S1 June 2016 Q3

EdexcelOld spec10 marksCorrelation & Regression Data Processing

3. Before going on holiday to Seapron, Tania records the weekly rainfall (\(x\) mm) at Seapron for 8 weeks during the summer. Her results are summarised as

\[\sum x = 86.8 \qquad \sum x^2 = 985.88\]

(a) Find the standard deviation, \(\sigma_x\), for these data. (3)

Tania also records the number of hours of sunshine (\(y\) hours) per week at Seapron for these 8 weeks and obtains the following

\[\bar{y} = 58 \qquad \sigma_y = 9.461 \text{ (correct to 4 significant figures)} \qquad \sum xy = 4900.5\]

(b) Show that \(\mathrm{S}_{yy} = 716\) (correct to 3 significant figures) (1)

(d) Calculate the product moment correlation coefficient, \(r\), for these data. (2)

During Tania’s week-long holiday at Seapron there are 14 mm of rain and 70 hours of sunshine.

(e) State, giving a reason, what the effect of adding this information to the above data would be on the value of the product moment correlation coefficient. (2)

Mark scheme (a)

Scheme	Marks
\([\sigma_x^2 =]\ \dfrac{985.88}{8} - \left(\dfrac{86.8}{8}\right)^2 = \dfrac{985.88}{8} - 10.85^2\)	M1
\(\sigma_x = \sqrt{\dfrac{985.88}{8} - \left(\dfrac{86.8}{8}\right)^2} = \sqrt{123.235 - 117.7225} = \sqrt{5.5125}\) or \(\sqrt{\dfrac{44.1}{8}}\)	A1
\(= 2.3478\ldots = \) awrt 2.35	A1
	(3)

Notes

M1 for a correct expr’ for st. dev or variance (ignore label)[may be implied by 2.35 or 5.5125]

1st A1 for a correct expression for st. dev (must have square root) can ignore label

2nd A1 for awrt 2.35 (allow \(s = 2.5099\ldots\) or awrt 2.51). If they have \(\sigma^2 = 2.35\) score A0 but condone no label

Mark scheme (b)

Scheme	Marks
\(\mathrm{S}_{yy} = 8 \times \sigma_y^{\,2} = 716\) (3 sf) but may see \(1136.584 - \dfrac{58^2}{8}\) or \(27628(.084168) - \dfrac{464^2}{8}\) or 716.08... (= 716 to 3 sf) (*)	B1cso
	(1)

Notes

B1cso for a correct expression or sight of at least 716.08… (NB limits: 716.00~716.16)
Do not allow verification. Beware circular arguments: \(716 \to \Sigma y^2 \to\) expr’ \(\to 716\)

Mark scheme (c)

Scheme	Marks
\(\mathrm{S}_{xy} = 4900.5 - 58 \times 86.8\) or \(4900.5 - \dfrac{86.8 \times 464}{8}\)	M1
\(= -\)133.9 (Allow \(-134\))	A1
	(2)

Notes

M1 for a correct expression for \(\mathrm{S}_{xy}\) (NB \(\Sigma y = 464\))

A1 for \(-133.9\) or awrt \(-134\) [No fractions] (Answer only 2/2)

Mark scheme (d)

Scheme	Marks
\(r = \dfrac{-133.9/8}{\sigma_x \times \sigma_y}\) or \(\dfrac{-133.9}{\sqrt{44.1 \times 716}}\)	M1
\(= \) awrt \(-\)0.753 or \(-\)0.754	A1
	(2)

Notes

M1 for a correct expression for \(r\) (ft their values for \(\mathrm{S}_{xy}\) and \(\sigma_x\) or \(\mathrm{S}_{xx}\))[Allow ft of \(\mathrm{S}_{yy}\)]

A1 for awrt \(-0.753\) or \(-0.754\) (Answer only 2/2)

Mark scheme (e)

Scheme	Marks
\(r \lt 0\) means high sunshine and low rain; this is high sunshine high rain	B1
[this is not in keeping with the trend so] \(r\) is closer to 0 or \(\|r\|\) decreases	B1
	(2)
	(10 marks)

Notes

If they do not have an answer to (d) or their value of \(r\) is > 0 or \(|r| \gt 1\) score B0B0 here

1st B1 for a suitable reason contradicting \(r \lt 0\) e.g. new value is not in keeping with trend or both \(14 \gt \bar{x}\) and \(70 \gt \bar{y}\) or saying both above average. Allow for \(-0.48 \lt\) new \(r \lt -0.47\)

2nd B1 for a correct statement about \(r\) getting closer to zero e.g. \(|r|\) decreases
A comment that \(r\) decreases or \(r\) is smaller or \(r\) is “less negative” is B0
“\(r\) increases” is B0 unless they also say that it gets closer to 0