AS June 2025 Q3

EdexcelAS paperCurrent spec14 marksIncludes hypothesis testingCorrelation Linear Regression

3. Dina is investigating the relationship between a marathon runner’s body mass index (BMI), \(x\,\mathrm{kg\,m^{-2}}\), and the runner’s average speed when running a marathon, \(v\,\mathrm{m\,s^{-1}}\)

She collects data from 20 marathon runners.

Some summary statistics are given below.

\[\bar{x} = 19.8 \qquad \bar{v} = 5.2 \qquad \sum xv = 2056.63 \qquad \mathrm{S}_{xx} = 15.78 \qquad \mathrm{S}_{vv} = 0.94\]

(a) Show that \(\mathrm{S}_{xv} = -2.57\) (1)

(b) Find the value of the product moment correlation coefficient between \(v\) and \(x\) (2)

(c) Use a suitable test, with a 5% level of significance, to assess whether or not there is evidence of a negative correlation between BMI and average speed.
State the hypotheses and the critical value used. (3)

(d) Find the equation of the regression line of \(v\) on \(x\) in the form \(v = a + bx\) giving the values of \(a\) and \(b\) to 2 decimal places. (3)

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

Information about 2 of the 20 marathon runners is listed in the table below.

Runner	BMI	Residual
\(A\)	20.9	0.18
\(B\)	19.7	–0.22

(f) Determine which of these runners took the shorter amount of time to run the marathon.
You must show your working. (3)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{S}_{xv} = \sum xv - \dfrac{\left(\sum x \sum v\right)}{20} = 2056.63 - \dfrac{(19.8 \times 20) \times (5.2 \times 20)}{20} = -2.57\)*	B1cso*	1.1b
	(1)

Notes

B1*: complete calculation (oe) for \(\mathrm{S}_{xv}\) leading to given answer. Allow … \(-\dfrac{396 \times 104}{20}\)

Mark scheme (b)

Scheme	Marks	AO
\(r = \dfrac{-2.57}{\sqrt{15.78 \times 0.94}},\ = -0.667291\ldots\) = awrt \(-\)0.667	M1, A1	1.1b 1.1b
	(2)

Notes

M1: use of formula for \(r\) with all given values

A1: awrt –0.667

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{H}_0 : \rho = 0 \qquad \mathrm{H}_1 : \rho \lt 0\)	B1	2.5
5% one-tail cv for \(r\) is: (\(\pm\))0.3783	M1	1.1b
Significant result so there is evidence of negative correlation (between BMI and speed)	A1	2.2b
	(3)

Notes

B1: for both hypotheses correct in terms of \(\rho\)

M1: for use of tables to find the cv of 0.3783 (Condone 0.378)

A1: for a correct conclusion that mentions negative correlation. (B0M1A1 possible)

NB A comparison is not required but if seen it must be correct for the A1

Mark scheme (d)

Scheme	Marks	AO
\(b = \dfrac{\text{‘}-2.57\text{’}}{15.78}\ [= -0.16286\ldots]\)	M1	3.3
\(a = 5.2 - \text{‘}b\text{’}(19.8)\)	M1	1.1b
\(v = 8.42 - 0.16x\)	A1	1.1b
	(3)

Notes

M1: setting up linear model by finding gradient

M1: attempting \(v\)-intercept of linear model

A1: correct model with \(b\) = awrt –0.16 and \(a\) = awrt 8.42 (NB 8.43 is A0 here)

Mark scheme (e)

Scheme	Marks	AO
\(\text{RSS} = \mathrm{S}_{vv} \times (1 - r^2)\) or \(\mathrm{S}_{vv} - \dfrac{(\mathrm{S}_{xv})^2}{\mathrm{S}_{xx}}\), \(= 0.521\ldots\) = awrt 0.52	M1, A1	1.1b 1.1b
	(2)

Notes

M1: attempt at either correct expression

A1: awrt 0.52

Mark scheme (f)

Scheme	Marks	AO
\(v_\mathrm{A} = 8.42 - 0.16(20.9) + 0.18\) \(v_\mathrm{B} = 8.42 - 0.16(19.7) - 0.22\)	M1	3.4
\(v_\mathrm{A}\) = awrt 5.2 or awrt 5.3 \(v_\mathrm{B}\) = awrt 5.0	A1	1.1b
\(v_\mathrm{A} \gt v_\mathrm{B}\) therefore \(A\) took the shorter amount of time.	A1	2.2a
	(3)
	(14 marks)

Notes

M1: using the linear model to find the speed of either runner.
Must see use of residuals attempted but condone wrong sign.

A1: both speeds found correctly. Can allow 2^nd A1 for slightly incorrect speeds M1A0A1

A1: deducing that \(A\) took the shortest time (dep on M1 and consistent with speeds found)