shells together the product moment correlation coefficient is 0.946. For the 6 shells of species Q: Σx = 266, Σy = 92, Σx² = 11848, Σy² = 1418, Σxy = 4067.
(a)
Describe the pattern shown by the points for species Q alone.
(1)
(b)
Explain why the value 0.946 gives a misleading picture of the relationship between length and mass for shells of one species.
(2)
(c)
Calculate the product moment correlation coefficient for the shells of species Q.
(2)
(d)
Interpret your value in context.
(1)
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Answer
Mark
Notes
Part (a)
no clear positive trend: if anything slight negative correlation (longer shells of Q are not heavier)
B1
Part (b)
the high value comes from the two separate groups: species Q shells are all longer and heavier than species P shells
B1
within each species there is little or no correlation, so 0.946 does not describe either species
weak to moderate negative correlation: for species Q, longer shells do not tend to be heavier
B1
Common error: Trusting a single correlation coefficient when a scatter diagram shows two separate groups. Always look at the diagram: two clusters can produce a large value of r even when neither group shows a trend.
Six adults who are learning to touch-type each record the number of hours of practice they have done, x, and their typing speed in words per minute, y.
hours of practice, x
2
4
5
7
9
12
typing speed, y
28
33
37
40
46
52
Figure: scatter diagram of the typing data
(a)
Find Sxx, Syy and Sxy.
(4)
(b)
Calculate the product moment correlation coefficient.
(2)
(c)
Find the equation of the regression line of y on x.
(3)
(d)
Estimate the typing speed of an adult in this group after 10 hours of practice.
Exam tip: Write down the five sums before using the formulae. If one sum is wrong, the working shows exactly where, and the later parts can still follow through.
An estate agent records the floor area x m² and the selling price £y thousand of 10 flats. The data are coded using u = (x − 100)⁄10 and v = (y − 300)⁄20, giving n = 10, Σu = 19.2, Σv = 18.45, Σu² = 129.8, Σuv = 135.955.
(a)
Find Suu and Suv.
(2)
(b)
Find the equation of the regression line of v on u.
(3)
(c)
Hence find the equation of the regression line of y on x, giving your answer in the form y = a + bx.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
Suu = 129.8 − 19.2²⁄10 = 92.936
B1
Suv = 135.955 − 19.2 × 18.45⁄10 = 100.531
B1
Part (b)
gradient = 100.531⁄92.936 = 1.082…
M1
intercept = 1.845 − 1.082… × 1.92 = −0.2319
M1
v = −0.232 + 1.08u
A1
Part (c)
(y − 300)⁄20 = −0.2319 + 1.082 × (x − 100)⁄10
M1
y = 300 + 20 × (−0.2319) + 2 × 1.082(x − 100)
M1
y = 79.0 + 2.16x
A1
Exam tip: Substitute the coding expressions for u and v into the coded line, then make y the subject. Check: the line must pass through (x̄, ȳ) = (119.2, 336.9).
A dealer records the age x years and the price £y thousand of 10 used cars of the same model, one of each age from 1 to 10 years. The data give Σx = 55, Σy = 142, Σx² = 385, Σy² = 2294.76, Σxy = 648.5. It is later found that the price of the 6-year-old car was entered as 21.0 when it should have been 12.0.
(a)
Find the correct values of Σy, Σy² and Σxy.
(3)
(b)
Calculate the correct product moment correlation coefficient.
(4)
(c)
The incorrect data gave r = −0.874. Comment on the effect of the error.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
Σy = 142 − 21.0 + 12.0 = 133
B1
Σy² = 2294.76 − 21.0² + 12.0² = 1997.76
B1
Σxy = 648.5 − 6 × 21.0 + 6 × 12.0 = 594.5
B1
Part (b)
Sxx = 385 − 55²⁄10 = 82.5
B1
Syy = 1997.76 − 133²⁄10 = 228.86 and Sxy = 594.5 − 55 × 133⁄10 = −137
M1
r = −137⁄√(82.5 × 228.86)
M1
= −0.997
A1
Part (c)
one wrong value weakened the correlation a lot: the corrected data show strong negative correlation (older cars tend to be cheaper), close to a straight line
B1
Common error: Correcting Σy but forgetting Σy² and Σxy. Every sum that contains the wrong value must be corrected; Σx and Σx² do not change.
