S1 June 2014 Q3

EdexcelOld spec13 marksCorrelation & Regression

3. The table shows data on the number of visitors to the UK in a month, \(v\) (1000s), and the amount of money they spent, \(m\) (£ millions), for each of 8 months.

Number of visitors \(v\) (1000s)	2450	2480	2540	2420	2350	2290	2400	2460
Amount of money spent \(m\) (£ millions)	1370	1350	1400	1330	1270	1210	1330	1350

You may use

\[S_{vv} = 42587.5 \quad S_{vm} = 31512.5 \quad S_{mm} = 25187.5 \quad \sum v = 19390 \quad \sum m = 10610\]

(a) Find the product moment correlation coefficient between \(m\) and \(v\). (2)

(b) Give a reason to support fitting a regression model of the form \(m = a + bv\) to these data. (1)

(d) Find the equation of the regression line of \(m\) on \(v\). (2)

(e) Interpret your value of \(b\). (2)

(f) Use your answer to part (d) to estimate the amount of money spent when the number of visitors to the UK in a month is 2 500 000 (2)

(g) Comment on the reliability of your estimate in part (f). Give a reason for your answer. (2)

Mark scheme (a)

Scheme	Marks
\(r = \dfrac{31512.5}{\sqrt{42587.5\times 25187.5}} = 0.962\) awrt 0.962	M1 A1
	(2)

Notes

M1 for a correct expression for \(r\). Ans only of 0.96 or awrt 0.96 is M1A0. Ans only of 0.962 or awrt 0.962 is M1A1. Do not allow fractions for A1

Mark scheme (b)

Scheme	Marks
\(r\) is close to 1 or a strong correlation. [“points are close to a straight line” isB0] [Just “positive” correlation is B0] [Use of “relationship” or “skew” not “correlation” is B0]	B1
	(1)

Notes

B1 for comment implying strong correlation. (e.g. big/high/clear etc) B0 if \(|r| \gt 1\)

Mark scheme (c)

Scheme	Marks
\(b = \dfrac{31512.5}{42587.5} = 0.739947\ldots = 0.740\) (3 dp) 0.740 (only)	M1 A1cao
	(2)

Notes

M1 for a correct expression for \(b\) (may be implied by 0.74 or better in regression equation)

A1 A1 for 0.740 only in (c) or \(b = 0.740\) seen elsewhere (M1A0 for \(\dfrac{2521}{3407}\) or awrt 0.74 here)

Mark scheme (d)

Scheme	Marks
\(a = 1326.25 - (0.7399\ldots\times 2423.75)\) \([= -467.2\) or awrt \(-467]\)	M1
So \(\boldsymbol{m = -467 + 0.74v}\)	A1
	(2)

Notes

M1 for 1326.25 – (‘their \(b\)’ \(\times\) 2423.75) Condone fractions or awrt 1330 for \(\bar{m}\) and awrt 2420 for \(\bar{v}\)

A1 for a correct equation in \(m\) and \(v\) with \(a\) = awrt \(-467\) and \(b\) = awrt 0.74. Condone \(\frac{2521}{3407}\) for \(b\) and \(\frac{-1591740}{3407}\) for \(a\). [Equation in \(y\) and \(x\) is A0]

Mark scheme (e)

Scheme	Marks
\(b\) is the money (spent) per visitor. (i.e. definition of a rate in words.)[ignore values]	B1
So each 1000 visitors generates an extra £0.74 million or each visitor spends £740 oe	B1ft
	(2)

Notes

1st B1 for a correct definition of the rate in words. Must state or imply “money per visitor”. Allow alternative words or symbols e.g. £ or “pounds” for money, “people” for visitors etc

2nd B1ft for a correct numerical rate (ft their value of \(b\))

e.g. “each visitor spends £740” is B1B1, “\(b\) is the extra money spent per visitor” is B1B0 [no values]
“\(b\) is increase of £0.74 million in \(m\) as \(v\) increases by 1000” is B0B1[£ for money but no “visitors”]
“increase in \(m\) as \(v\) increases” is B0B0 [Idea of rate but letters not words and no numerical value of rate]

Mark scheme (f)

Scheme	Marks
\(m = -467 + 0.74\times 2500\)	M1
\(m = 1383\) (£ million) awrt 1380	A1
	(2)

Notes

M1 sub. \(v = 2500\) into their equation. Simply substituting 2 500 000 is M0 (unless adjusted eqn)

A1 awrt 1380 units (£ and million not required)

Mark scheme (g)

Scheme	Marks
As 2500 is within the range of the data set or it involves interpolation.	B1
The value of money spent is reliable	dB1
	(2)
	(13 marks)

Notes

1st B1 for 2500 or 2 500 000 or visitors or \(v\) is in range. “it” is B0 unless \(v\) clearly implied

2nd dB1 for stating it is reliable. Dependent on previous B mark being awarded

“both \(v\) and \(m\) in range” or “1380 in range” is B0 but use ISW so “interpolation since both in range” scores B1 for the “interpolation”. “Not extrapolation” counts as “interpolation”