A-level · Data presentation

Correlation and regressionExam-style questions

Bivariate data and scatter diagrams; describing correlation by its direction (positive/negative) and strength; why correlation does not imply causation; and using a given regression line y = a + bx to make and interpret predictions — including why prediction inside the data range (interpolation) is safe but outside it (extrapolation) is not. Interpreting the PMCC is flagged as A-level.

12 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAllLevel 2Level 3Level 4Level 5
  1. 1
    A running coach is studying the runners at her athletics club. For each of 1010 runners, she records the number of hours, hh, that the runner spends training in a typical week and the time, mm minutes, that the runner takes to complete a 55 km race. The values of hh for these runners range from 33 to 1414

    Her results are summarised below.

    ∑h=84∑m=243.4∑h2=840∑hm=1945.8Smm=79.904\sum h = 84 \qquad \sum m = 243.4 \qquad \sum h^2 = 840 \qquad \sum hm = 1945.8 \qquad S_{mm} = 79.904
    (a) Find the exact value of ShhS_{hh} and the exact value of ShmS_{hm}

    (3 marks)
    (b) Calculate the product moment correlation coefficient between hh and mm

    (2 marks)
    (c) Interpret your value of the product moment correlation coefficient in the context of these data.

    (1 mark)
    (d) Show that the equation of the regression line of mm on hh can be written as

    m=30.5−0.735hm = 30.5 - 0.735h

    where the values of the intercept and gradient are given to 3 significant figures.

    (3 marks)
    (e) Give an interpretation, in context, of the gradient of this regression line.

    (1 mark)
    (f) Using the equation in part (d),

    (i) estimate the time taken to complete a 55 km race by a runner who trains for 99 hours each week,

    (ii) explain why it would not be sensible to use the equation to estimate the time taken by a runner who trains for 2525 hours each week.

    (2 marks)
    Level 4 · 12 marksTry it
  2. 2
    An energy consultant visits 1111 houses in a town. For each house, she records the floor area, x m2x\ \text{m}^2, and the amount of electricity used by the household in the previous year, yy MWh. Her results are shown in the table below.
    HouseABCDEFGHIJK
    Floor area (xx m2\text{m}^2)6275849098110118130140156172
    Electricity used (yy MWh)2.32.63.13.03.53.64.09.64.44.95.3
    The household in house H heats an outdoor swimming pool.

    An outlier is defined as a value that is greater than Q3+1.5×(Q3−Q1)Q_3 + 1.5 \times (Q_3 - Q_1) or smaller than Q1−1.5×(Q3−Q1)Q_1 - 1.5 \times (Q_3 - Q_1)
    (a) Show that 9.69.6 is an outlier for the amounts of electricity used.

    (3 marks)

    The consultant removes house H from the data. For the remaining 1010 houses,

    ∑x=1105∑y=36.7Sxx=11 670.5Sxy=315.65\sum x = 1105 \qquad \sum y = 36.7 \qquad S_{xx} = 11\,670.5 \qquad S_{xy} = 315.65
    (b) Show that, for these 1010 houses, the regression line of yy on xx has equation

    y=0.681+0.0270xy = 0.681 + 0.0270x

    where the values of the intercept and gradient are given to 3 significant figures. You must show your working.

    (3 marks)
    (c) Give an interpretation of the gradient of this regression line.

    (1 mark)

    Two other houses in the town have floor areas of 125 m2125\ \text{m}^2 and 250 m2250\ \text{m}^2
    (d) Use the equation in part (b) to estimate the amount of electricity used in a year by

    (i) the house with a floor area of 125 m2125\ \text{m}^2,

    (ii) the house with a floor area of 250 m2250\ \text{m}^2.

    (3 marks)
    (e) State, giving a reason, which of your two estimates in part (d) is likely to be more reliable.

    (2 marks)
    Level 4 · 12 marksTry it
  3. 3
    A courier firm records the distance, xx miles, and the time taken, yy minutes, for each of 1515 parcel deliveries.

    The following summary statistics are available.

    ∑x=81.4∑x2=545.46Syy=1110Sxy=322.7n=15\sum x = 81.4 \qquad \sum x^2 = 545.46 \qquad S_{yy} = 1110 \qquad S_{xy} = 322.7 \qquad n = 15
    (a) (i) Find the mean distance of these 1515 deliveries.

    (ii) Show that the standard deviation of the distances is 2.632.63 miles, correct to 3 significant figures.

    (3 marks)
    (b) Calculate the product moment correlation coefficient between xx and yy

    (3 marks)

    The charge, £CC, for delivering a parcel a distance of xx miles is worked out using the formula

    C=4.50+0.90xC = 4.50 + 0.90x
    (c) Using your answers to part (a), find the mean and the standard deviation of the charges for these 1515 deliveries. Give your answers to 3 significant figures.

    (4 marks)
    (d) State, giving a reason, the value of the product moment correlation coefficient between CC and yy

    (2 marks)
    Level 4 · 12 marksTry it
  4. 4
    A car dealer records the age, xx years, and the price, £yy thousand, of each car in a random sample of 1616 used cars of the same model. Using these data, the dealer finds that
    • the regression line of yy on xx is y=18.4−1.6x\quad y = 18.4 - 1.6x
    • the regression line of xx on yy is x=10.4−0.5y\quad x = 10.4 - 0.5y
    (a) (i) Find the coordinates of the point where these two lines meet.

    (ii) Hence show that ∑y=140.8\sum y = 140.8

    (4 marks)

    The dealer also calculates that ∑x2=676\sum x^2 = 676
    (b) Find the value of SxyS_{xy}

    (4 marks)
    (c) Find the product moment correlation coefficient between xx and yy

    (4 marks)
    Level 4 · 12 marksTry it
  5. 5
    A meteorologist believes that windier days tend to be colder. She records the daily mean windspeed, ww knots, and the daily mean air temperature, T ∘T\,^\circC, on 1010 randomly chosen days at a weather station. The results are shown in the table.

    w4679101214151820T16.213.814.917.013.114.411.915.510.812.6\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}\hline w & 4 & 6 & 7 & 9 & 10 & 12 & 14 & 15 & 18 & 20 \\ \hline T & 16.2 & 13.8 & 14.9 & 17.0 & 13.1 & 14.4 & 11.9 & 15.5 & 10.8 & 12.6 \\ \hline \end{array}

    For these data the product moment correlation coefficient is r=−0.627r = -0.627 and the equation of the regression line of TT on ww is T=16.7−0.234wT = 16.7 - 0.234w.
    (a) Test, at the 5%5\% level of significance, whether these data provide evidence to support the meteorologist's belief. State your hypotheses clearly. (4 marks)
    (b) Give an interpretation of the gradient of the regression line in this context. (1 mark)
    (c) Use the regression line to estimate the daily mean air temperature on a day when the daily mean windspeed is 12.512.5 knots, giving your answer to 11 decimal place. (2 marks)
    (d) Explain why it would be unreliable to use this regression line to estimate (i) the daily mean air temperature on a day when the daily mean windspeed is 3535 knots, (ii) the daily mean windspeed on a day when the daily mean air temperature is 12 ∘12\,^\circC. (2 marks)
    Level 4 · 9 marksTry it

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