In an experiment, different loads x newtons are hung from a spring and its length y cm is measured. Eight loads from 0.5 N to 4.0 N are used. Σx = 18, Σy = 132.1, Sxx = 10.5 and Sxy = 9.075.
(a)
Explain why the load is the explanatory variable.
(1)
(b)
Find the equation of the regression line of y on x.
(3)
(c)
Interpret the values of a and b in this context.
(2)
(d)
Give a reason why the line should not be used to estimate the length of the spring with a load of 40 N.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the load is chosen (controlled) by the experimenter, and the length of the spring depends on it
B1
Part (b)
b = 9.075⁄10.5 = 0.8643…
M1
a = 16.5125 − 0.8643… × 2.25
M1
y = 14.6 + 0.864x
A1
Part (c)
a: the length of the spring with no load (its natural length) is about 14.6 cm
B1
b: the spring stretches by about 0.864 cm for each extra newton of load
B1
Part (d)
40 N is far outside the loads used (extrapolation): the spring may stop stretching in a straight-line way (or be permanently damaged) at large loads
B1
Exam tip: Here x = 0 is meaningful and close to the data, so a has a sensible interpretation. That is not always true — always ask whether x = 0 makes sense before interpreting a.
For 15 farms in one region, a researcher records the rainfall x mm during the growing season and the wheat yield y tonnes per hectare. The rainfall ranged from 38 mm to 91 mm. Sxx = 1600, Syy = 4.84, x̄ = 62.5, ȳ = 4.2, and the product moment correlation coefficient is exactly 0.6.
(a)
Show that Sxy = 52.8.
(2)
(b)
Find the equation of the regression line of y on x.
(3)
(c)
Use your line to estimate the yield for a farm with 70 mm of rainfall, and comment on the reliability of the estimate.
The boiling point of water, y °C, is measured at 8 places at different altitudes x metres. The data are coded using u = (2000 − x)⁄400 and v = 10(y − 90), giving Suu = 42, Svv = 7288, Suv = 552, ū = 1.5 and v̄ = 53.5.
(a)
Calculate the product moment correlation coefficient between u and v.
(2)
(b)
Write down the product moment correlation coefficient between x and y. Explain your answer.
(2)
(c)
Find the regression line of v on u, and hence the regression line of y on x.
(4)
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Answer
Mark
Notes
Part (a)
552⁄√(42 × 7288)
M1
= 0.998
A1
Part (b)
−0.998
B1
u decreases as x increases (the coding multiplies x by a negative number), so the sign changes; the size is unchanged
Common error: Assuming coding never changes r. Shifting and dividing by a positive number leave it unchanged, but u = (2000 − x)/400 reverses the order of the x values, so r changes sign.
A post office records the number of staff on duty, x, and the mean waiting time of customers, y minutes, for 6 sessions. Figure 1 shows the data; every point lies on a gridline intersection. You may use Σx² = 139 and Σxy = 266.
(a)
Use Figure 1 to find Σx and Σy.
(2)
(b)
Find Sxx and Sxy.
(2)
(c)
Find the equation of the regression line of y on x, giving your answer in the form y = a + bx.
(3)
(d)
Interpret the value of b in context.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
Σx = 2 + 3 + 4 + 5 + 6 + 7 = 27
B1
Σy = 17 + 15 + 12 + 11 + 7 + 6 = 68
B1
Part (b)
Sxx = 139 − 27²⁄6 = 17.5
B1
Sxy = 266 − 27 × 68⁄6 = −40
B1
Part (c)
b = −40⁄17.5 = −2.2857…
M1
a = 68⁄6 − (−2.2857…) × 27⁄6 = 21.619…
M1
y = 21.6 − 2.29x
A1
Part (d)
each extra member of staff on duty reduces the mean waiting time by about 2.29 minutes
B1
Exam tip: Before using the formulae, list the coordinates you read from the graph. Then Σx and Σy are easy to check, and a misread point is easy to spot.
