AS June 2023 Q3

EdexcelAS paperCurrent spec10 marksCorrelation Linear Regression

3. Pat is investigating the relationship between the height of professional tennis players and the speed of their serve. Data from 9 randomly selected professional male tennis players were collected. The variables recorded were the height of each player, \(h\) metres, and the maximum speed of their serve, \(v\) km/h.

Pat summarised these data as follows

\[\sum h = 17.63 \qquad \sum v = 2174.9 \qquad \sum v^2 = 526\,407.8 \qquad S_{hh} = 0.0487 \qquad S_{hv} = 5.1376\]

(a) Calculate the product moment correlation coefficient between \(h\) and \(v\) (2)

(b) Explain whether the answer to part (a) is consistent with a linear model for these data. (1)

(c) Find the equation of the regression line of \(v\) on \(h\) in the form \(v = a + bh\)
where \(a\) and \(b\) are to be given to one decimal place. (3)

Pat calculated the sum of the residuals for the 9 tennis players as 1.04

(d) Without doing a calculation, explain how you know Pat has made a mistake. (1)

Pat made one mistake in the calculation. For the tennis player of height 1.96 m Pat misread the residual as 2.27

(e) Find the maximum speed of serve, in km/h, for the tennis player of height 1.96 m (3)

Mark scheme (a)

Scheme	Marks	AO
\(\left[S_{vv} = 526407.8 - \dfrac{2174.9^2}{9} = 831.132\ldots\right]\ S_{vv} = 831.132\)
\(r = \dfrac{5.1376}{\sqrt{\text{“}831.132\text{”} \times 0.0487}}\)	M1	1.1b
\(r = 0.80753\ldots\) awrt 0.808	A1	1.1b
	(2)

Notes

M1: for a complete correct method to find \(r\). Correct expressions for \(\mathrm{S}_{vv}\) and \(r\)

A1: for awrt 0.808

Mark scheme (b)

Scheme	Marks	AO
[Positive] correlation reasonably close to 1 is consistent with a linear relationship or strong (oe eg “high”) positive correlation, so is consistent	B1ft	2.4
	(1)

Notes

B1ft ft their answer to part (a). For a correct reason.
If \(r \lt 0.58\) allow “weak” correlation or correlation is close to 0 so is not consistent

Mark scheme (c)

Scheme	Marks	AO
\(b = \dfrac{5.1376}{0.0487}\ [= 105.49\ldots]\)	M1	3.3
\(a = \dfrac{2174.9}{9} - \text{“}105.49\ldots\text{”} \times \dfrac{17.63}{9}\) or \(a = 241.655\ldots - \text{“}105.49\ldots\text{”} \times 1.958\ldots\)	M1	1.1b
\(v = 105.5h + 35.0\)	A1	1.1b
	(3)

Notes

M1: for use of a correct model. i.e. a correct expression for \(b\) (or 105 or better)

M1: for use of a correct model i.e. a correct expression (ft) for \(a\) (or 35 or better)

A1: for the equation on the regression line with \(b = 105.5\) or awrt 105 and \(a\) = awrt 35.0 [ condone \(a = 35\) but not awrt 35]

Scheme	Marks	AO
The sum of the residuals should be zero	B1	2.4
	(1)

Mark scheme (e)

Scheme	Marks	AO
Residual \(= 2.27 - 1.04\ [=1.23]\)	M1	3.1b
\(v = \text{“}105.5\text{”} \times 1.96 + \text{“}35.0\text{”} + \text{“}1.23\text{”}\)	M1	3.4
\(= 243.01\) awrt 243	A1ft	1.1b
	(3)
	(10 marks)

Notes

M1: for a correct method to calculate the actual residual

M1: for using the model to estimate the speed. Must be \(\hat{v}\) + their actual residual (using 2.27…)

A1ft for awrt 243 or ft their values of \(a\) and \(b\)