Correlation and regression lines – IB Maths exam questions with mark scheme

1StandardCalculator7 marks

Six students recorded the number of hours of revision, $x$, and their test score, $y$. The data are shown in the table and on the scatter diagram.

$x$	2	4	5	7	8	10
$y$	14	19	21	28	29	36

(a) Use your graphic display calculator to find the Pearson correlation coefficient $r$ and the equation of the regression line of $y$ on $x$. [3]

(b) Describe the correlation and interpret the gradient of the line in context. [2]

(d) Explain why the line should not be used to predict the score of a student who revises for 20 hours. [1]

$r=0.996$ (accept $0.99586\ldots$ or any more accurate value)A1
Gradient $a=2.74$ ($2.73809\ldots$); accept $\frac{115}{42}$A1
Intercept $b=8.07$ ($8.07142\ldots$), so $y=2.74x+8.07$ (award only if written as an equation in $y$ and $x$)A1
Strong positive (linear) correlation (accept 'very strong positive'; do not accept 'positive' alone)A1
Each extra hour of revision is associated with about $2.74$ (accept $2.7$) more marks on average, in context (do not accept a causal claim such as 'revision causes')A1
$y\approx24.5$ (accept $24.51$ from the 3 s.f. coefficients)A1
$x=20$ is outside the data range ($2$ to $10$ hours), so this is extrapolation and the prediction is unreliableR1
Answer: (a) $r=0.996$, $y=2.74x+8.07$ (b) very strong positive; $\approx2.7$ marks per hour (c) $\approx24.5$ (d) extrapolation
Examiner tip: Interpretation must be in context and say 'on average' or 'associated with'; do not claim causation.

2StandardCalculator5 marks

A café owner records the midday temperature, $x\,^\circ$C, and the sales of cold drinks, $y$ (in pounds), on 25 days with temperatures between $14\,^\circ$C and $30\,^\circ$C. The regression line of $y$ on $x$ is $y=12.5x-90$ and $r=0.94$.

(a) Interpret the value $12.5$ in context. [1]

(b) Describe the correlation. [1]

(d) Explain why the model should not be used at $5\,^\circ$C. [1]

(e) The owner concludes that higher temperatures cause higher sales. Comment on this claim. [1]

For each extra $1\,^\circ$C, sales increase by about £12.50 on average (must be in context)A1
Strong positive (linear) correlation, as $r=0.94$ is close to 1 (do not accept 'positive' alone)A1
$12.5(25)-90=$ £222.50 (accept $222.5$ or £223 to 3 s.f.)A1
$5\,^\circ$C is outside the range of the data ($14$ to $30\,^\circ$C), so this would be extrapolation and unreliable (also accept: the model gives negative sales, $12.5(5)-90=-27.5$, which is impossible; either reason scores)R1
Correlation does not imply causation: the data show only an association (naming a possible third factor, e.g. time of year or sunshine, is not required but may be credited instead)R1
Answer: (a) +£12.50 per °C (b) strong positive (c) £222.50 (d) extrapolation (e) correlation is not causation
Examiner tip: 'Positive correlation' alone is not enough for the description mark: say how strong it is (e.g. strong positive).

3HardCalculator6 marks

The table shows five pairs of values. One value, $k$, is missing. The regression line of $y$ on $x$ is $y=3x+5$.

$x$	2	4	6	8	10
$y$	12	15	$k$	27	36

(a) Write down the coordinates of the mean point $(\bar x,\bar y)$ and find the value of $k$. [4]

(b) Find the residual (actual $-$ predicted) at $x=6$. [2]

$\bar x=6$A1
The regression line passes through $(\bar x,\bar y)$: $\bar y=3(6)+5$M1
$\bar y=23$A1
$\dfrac{12+15+k+27+36}{5}=23\Rightarrow k=25$A1
Predicted value at $x=6$ is $23$M1
Residual $=25-23=2$A1
Answer: (a) $(6,23)$, $k=25$ (b) $+2$
Examiner tip: Remember that the least-squares line always passes through $(\bar x,\bar y)$: this lets you find a missing value without calculating a gradient.

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4CoreNon-calculator4 marks

The diagram shows four scatter diagrams, labelled A, B, C and D, each drawn on the same scales.

The values of Pearson's product-moment correlation coefficient $r$ for the four sets of data are, in no particular order,

$$0.86,\qquad -0.96,\qquad 0.04,\qquad -0.45.$$

Write down the value of $r$ that corresponds to each scatter diagram. [4]

A: $r=-0.45$A1
B: $r=0.86$A1
C: $r=-0.96$A1
D: $r=0.04$A1
Answer: A: $-0.45$; B: $0.86$; C: $-0.96$; D: $0.04$
Examiner tip: Judge strength by how tightly the points hug a straight line, not by how steep the cloud looks; the sign comes from the direction of the trend.

5StandardNon-calculator7 marks

For a group of adults, the arm span, $x$ cm, and the height, $y$ cm, were measured.

The regression line of $y$ on $x$ is $y=0.8x+34$.