A gardener applies x kg of fertiliser to each of 6 plots and records the yield of tomatoes, y kg. Figure 1 shows the data, with every point on a gridline intersection. Two students each drew a line on the diagram, labelled L1 and L2. One of the lines is the regression line of y on x.
(a)
Use Figure 1 to find x̄ and ȳ.
(2)
(b)
Explain why L2 cannot be the regression line of y on x, and show that L1 could be.
(2)
(c)
L1 passes through gridline intersections at x = 2 and x = 6. Use Figure 1 to find the equation of L1 in the form y = a + bx.
(2)
(d)
Estimate the yield from a plot given 4.5 kg of fertiliser.
(1)
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Answer
Mark
Notes
Part (a)
x̄ = 24⁄6 = 4
B1
ȳ = 132⁄6 = 22
B1
Part (b)
the regression line must pass through (x̄, ȳ) = (4, 22)
B1
at x = 4, L2 gives 19 (not 22), but L1 passes through (4, 22)
B1
Part (c)
L1 passes through (2, 16) and (6, 28): gradient = 12⁄4 = 3
M1
y = 10 + 3x
A1
Part (d)
10 + 3 × 4.5 = 23.5 kg, ft their line
B1
Exam tip: Every least-squares regression line passes through the mean point (x̄, ȳ). This is a quick check on any line you calculate or are given.
A chemist measures the time, y seconds, for a reaction to finish at 8 different temperatures, x °C, from 22.5 °C to 47.5 °C. The data are coded using u = x − 20⁄5 and v = y − 60⁄4. Figure 1 shows the coded data and the regression line of v on u. The ringed points lie on the line at gridline intersections. The product moment correlation coefficient between u and v is −0.979.
(a)
Use Figure 1 to find the equation of the regression line of v on u.
(2)
(b)
Hence find the equation of the regression line of y on x, in the form y = a + bx.
(3)
(c)
Estimate the time for the reaction at 32 °C.
(1)
(d)
Write down the product moment correlation coefficient between x and y, giving a reason.
(1)
(e)
Interpret the gradient of the regression line of y on x in context.
(1)
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Answer
Mark
Notes
Part (a)
gradient = 3 − 6⁄4 − 2 = −1.5
M1
v = 9 − 1.5u
A1
Part (b)
y − 60⁄4 = 9 − 1.5 × x − 20⁄5
M1
y = 60 + 36 − 1.2(x − 20)
M1
y = 120 − 1.2x
A1
Part (c)
120 − 1.2 × 32 = 81.6 s, ft their line
B1
Part (d)
−0.979: the coding subtracts constants and divides by positive numbers, which does not change r
B1
Part (e)
for each 1 °C rise in temperature, the reaction takes about 1.2 seconds less
B1
Common error: Reading the coded gradient −1.5 as the gradient of y on x. Substitute the coding into the coded line first: here y changes by 4 × (−1.5) ÷ 5 = −1.2 for each 1 °C.
A student measures the period, y seconds, of a pendulum for 10 different lengths, x cm, from 10 cm to 100 cm. Figure 1 shows the data. Σx = 550, Σy = 14.25, Sxx = 8250, Syy = 1.81905 and Sxy = 121.15.
(a)
Calculate the product moment correlation coefficient.
(2)
(b)
Find the equation of the regression line of y on x, in the form y = a + bx.
(3)
(c)
Use your line to estimate the period for a length of 10 cm, and compare it with the value shown in Figure 1.
