AS June 2019 Q3

EdexcelAS paperCurrent spec11 marksCorrelation Linear Regression

3. Two students, Jim and Dora, collected data on the mean annual rainfall, \(w\) cm, and the annual yield of leeks, \(l\) tonnes per hectare, for 10 years.

Jim summarised the data as follows

\[\mathrm{S}_{wl} = 42.786 \qquad \mathrm{S}_{ww} = 9936.9 \qquad \sum l^2 = 26.2326 \qquad \sum l = 16.06\]

(a) Find the product moment correlation coefficient between \(l\) and \(w\) (2)

Dora decided to code the data first using \(s = w - 6\) and \(t = l - 20\)

(b) Write down the value of the product moment correlation coefficient between \(s\) and \(t\).
Give a justification for your answer. (1)

Dora calculates the equation of the regression line of \(t\) on \(s\) to be \(t = 0.00431s - 18.87\)

(c) Find the equation of the regression line of \(l\) on \(w\) in the form \(l = a + bw\), giving the values of \(a\) and \(b\) to 3 significant figures. (3)

(d) Use your equation to estimate the yield of leeks when \(w\) is 100 cm. (1)

(e) Calculate the residual sum of squares. (2)

The graph shows the residual for each value of \(l\)

Residual plot: residual (from –0.2 to 0.5) against w (from 70 to 200) for 10 points; one residual of about 0.42 at w = 111, the others between about –0.15 and 0.06

(f)

(i) State whether this graph suggests that the use of a linear regression model is suitable for these data. Give a reason for your answer.

(ii) Other than collecting more data, suggest how to improve the fit of the model in part (c) to the data.

(2)

Mark scheme (a)

Scheme	Marks	AO
\(\left[S_{ll} = 26.2326 - \dfrac{16.06^2}{10} = 0.44024\right]\)
\(r = \dfrac{42.786}{\sqrt{9936.9 \times \text{“}0.44024\text{”}}}\)	M1	1.1b
\(r = 0.64689\ldots\) awrt 0.647	A1	1.1b
	(2)

Notes

M1: For a complete correct method to find \(r\)

A1: for awrt 0.647

Scheme	Marks	AO
\(\text{“}0.647\text{”}\) coding has no effect on the pmcc	B1ft	1.1b
	(1)

Mark scheme (c)

Scheme	Marks	AO
\(l - 20 = 0.00431(w - 6) - 18.87\)	M1	3.1a
\(l = 0.00431w + \ldots.\)	M1	1.1b
\(l = 0.00431w + 1.10\,414\)	A1	1.1b
	(3)

Notes

M1: for use of a correct model. i.e. a correct expression for \(b\)

M1: for use of a correct model i.e. a correct expression (ft) for \(a\)

A1: for correct model \(l = 0.00431w + 1.10\) with awrt 0.00431 and awrt 1.10

Mark scheme (d)

Scheme	Marks	AO
\(l = 0.00431 \times 100 + 1.10 = 1.53\)	B1ft	3.4
	(1)

Notes

B1ft: correct answer using their equation and \(w = 100\) or using \(t = 0.00431s - 18.87\) and \(s = 94\)
Allow awrt 1.53/1.54

Mark scheme (e)

Scheme	Marks	AO
\(\text{RSS} = \text{“}0.44024\text{”} - \dfrac{(42.786)^2}{9936.9}\) or \(\text{“}0.44024\text{”}\left(1 - \text{“}0.647\text{”}^2\right)\)	M1	1.1b
RSS = 0.2560	A1	1.1b
	(2)

Notes

M1: for a correct expression for RSS

A1: awrt 0.256

Mark scheme (f)

Scheme	Marks	AO
(i) The points appear randomly scattered above and below zero giving us no reason to doubt the suitability of the linear model.	B1	3.5a
(ii) There is a possible outlier that could be removed (and the regression line recalculated).	B1	3.5c
	(2)
	(11 marks)

Notes

(i) B1: For explaining why the model may be suitable. Allow randomly scattered around \(w\) (\(x\)) axis.
Do not allow most residuals close to zero or not suitable as not randomly scattered.

(ii) B1: For explaining how the fit of the model might be improved.