S1 January 2012 Q5

EdexcelOld spec15 marksCorrelation & Regression

5. The age, \(t\) years, and weight, \(w\) grams, of each of 10 coins were recorded. These data are summarised below.

\[\sum t^2 = 2688 \qquad \sum tw = 1760.62 \qquad \sum t = 158 \qquad \sum w = 111.75 \qquad S_{ww} = 0.16\]

(a) Find \(S_{tt}\) and \(S_{tw}\) for these data. (3)

(b) Calculate, to 3 significant figures, the product moment correlation coefficient between \(t\) and \(w\). (2)

(d) State, with a reason, which variable is the explanatory variable. (2)

(e) Using this model, estimate

(i) the weight of a coin which is 5 years old,

(ii) the effect of an increase of 4 years in age on the weight of a coin.

(2)

It was discovered that a coin in the original sample, which was 5 years old and weighed 20 grams, was a fake.

(f) State, without any further calculations, whether the exclusion of this coin would increase or decrease the value of the product moment correlation coefficient. Give a reason for your answer. (2)

Mark scheme (a)

Scheme	Marks
\(S_{tt} = 2688 - \dfrac{158^2}{10} = 191.6\) awrt 192	M1 A1
\(S_{tw} = 1760.62 - \dfrac{158 \times 111.75}{10} = -5.03\) awrt -5.03	A1
	(3)

Notes

M1 for correct attempt at either method,
A1 awrt 192
A1 awrt -5.03

Mark scheme (b)

Scheme	Marks
\(r = \dfrac{-5.03}{\sqrt{191.6 \times 0.16}} = -0.908469\ldots\) awrt -0.908(5)	M1A1
	(2)

Notes

M1 for correct attempt at use of formula, square root required.
A1 awrt -0.908(5)

Mark scheme (c)

Scheme	Marks
\(b = \dfrac{-5.03}{191.6} = -0.0263\) awrt -0.026	M1 A1
\(a = 11.175 + 0.0263 \times 15.8\) \(= 11.59\)	M1
\(w = 11.6 - 0.0263t\)	A1
	(4)

Notes

M1 require ‘their -5.03’ as numerator and /their 191.6’ as denominator.
A1 awrt -0.026
M1 for use of correct formula with \(b\) or ‘their \(b\)’; require \(- -\) or + and values in the correct place.
A1 for equation as written with values awrt 3 sf. with \(w\) and \(t\).
Accept fractional answers that are accurate to 3sf when evaluated as decimals

Mark scheme (d)

Scheme	Marks
The explanatory variable is the age of each coin. This is because the age is set and the weight varies.	B1 B1
	(2)

Notes

B1 for ‘Age’ or \(t\) or ‘years’
B1 for ‘you use age / t to predict w’ or ‘you can control t/ age’ or ‘weight depends on age’ or similar

Scheme	Marks
(i) awrt 11.5	B1
(ii) Decrease(in weight of coin of 0.1052 g) = 0.1 or -0.1 or increase of -0.1 awrt(-0.1)	B1
	(2)

Mark scheme (f)

Scheme	Marks
Decrease; removing the fake will result in a better linear fit so \(r\) will be closer to -1	B1;B1
	(2)
	(15 marks)

Notes

B1 for Decrease only but ‘mod r increases’ explicitly stated in words or symbols award B1.
B1 accept ‘stronger correlation’ or ‘increase in correlation’ or ‘better linear fit’ or ‘ \(r\) closer to -1’ or ‘points are closer to a straight line’ or ‘point is an outlier’ or equivalent

Special Case 1 Attempt to calculate \(S_{tw}\)
\(\sum tw = 1660.62,\ \sum t = 153,\ \sum w = 91.75\) or \(S_{tw} = 1660.62 - \dfrac{153 \times 91.75}{9}\) or awrt 101 (corrected from the printed mark scheme: \(\sum tw = 1669.62\))
or \(S_{tw} \gt 0\) with some calculation B1
“Increase” B1 (2)

Special Case 2 Attempt to calculate \(S_{ww}\)
\(\sum w^2 = 1248.96625 - 400 = 848.96625\) or awrt 849 or \(S_{ww} = 848.96625 - \dfrac{91.75^2}{9}\)
or awrt -86.4 or \(S_{ww} \lt 0\) B2 (2)

Special Case 3 Argument based on standard deviation.
e.g. \(\sigma_w \approx 0.126\) and \(\bar{w} = 11.175\) so fake coin is over 69 sds away from the mean B1
‘(very) unlikely’ or ‘impossible’ B1 (2)