(1)
(d)
The student says that, because the value of the correlation coefficient is close to 1, a straight line is a good model for these data. Use Figure 1 to comment on this claim.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
r = 121.15⁄√(8250 × 1.81905)
M1
= 0.989
A1
Part (b)
b = 121.15⁄8250 = 0.014684…
M1
a = 1.425 − 0.014684… × 55 = 0.6173…
M1
y = 0.617 + 0.0147x
A1
Part (c)
0.6173 + 0.014684 × 10 = 0.764 s, which is greater than the measured value of 0.63 s
B1
Part (d)
the points lie on a curve: the period increases more and more slowly as the length increases
B1
so a straight line is not suitable even though r is close to 1: it overestimates at the ends of the range and underestimates in the middle (e.g. 1.35 s against the measured 1.42 s at 50 cm)
B1
Common error: Treating a value of r close to 1 as proof that the relationship is linear. A smooth curve can also give a high r; the scatter diagram shows whether a straight line really fits.
An office tests two printers, A and B. For each print job it records the number of pages, x, and the time taken, y seconds. Figure 1 shows the data for each printer and the regression line of y on x for each printer, drawn over the range of that printer's data. The ringed points lie on the lines at gridline intersections.
(a)
Use Figure 1 to find the equation of each regression line in the form y = a + bx.
(4)
(b)
Compare the gradients of the two lines in context.
(1)
(c)
Find the number of pages for which the two lines predict the same time.
(2)
(d)
Comment on the reliability of your answer to the previous part.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
A: gradient = 32 − 22⁄8 − 4 = 2.5
M1
y = 12 + 2.5x
A1
B: gradient = 48 − 36⁄12 − 4 = 1.5
M1
y = 30 + 1.5x
A1
Part (b)
each extra page takes about 2.5 s on printer A but only about 1.5 s on printer B, so B prints each extra page faster
B1
Part (c)
12 + 2.5x = 30 + 1.5x
M1
x = 18 pages
A1
Part (d)
unreliable: 18 pages is outside the range of the data for both printers (A up to 13, B up to 15), so it is extrapolation
B1
Exam tip: When two regression lines are compared, check where each one is valid. A crossing point outside the range of the data is extrapolation for both lines.
A researcher records the number of cafés, x, and the number of dentists, y, in each of 6 towns. Figure 1 shows the data; every point lies on a gridline intersection.
(a)
Use Figure 1 to find Sxx, Syy and Sxy.
(4)
(b)
Calculate the product moment correlation coefficient.
(2)
(c)
A councillor says: “The value of the correlation coefficient shows that opening more cafés in a town will bring more dentists to it.” Comment on this statement.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
Σx = 60, Σy = 42, Σx² = 706, Σy² = 336, Σxy = 486
B1
Sxx = 706 − 60²⁄6 = 106
A1
Syy = 336 − 42²⁄6 = 42
A1
Sxy = 486 − 60 × 42⁄6 = 66
A1
Part (b)
66⁄√(106 × 42)
M1
= 0.989
A1
Part (c)
the correlation coefficient shows only an association, not that one variable causes the other
B1
both are probably linked to a third factor, such as the population of the town: bigger towns have more of both
B1
Exam tip: Strong correlation does not show cause and effect. When asked to comment on a causal claim, say so, and suggest a third variable that could affect both.
A small online shop records its spending on advertising, £x hundred, and its sales, £y thousand, in each of 10 months.
advertising, x
8
12
15
18
20
24
27
30
35
40
sales, y
42
51
49
60
58
66
70
68
79
85
You may use Σx = 229, Σy = 628, Σx² = 6187, Σy² = 41076, Σxy = 15602.Figure: scatter diagram of advertising spend and sales
(a)
Find Sxx, Syy and Sxy.
(3)
(b)
Calculate the product moment correlation coefficient, and interpret it in context.
(3)
(c)
Find the equation of the regression line of y on x.
(3)
(d)
Interpret the value of b in context.
