Edexcel A level Maths (9MA0) · Statistics › Data presentation and interpretation
Practise Correlation and regression. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
A shop records the daily maximum temperature, \(x\) °C, and the number of ice creams sold, \(y\), on 30 days. The temperatures ranged from 8 °C to 26 °C. The equation of the regression line of \(y\) on \(x\) is \(y = 12.4 + 3.1x\), and the product moment correlation coefficient is 0.83.
(a) Interpret the value 3.1 in the context of the question.[1]
(b) Use the regression line to estimate the number of ice creams sold on a day with a maximum temperature of 20 °C.[1]
(c) A student uses the line to estimate the sales on a day with a maximum temperature of 40 °C. Give a reason why this estimate is unreliable.[1]
(d) The student also uses the line to estimate the maximum temperature on a day when 60 ice creams were sold, obtaining \(x = \frac{60 - 12.4}{3.1} = 15.4\) °C. Explain why this method is not appropriate.[1]
(e) The student concludes: “The correlation coefficient shows that hot weather causes people to buy more ice creams.” Comment on this conclusion.[1]
Show the answer and mark scheme
(a)Answer: For each 1 °C rise in maximum temperature, about 3.1 more ice creams are sold (on average).
B1 for about 3.1 more ice creams sold for each extra 1 °C of maximum temperature
Worked solution: The gradient 3.1 means that, on average, sales rise by about 3.1 ice creams for each 1 °C increase in the daily maximum temperature.
(b)Answer: 74.4, so about 74
B1 for 74.4 (accept 74)
Worked solution: \(y = 12.4 + 3.1 \times 20 = 74.4\), so about 74 ice creams.
(c)Answer: 40 °C is outside the range of the data (8 °C to 26 °C): extrapolation.
B1 for extrapolation: 40 °C is outside the range of temperatures in the data
Worked solution: The data only cover temperatures from 8 °C to 26 °C. Using the line at 40 °C is extrapolation, and the relationship may not continue (for example, very hot days might keep people indoors).
(d)Answer: The regression line of \(y\) on \(x\) should only be used to estimate \(y\) from \(x\); estimating \(x\) needs the regression line of \(x\) on \(y\).
B1 for the regression line of \(y\) on \(x\) is designed to predict \(y\) (sales) for a given \(x\) (temperature), not \(x\) from \(y\)
Worked solution: The line of \(y\) on \(x\) minimises errors in \(y\), so it is only suitable for estimating sales from temperature. Temperature is the explanatory variable; estimating it from sales would need a different regression line (of \(x\) on \(y\)).
(e)Answer: Correlation shows a strong positive linear association, but it does not by itself prove causation (although a causal link is plausible here).
B1 for correlation does not imply causation: the value 0.83 only shows a strong positive linear relationship
Worked solution: A correlation coefficient of 0.83 shows strong positive linear correlation between temperature and sales, but correlation alone does not show that one causes the other. A causal link is plausible here, but it would need other evidence.
Question 2Medium4 marks
The table shows the age of a used car, \(a\) years, and its price, £\(p\) thousand, for a random sample of 7 used cars of the same model.
[object Object]
(a) Use your calculator to find the product moment correlation coefficient, \(r\), between \(a\) and \(p\). Give your answer to 3 significant figures.[1]
(b) Interpret your value of \(r\) in the context of the question.[1]
(c) A newspaper states that the value of \(r\) proves that changes in age of a used car cause changes in price of the car. Comment on this statement.[2]
Show the answer and mark scheme
(a)Answer: −0.817
B1 for awrt −0.817
(b)Answer: Strong negative correlation: as the age of a used car increases, the price of the car tends to decrease.
B1 for negative correlation described in context (both variables mentioned)
(c)Answer: Correlation does not imply causation: the value of r only shows a linear association in this sample. Another factor could affect both variables, or the association could be due to chance in a small sample (n = 7).
B1 for correlation does not imply causation
B1 for a further valid point, e.g. a third variable could affect both, or the sample of 7 is small
Question 3Hard6 marks
A student takes a random sample of 25 days from the large data set for Hurn in 2015. For each day, the student records the daily mean windspeed, \(w\) knots, and the daily maximum gust, \(g\) knots. The values of \(w\) range from 3 to 15. The regression line of \(g\) on \(w\) is \(g = 4.8 + 1.96w\). (The values are illustrative.)
(a) Give an interpretation of the gradient of the regression line.[1]
(b) Estimate the daily maximum gust, in mph, on a day with daily mean windspeed 9 knots. (Use 1 knot = 1.15 mph, as stated in the large data set.)[3]
(c) On one day at Hurn the daily maximum gust was 44 knots. A student uses the regression line to estimate the daily mean windspeed that day as \(\dfrac{44 - 4.8}{1.96} = 20.0\) knots. Give two reasons why this estimate is unreliable.[2]
Show the answer and mark scheme
(a)Answer: For each increase of 1 knot in the daily mean windspeed, the daily maximum gust increases by about 1.96 knots, on average.
B1 for a correct interpretation in context, including the rate (1.96 knots per 1 knot)
(b)Answer: 25.8 mph
M1 for substituting \(w = 9\): \(g = 22.44\) knots
M1 for multiplying their gust in knots by 1.15
A1 for awrt 25.8 mph
(c)Answer: The regression line of g on w is for estimating g from a given value of w, not w from g. Also, 20.0 knots is outside the range of daily mean windspeeds in the sample (3 to 15 knots), so this is extrapolation.
B1 for the regression line of g on w being for estimating g (not w)
B1 for extrapolation: 20.0 knots (or a gust of 44 knots) is outside the range of the data used