The regression line of $x$ on $y$ is $x=1.1y-17$.

(a) Estimate the height of an adult from this group whose arm span is $160$ cm. [2]

(b) Estimate the arm span of an adult from this group whose height is $180$ cm. [2]

Uses the line of $y$ on $x$ (since $x$ is known): $y=0.8(160)+34$M1
$y=162$ cmA1
Uses the line of $x$ on $y$ (since $y$ is known): $x=1.1(180)-17$. M0 if they rearrange the $y$ on $x$ line ($x=(180-34)/0.8=182.5$)M1
$x=181$ cmA1
Recognises both lines pass through $(\bar x,\bar y)$ and attempts to solve simultaneously, e.g. $y=0.8(1.1y-17)+34$M1
$0.12y=20.4$ or equivalent, giving $\bar y=170$A1
$\bar x=170$ (mean arm span $170$ cm, mean height $170$ cm)A1
Answer: (a) $162$ cm (b) $181$ cm (c) $\bar x=170$ cm, $\bar y=170$ cm
Examiner tip: To estimate $x$ from a given $y$ you must use the $x$ on $y$ line; rearranging the $y$ on $x$ line loses the method mark.

6StandardCalculator7 marks

A dealer records the age, $x$ years, and the value, $y$ thousand pounds, of eight used cars of the same model. The data are shown in the table and on the scatter diagram.

$x$	1	2	3	4	5	6	7	8
$y$	15.8	14.6	12.6	11.4	9.4	18.5	6.9	5.9

(a) Find the value of Pearson's product-moment correlation coefficient $r$ for all eight cars. [1]

(b) One of the values was recorded incorrectly. Identify the car, giving a reason. [1]

The incorrectly recorded car is removed from the data.

(c) For the remaining seven cars, find
(i) the new value of $r$;
(ii) the equation of the regression line of $y$ on $x$. [3]

(d) Use your answer to (c)(ii) to estimate the value of a car of this model that is $6$ years old. [1]

(e) Comment on the effect that the incorrect value had on $r$. [1]

$r=-0.614$ (accept $-0.6137\ldots$)A1
The car aged $6$ years (value $18.5$): it does not fit the trend: it is worth more than every younger car and lies far above the line of the other points (accept 'outlier' with reference to the pattern)R1
(i) $r=-0.997$A1
(ii) gradient $-1.46$ (accept $-1.455$); award for $y=-1.46x+\ldots$A1
intercept $17.2$: $y=-1.46x+17.2$ (accept $-1.4554x+17.180$). Award A1A0 if not written as an equationA1
$-1.4554(6)+17.180=8.45$ (thousand pounds), i.e. about £$8450$ (accept $8.44$ to $8.46$ from rounded coefficients; follow through from their line)A1
The incorrect value greatly weakened the correlation: $|r|$ rose from $0.614$ (moderate) to $0.997$ (very strong) when it was removed (a statement that it made the correlation appear weaker is sufficient)R1
Answer: (a) $-0.614$ (b) car aged 6 years (c) $r=-0.997$, $y=-1.46x+17.2$ (d) $8.45$ thousand pounds (e) it made the correlation look much weaker
Examiner tip: Give $r$ and regression coefficients to 3 significant figures, and write the line as a full equation $y=\ldots$, not just the values of $a$ and $b$.

7CoreNon-calculator4 marks

A survey of $40$ teenagers recorded the daily screen time, $x$ hours, and the hours of sleep, $y$, on one night. The values of $x$ ranged from $1$ to $7$ hours.

For these data $r=-0.28$ and the regression line of $y$ on $x$ is $y=-0.15x+8.6$.

(a) Describe the correlation between $x$ and $y$. [1]

(b) Ali uses the line to predict that a teenager with $4$ hours of screen time sleeps for $8$ hours. Comment on the reliability of this prediction. [1]

(c) Bella wants to estimate the screen time of a teenager who slept for $6$ hours. She solves $6=-0.15x+8.6$ and obtains $x=17.3$. Give two reasons why Bella's method is not valid. [2]

Weak negative (linear) correlation. Both 'weak' and 'negative' neededA1
Unreliable, because the correlation is weak ($r=-0.28$ is close to $0$), even though $4$ hours is within the data range. Do not accept 'reliable because it is interpolation' aloneR1
She should use the regression line of $x$ on $y$ to estimate $x$ from a given $y$ (not rearrange the $y$ on $x$ line)R1
Her answer $17.3$ hours is outside the range of the data ($1$ to $7$ hours), i.e. extrapolation (also accept: more than any possible daily total sensibly; or the weak correlation makes any estimate unreliable, if not already used in (b))R1
Answer: (a) weak negative (b) unreliable: weak correlation (c) wrong line (needs $x$ on $y$); extrapolation outside $1$–$7$
Examiner tip: Interpolation is only reliable when the correlation is strong; always check the size of $r$ as well as the data range.

8HardCalculator6 marks

A household records the mean daily outside temperature, $x\,^\circ$C, and the gas used, $y$ kWh, on $30$ days. The regression line of $y$ on $x$ is $y=-2.4x+61$ and $r=-0.87$.