(1)
(e)
Estimate the sales in a month when £2500 is spent on advertising. The owner also wants an estimate for a month with £6000 of advertising. Comment on the reliability of each estimate.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
Sxx = 6187 − 229²⁄10 = 942.9
B1
Syy = 41076 − 628²⁄10 = 1637.6
B1
Sxy = 15602 − 229 × 628⁄10 = 1220.8
B1
Part (b)
r = 1220.8⁄√(942.9 × 1637.6)
M1
= 0.982
A1
strong positive correlation: months with more advertising tend to have higher sales
B1
Part (c)
b = 1220.8⁄942.9 = 1.295…
M1
a = 62.8 − 1.295… × 22.9 = 33.15
M1
y = 33.2 + 1.29x
A1
Part (d)
each extra £100 spent on advertising is associated with about £1290 more sales
B1
Part (e)
x = 25: 33.15 + 1.295 × 25 = 65.5, so about £65500
B1
reliable: 25 is within the data (interpolation) and the correlation is strong
B1
x = 60 is outside the data (extrapolation), so an estimate there is unreliable
B1
Exam tip: Watch the units: x is in hundreds of pounds and y in thousands, so £2500 means x = 25 and an answer of 65.5 means about £65500.
A test driver measures the fuel economy y km per litre of a car driven at 8 different steady speeds x km h−1, from 50 km h−1 to 110 km h−1. The data are coded using u = (x − 80)⁄10 and v = 2(y − 15), giving Σu = −1.5, Σv = 11.4, Σu² = 30.25, Σv² = 96.6, Σuv = −51.1.
(a)
Find Suu, Svv and Suv.
(3)
(b)
Calculate the product moment correlation coefficient between x and y, explaining why it can be found from the coded values.
(3)
(c)
Find the equation of the regression line of v on u.
(2)
(d)
Hence find the equation of the regression line of y on x, and estimate the fuel economy at 75 km h−1.
(3)
(e)
Explain why the value of a in your line has no practical meaning.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
Suu = 30.25 − (−1.5)²⁄8 = 29.96875
B1
Svv = 96.6 − 11.4²⁄8 = 80.355
B1
Suv = −51.1 − (−1.5) × 11.4⁄8 = −48.9625
B1
Part (b)
−48.9625⁄√(29.96875 × 80.355)
M1
= −0.998
A1
the coding only subtracts constants and multiplies by positive constants, so r is the same for x and y
On 10 summer days, a power company records the maximum temperature x °C in a town and the electricity demand y MWh. The temperatures range from 18 °C to 33 °C. The recorded data give Σx = 255, Σy = 4081, Σx² = 6713, Σy² = 1680103, Σxy = 105778. It is then found that the demand on the day with a maximum temperature of 29 °C was recorded as 414 MWh instead of 441 MWh.
(a)
Find the correct values of Σy, Σy² and Σxy.
(3)
(b)
Using the correct values, find Sxx, Syy and Sxy.
(3)
(c)
Calculate the product moment correlation coefficient, and interpret it in context.
(2)
(d)
Find the equation of the regression line of y on x.
(3)
(e)
Estimate the demand on a day with a maximum temperature of 27 °C, and state, with a reason, whether you would expect it to be reliable.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
Σy = 4081 − 414 + 441 = 4108
B1
Σy² = 1680103 − 414² + 441² = 1703188
B1
Σxy = 105778 + 29 × (441 − 414) = 106561
B1
Part (b)
Sxx = 6713 − 255²⁄10 = 210.5
B1
Syy = 1703188 − 4108²⁄10 = 15621.6
B1
Sxy = 106561 − 255 × 4108⁄10 = 1807
B1
Part (c)
1807⁄√(210.5 × 15621.6) = 0.996
B1
strong positive correlation: hotter days tend to have higher electricity demand
B1
Part (d)
b = 1807⁄210.5 = 8.584…
M1
a = 410.8 − 8.584… × 25.5
M1
y = 192 + 8.58x
A1
Part (e)
191.9 + 8.584 × 27 = 424 MWh
B1
reliable: 27 °C is inside the range of the data and the correlation is very strong
B1
Common error: Correcting Σxy by adding 441 − 414 instead of 29 × (441 − 414). The error is in a product xy, so the change is multiplied by the x-value of that day.
