S1 June 2014 (R) Q3

EdexcelOld spec16 marksCorrelation & Regression

3. A large company is analysing how much money it spends on paper in its offices every year. The number of employees, \(x\), and the amount of money spent on paper, \(p\) (£ hundreds), in 8 randomly selected offices are given in the table below.

\(x\)	8	9	12	14	7	3	16	19
\(p\) (£ hundreds)	40.5	36.1	30.4	39.4	32.6	31.1	43.4	45.7

(You may use \(\sum x^2 = 1160 \qquad \sum p = 299.2 \qquad \sum p^2 = 11\,422 \qquad \sum xp = 3449.5\))

(a) Show that \(S_{pp} = 231.92\) and find the value of \(S_{xx}\) and the value of \(S_{xp}\) (5)

(b) Calculate the product moment correlation coefficient between \(x\) and \(p\). (2)

The equation of the regression line of \(p\) on \(x\) is given in the form \(p = a + bx\).

(d) Estimate the amount of money spent on paper in an office with 10 employees. (2)

(e) Explain the effect each additional employee has on the amount of money spent on paper. (1)

Later the company realised it had made a mistake in adding up its costs, \(p\). The true costs were actually half of the values recorded. The product moment correlation coefficient and the equation of the linear regression line are recalculated using this information.

(f) Write down the new value of

(i) the product moment correlation coefficient,

(ii) the gradient of the regression line. (2)

Mark scheme (a)

Scheme	Marks
\(\sum x = 88\)	B1
\(S_{pp} = 11422 - \dfrac{299.2^2}{8} = [231.92]\) (*)	B1cso
\(S_{xx} = 1160 - \dfrac{\text{'}88\text{'}^2}{8} = 192\)	M1 A1
\(S_{xp} = 3449.5 - \dfrac{\text{'}88\text{'} \times 299.2}{8} = 158.3\) awrt 158	A1
	(5)

Notes

1st B1 for \(\sum x = 88\) seen. May be in a correct formula or implied by 192 or 158.3

2nd B1cso for a correct expression for \(S_{pp}\)

M1 for a correct expression for \(S_{xx}\) or \(S_{xp}\) (ft their \(\Sigma x\)). If we don’t see an explicit \(\Sigma x = k\) but consistent use of \(k\) instead of 88 in \(S_{xp}\) and \(S_{xx}\) then award M1

1st A1 for \(S_{xx} = 192\) 2nd A1 for \(S_{xp}\) = awrt 158

Mark scheme (b)

Scheme	Marks
\(r = \left[\dfrac{S_{xp}}{\sqrt{S_{xx}S_{pp}}}\right] = \dfrac{\text{'}158.3\text{'}}{\sqrt{\text{'}192\text{'} \times 231.92}}\)	M1
\(r = 0.7501726031\ldots\) awrt 0.750	A1
	(2)

Notes

M1 for correct expression for \(r\) ft their 192 and 158.3 May be implied by \(r = 0.75\)

A1 for awrt 0.750 Allow A1 for \(r = 0.75\) if a correct expr’ is seen (since 3rd sf is 0)

Mark scheme (c)

Scheme	Marks
\(b = \left[\dfrac{S_{xp}}{S_{xx}}\right] = \dfrac{\text{'}158.3\text{'}}{\text{'}192\text{'}} = 0.824(479166\ldots)\) (*)	M1 A1cso
\(a = \bar{p} - b\bar{x} = \dfrac{299.2}{8} - 0.824\ldots \times \dfrac{\text{"}88\text{"}}{8} = 28.330729\ldots\) awrt 28.3	M1 A1
	(4)

Notes

1st M1 for a correct expression for \(b\) using their values NB. use of 158 gives 0.8229

1st A1 cso for \(b\) = awrt 0.824

SC If there is no expression but 0.8244... or better is seen award 1 mark as M0A1

2nd M1 for a correct expression for \(a\) ft their \(\Sigma x\)

2nd A1 for \(a\) = awrt 28.3

Mark scheme (d)

Scheme	Marks
\(p = 28.3\ldots + 0.824\ldots \times 10 = 36.57552\ldots\) awrt £3700	M1 A1
	(2)

Notes

M1 for substituting \(x = 10\) into their equation

A1 for awrt £3700 (£ 36.58 or £36.58 (hundreds) is A0)

Scheme	Marks
Goes up £82.40	B1
	(1)