The temperatures are converted to degrees Fahrenheit, $F$, using $F=1.8x+32$.

(a) Write down the value of the correlation coefficient between $F$ and $y$. Give a reason for your answer. [2]

(b) Find the equation of the regression line of $y$ on $F$, giving it in the form $y=aF+b$ where $a,b\in\mathbb{Q}$. [3]

$r=-0.87$A1
A linear change of scale with a positive multiplier ($1.8>0$) does not change $r$ (the pattern of points is just stretched and shifted)R1
Substitutes $x=\dfrac{F-32}{1.8}$ into the line: $y=-2.4\left(\dfrac{F-32}{1.8}\right)+61$M1
$a=-\dfrac{4}{3}$ (accept $-1.33$)A1
$y=-\dfrac{4}{3}F+\dfrac{311}{3}$ (accept $103\tfrac{2}{3}$ or $103.7$; FT on their $a$ only if method correct)A1
$F=1.8(10)+32=50$; $-\tfrac{4}{3}(50)+\tfrac{311}{3}=37$ and $-2.4(10)+61=37$ kWh (AG). Both values neededA1
Answer: (a) $-0.87$; positive linear rescaling leaves $r$ unchanged (b) $y=-\tfrac43F+\tfrac{311}{3}$ (c) both give $37$ kWh
Examiner tip: Changing units by a positive linear rule never changes $r$, but it does change the gradient and intercept of the regression line.

9StretchNon-calculator8 marks

For a set of bivariate data, the regression line of $y$ on $x$ is $y=-0.5x+15$ and the regression line of $x$ on $y$ is $x=-1.28y+23.52$.

(a) Find the coordinates of the point where the two lines intersect, and state what this point represents. [3]

It can be shown that $r^2$ is equal to the product of the gradients $b$ and $d$, where the lines are written as $y=bx+c$ and $x=dy+e$.

(b) Find the value of $r$, justifying its sign. [3]

It can also be shown that $b=r\,\dfrac{\sigma_y}{\sigma_x}$, where $\sigma_x$ and $\sigma_y$ are the standard deviations of $x$ and $y$.

Attempts to solve simultaneously, e.g. $y=-0.5(-1.28y+23.52)+15$ giving $0.36y=3.24$M1
$(12,\,9)$A1
It is the mean point $(\bar x,\bar y)$; both regression lines pass through itR1
$r^2=(-0.5)(-1.28)$M1
$r^2=0.64\Rightarrow r=\pm0.8$A1
$r=-0.8$, because the gradients of the regression lines are negative (negative correlation). The value $r=-0.8$ with no reason scores A1R0R1
$-0.5=-0.8\times\dfrac{\sigma_y}{4}$ (FT their $r$)M1
$\sigma_y=2.5$A1
Answer: (a) $(12,9)$, the mean point $(\bar x,\bar y)$ (b) $r=-0.8$ (c) $\sigma_y=2.5$
Examiner tip: When you take a square root to find $r$, you must choose the sign from the direction of the regression lines; leaving $\pm0.8$ loses the reasoning mark.

10StandardCalculator7 marks

A shop records its weekly spending on online advertising, $x$ (in hundreds of pounds), and the number of items sold, $y$, over eight weeks.

$x$	12	15	18	20	23	25	28	31
$y$	40	52	47	60	55	69	63	76

(a) Find the equation of the regression line of $y$ on $x$. [2]

(b) Use your equation to estimate the number of items sold in a week when $£2200$ is spent on advertising. [1]

(c) The manager says the $y$-intercept of the line shows how many items would be sold with no advertising. Explain why this interpretation is unreliable. [1]

(d) The shop wants to sell $70$ items in a week.
(i) Find the equation of the regression line of $x$ on $y$.
(ii) Hence estimate the amount the shop should spend on advertising. [3]

gradient $1.64$ (accept $1.6428\ldots$)A1
$y=1.64x+22.4$ (accept $22.43$); must be an equationA1
$x=22$: $y=58.6$ (accept $58.5$ from rounded coefficients); allow 'about $59$ items'A1
$x=0$ is outside the range of the data ($12$ to $31$), so using the intercept is extrapolation (accept: the relationship may not be linear down to zero spending)R1
(i) $x=0.501y-7.45$ (accept $0.5013y-7.450$)A1
(ii) Substitutes $y=70$ into the $x$ on $y$ line: $0.501(70)-7.45$. M0 for rearranging the $y$ on $x$ line (which gives $28.96$)M1
$x=27.6$, so about £$2760$ (accept £$2760$ to £$2770$)A1
Answer: (a) $y=1.64x+22.4$ (b) $58.6$ items (c) $x=0$ is outside $12$–$31$: extrapolation (d)(i) $x=0.501y-7.45$ (ii) about £$2760$
Examiner tip: Read the units: $x$ is in hundreds of pounds, so £2200 means $x=22$ and an answer of $x=27.6$ means £2760.

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