For 8 similar houses, a surveyor records the thickness of loft insulation x cm and the rate of heat loss through the roof y watts per square metre. The thicknesses range from 2 cm to 14 cm. Σx = 61, Σy = 221, Sxx = 119.875, Syy = 535.875 and Sxy = −252.125.
(a)
Calculate the product moment correlation coefficient and interpret it in context.
(3)
(b)
Find the equation of the regression line of y on x.
(3)
(c)
Interpret the gradient of the line in context, and estimate the rate of heat loss for a house with 9 cm of insulation.
(2)
(d)
A ninth house is surveyed. Its values of x and y are exactly x̄ and ȳ for the original 8 houses. Show that including it leaves Sxx and Sxy unchanged, and hence state the effect on the regression line.
(3)
(e)
Give one reason why the line should not be used to predict the heat loss of a house with no loft insulation.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
−252.125⁄√(119.875 × 535.875)
M1
= −0.995
A1
strong negative correlation: houses with thicker insulation tend to lose heat through the roof more slowly
B1
Part (b)
b = −252.125⁄119.875 = −2.1032…
M1
a = 27.625 − (−2.1032…)(7.625)
M1
y = 43.7 − 2.10x
A1
Part (c)
each extra centimetre of insulation reduces the heat loss by about 2.10 W per m²
B1
43.66 − 2.103 × 9 = 24.7 W per m²
B1
Part (d)
new Σx² = Σx² + x̄² and new Σx = 8x̄ + x̄ = 9x̄
M1
new Sxx = Σx² + x̄² − (9x̄)²⁄9 = Σx² − 8x̄² = old Sxx; similarly new Sxy = Σxy + x̄ȳ − 9x̄ȳ = old Sxy
A1
b is unchanged and the means are unchanged, so the regression line is unchanged
A1
Part (e)
x = 0 is outside the range of the data (2 to 14 cm): extrapolation
B1
Exam tip: Old Sxx = Σx² − 8x̄², because (Σx)²/8 = (8x̄)²/8. Writing Sxx this way makes the effect of adding a point easy to see.
At noon on a summer day, a geography student measures the soil temperature, y °C, at 8 depths, x cm, below the surface. Figure 1 shows the results; every point lies on a gridline intersection.
(a)
Explain why the depth is the explanatory variable.
(1)
(b)
Use Figure 1 to find Sxx, Syy and Sxy.
(4)
(c)
Calculate the product moment correlation coefficient, and interpret it in context.
(3)
(d)
Find the equation of the regression line of y on x, in the form y = a + bx.
(3)
(e)
Interpret the value of b in context, and give a reason why the line should not be used to estimate the temperature at a depth of 60 cm.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
the depths are chosen by the student, and the temperature depends on the depth
strong negative correlation: the deeper the soil, the lower the temperature tends to be
B1
Part (d)
b = −197.5⁄1050 = −0.18809…
M1
a = 20.875 − (−0.18809…) × 22.5 = 25.107…
M1
y = 25.1 − 0.188x
A1
Part (e)
the temperature falls by about 0.188 °C for each extra centimetre of depth
B1
60 cm is outside the depths measured (5 cm to 40 cm), so this is extrapolation: the linear trend may not continue
B1
Exam tip: Keep Sxx, Syy and Sxy exact and keep b unrounded in your calculator when finding a. Rounding early can change the third significant figure of a or r.
A bakery changes the price of a loaf each week. For 7 weeks it records the price, x pence, and the number of loaves sold, y. The data are coded using u = x − 150⁄10 and v = y − 400⁄20. Figure 1 shows the coded data; every point lies on a gridline intersection.
(a)
Use Figure 1 to find Suu, Svv and Suv.
(4)
(b)
Find the product moment correlation coefficient between x and y, explaining why it can be found from the coded values.
