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 teacher compares her students' results in a mock examination with their results in the final examination. For a random sample of 1212 of her students, she records the percentage mark in the mock examination, xx, and the percentage mark in the final examination, yy. The mock examination marks of these students range from 2828 to 8585

    The data are summarised as follows.

    ∑x=681∑y=723∑x2=42 055∑y2=45 711∑xy=43 629\sum x = 681 \qquad \sum y = 723 \qquad \sum x^2 = 42\,055 \qquad \sum y^2 = 45\,711 \qquad \sum xy = 43\,629
    (a) Find the values of SxxS_{xx}, SyyS_{yy} and SxyS_{xy}

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

    (2 marks)
    (c) Give an interpretation of your product moment correlation coefficient.

    (1 mark)
    (d) Show that the equation of the regression line of yy on xx is

    y=17.0+0.762xy = 17.0 + 0.762x

    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 the regression line.

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

    (i) estimate the final examination mark of a student who scored 6060 in the mock examination,

    (ii) explain why the equation should not be used to estimate the final examination mark of a student who scored 1515 in the mock examination.

    (2 marks)
    (g) Using the equation in part (d), find the range of values of xx for which a student is predicted to score a higher mark in the final examination than in the mock examination.

    (2 marks)
    Level 5 · 14 marksTry it
  2. 2
    Nadia is a letting agent in a city. She wants to model the relationship between the distance, dd km, of a one-bedroom flat from the city centre and the monthly rent, £pp, of the flat.
    (a) Give one advantage of using a statistical model to describe a real-world situation such as this.

    (1 mark)

    Nadia takes a random sample of 1212 one-bedroom flats that she lets. She codes the monthly rents using

    w=p−60010w = \frac{p - 600}{10}

    Her data are summarised below.

    ∑d=54∑w=589∑w2=33 549Sdd=70.12Sdw=−552.4\sum d = 54 \qquad \sum w = 589 \qquad \sum w^2 = 33\,549 \qquad S_{dd} = 70.12 \qquad S_{dw} = -552.4
    (b) Show that Sww=4638.92S_{ww} = 4638.92 to 2 decimal places.

    (1 mark)
    (c) Use the coding to find the value of SdpS_{dp} and the value of SppS_{pp}

    (3 marks)
    (d) Find the product moment correlation coefficient between dd and pp

    (2 marks)
    (e) Interpret this value in the context of the question.

    (1 mark)
    (f) Show that the equation of the regression line of ww on dd is

    w=84.5−7.88dw = 84.5 - 7.88d

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

    (3 marks)
    (g) Hence find the equation of the regression line of pp on dd, giving your answer in the form p=a+bdp = a + bd

    (1 mark)
    (h) Using your equation from part (g),

    (i) estimate the monthly rent of a one-bedroom flat that is 44 km from the city centre,

    (ii) state the change in monthly rent that the model predicts for each extra kilometre from the city centre.

    (2 marks)
    Level 5 · 14 marksTry it
  3. 3
    An engineer is investigating how the stopping distance of a car depends on its speed. She tests a car on a dry road at 1212 different speeds. For each test she records the speed of the car, x km h−1x\ \text{km h}^{-1}, and its stopping distance, yy metres, where 20⩽x⩽13020 \leqslant x \leqslant 130

    The equation of the regression line of yy on xx for her data is

    y=−22.2+1.01xy = -22.2 + 1.01x
    (a) Give an interpretation of the gradient of this regression line.

    (1 mark)
    (b) Use the regression line to estimate the stopping distance of the car when it is travelling at 150 km h−1150\ \text{km h}^{-1}

    (2 marks)
    (c) State, giving a reason, whether your estimate in part (b) is likely to be reliable.

    (2 marks)

    The engineer also calculates the following summary statistics.

    ∑y=551.7∑y2=40 583.05∑x2=69 150n=12\sum y = 551.7 \qquad \sum y^2 = 40\,583.05 \qquad \sum x^2 = 69\,150 \qquad n = 12
    (d) Using these summary statistics and the equation of the regression line, show that the product moment correlation coefficient between xx and yy is 0.9850.985 to 3 decimal places.

    (7 marks)

    The engineer's results are shown on the scatter diagram below.2040608010012014020406080100120Speed (x km/h)Stopping distance (y m)(e) Using the scatter diagram and the product moment correlation coefficient, discuss whether a linear regression model is suitable for describing how stopping distance depends on speed.

    (2 marks)
    Level 5 · 14 marksTry it