A2 June 2025 Q1

EdexcelCurrent spec10 marksLinear Regression

1. A medical researcher is exploring the relationship between the weight, \(x\) kg, and the head circumference, \(y\) cm, of newborn babies. A random sample of 15 newborn babies is taken and the data are summarised by the following statistics

\[\sum x = 50.46 \qquad \sum x^2 = 171.828 \qquad \sum y = 518.9 \qquad \sum y^2 = 18\,004.47 \qquad \mathrm{S}_{xy} = 7.8284\]

(a) Show that, to 3 decimal places, \(\mathrm{S}_{xx} = 2.081\) and \(\mathrm{S}_{yy} = 53.989\) (3)

(b) Find the equation of the regression line of \(y\) on \(x\), giving your answer in the form \(y = a + bx\) (3)

One of these 15 babies had a birth weight of 3.26 kg and a head circumference of 36.8 cm

(d) Find the residual for this baby. (2)

The researcher claims that this baby could be an outlier.

(e) Using your answers to part (c) and part (d), give a reason that might support this claim. (1)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{S}_{xx} = 171.828 - \dfrac{50.46^2}{15}\) or \(\mathrm{S}_{yy} = 18\,004.47 - \dfrac{518.9^2}{15}\)	M1	1.1b
\(= 2.08056 = \)2.081* (3dp) ; \(= 53.98933\ldots = \)53.989* (3dp)	A1; A1	1.1b(x2)
	(3)

Notes

M1: for either correct expression

1^st A1*: cso for 2.081 or better following a correct expression

2^nd A1*: cso for 53.989 or better following a correct expression

Mark scheme (b)

Scheme	Marks	AO
\(b = \left[\dfrac{\mathrm{S}_{xy}}{\mathrm{S}_{xx}} = \right] \dfrac{7.8284}{2.081} = (3.76\ldots)\)	M1	1.1b
\(a = \dfrac{518.9}{15} - b\dfrac{50.46}{15} = 34.593\ldots - \text{“}3.76\ldots\text{”} \times 3.364\ (= 21.9\ldots)\)	M1	1.1b
Equation of line is: \(y = 21.9 + 3.76x\)	A1	1.1b
	(3)

Notes

1^st M1: for a correct numerical expression for gradient (or ft their \(\mathrm{S}_{xx} \ne 2.081\), \(\mathrm{S}_{xy} \ne 7.8284\))

2^nd M1: for a correct expression for intercept (ft their \(b\)). May be implied by awrt 21.9

A1: for equation with \(a\) = awrt 21.9 and \(b\) = awrt 3.76

Note: A correct equation scores M1M1A1

Scheme	Marks	AO
\(\left[\text{RSS} = \mathrm{S}_{yy} - \dfrac{\left(\mathrm{S}_{xy}{}^2\right)}{\mathrm{S}_{xx}} = \right] 24.5397\ldots\) = awrt 24.5	B1	1.1b
	(1)

Mark scheme (d)

Scheme	Marks	AO
\(\hat{y} = \text{“}21.9\text{”} + \text{“}3.76\text{”} \times 3.26\ \ (= 34.1576)\)	M1	3.4
[Residual \(= y - \hat{y} = 36.8 - \text{“}34.15\ldots\text{”} = 2.6424\)] = awrt 2.6	A1	1.1b
	(2)

Notes

M1: for an expression for \(\hat{y}\) or the residual (ft their regression line equation). Implied by awrt 2.6

A1: for awrt 2.6 (exact figures give 2.5979…)

Mark scheme (e)

Scheme	Marks	AO
\(\text{“}2.64\ldots\text{”}^2 \approx 7\) which is a high proportion of RSS (so this value may be an outlier)	B1	2.4
	(1)
	(10 marks)

Notes

B1: Dependent on \(0.25 \lt \dfrac{\text{“}2.6\text{”}^2}{\text{“}24.5\text{”}} \lt 1\) for a suitable explanation based on their (d)² compared with their (c) or other suitable calculations which enable a comparison to be made.
e.g. this residual is 27.5% of the total which is a large proportion”.
References to size e.g. “large” requires quoting the numerical value as well.