(3)
(c)
Find the equation of the regression line of v on u.
(2)
(d)
Hence find the equation of the regression line of y on x, in the form y = a + bx.
(3)
(e)
Estimate the number of loaves sold in a week when the price is 175 pence.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
Σu = 1, Σv = 37, Σu² = 35, Σv² = 259, Σuv = −41
B1
Suu = 35 − 1²⁄7 = 34.857
A1
Svv = 259 − 37²⁄7 = 63.429
A1
Suv = −41 − 1 × 37⁄7 = −46.286
A1
Part (b)
−46.286⁄√(34.857 × 63.429)
M1
= −0.984
A1
the coding subtracts constants and divides by positive constants, so r is unchanged
907.87 − 2.6557 × 175 = 443.1, so about 443 loaves
B1
Exam tip: Read the coded values straight from the axes: there is no need to decode the points first. Decode only at the end, by substituting the coding into the line for v on u.
A running club records the weekly training distance, x km, and the time to run 5 km, y minutes, of 9 runners. Figure 1 shows the data. For all 9 runners, Σx = 302, Σy = 233, Σx² = 11908, Σy² = 6103, Σxy = 7538, and the product moment correlation coefficient is −0.791. The point K is a runner who has just returned from an injury.
(a)
Use Figure 1 to write down the values of x and y for runner K.
(1)
(b)
Runner K is removed from the data. Find the new values of Σx, Σy, Σx², Σy² and Σxy.
(3)
(c)
Calculate the product moment correlation coefficient for the other 8 runners.
(4)
(d)
Find the equation of the regression line of y on x for the 8 runners, in the form y = a + bx.
(3)
(e)
Comment on the effect of removing runner K, and on whether it was sensible to remove K.
Syy = 5203 − 203²⁄8 = 51.875 and Sxy = 6398 − 264 × 203⁄8 = −301
M1
r = −301⁄√(1752 × 51.875)
M1
= −0.998
A1
Part (d)
b = −301⁄1752 = −0.17180…
M1
a = 25.375 − (−0.17180…) × 33 = 31.044…
M1
y = 31.0 − 0.172x
A1
Part (e)
the correlation becomes much stronger (−0.791 to −0.998); sensible, because K's time was affected by the injury and does not represent the link between training and time for the other runners
B1
Common error: Removing a point from Σx² by subtracting 38 instead of 38², or from Σxy by subtracting 38 + 30 instead of 38 × 30. Remove exactly the term the point added to each sum.
A café records the number of hours of sunshine, x, and the number of cold drinks sold, y, on 10 days. Figure 1 shows the data and the regression line of y on x. The ringed points lie on the line at gridline intersections. For these data Σy = 590, Sxx = 63.6 and Syy = 3626.
(a)
Use Figure 1 to find the equation of the regression line in the form y = a + bx.
(3)
(b)
Find the value of x̄.
(2)
(c)
Find the value of Sxy.
(1)
(d)
Calculate the product moment correlation coefficient.
(2)
(e)
Find the values of Σx and Σxy.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
b = 80 − 20⁄8 − 0 = 7.5
M1
a = 20, where the line meets the y-axis
B1
y = 20 + 7.5x
A1
Part (b)
ȳ = 590⁄10 = 59 and the line passes through (x̄, ȳ)
M1
59 = 20 + 7.5x̄, so x̄ = 5.2
A1
Part (c)
Sxy = b × Sxx = 7.5 × 63.6 = 477
B1
Part (d)
477⁄√(63.6 × 3626)
M1
= 0.993
A1
Part (e)
Σx = 10 × 5.2 = 52
B1
Σxy = Sxy + Σx Σy⁄n = 477 + 52 × 590⁄10 = 3545
B1
Exam tip: Since b = Sxy ÷ Sxx, knowing the gradient and Sxx gives Sxy at once; then r = Sxy ÷ √(SxxSyy). Work backwards through the formulae as well as forwards.
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