S1 June 2012 Q3

EdexcelOld spec15 marksCorrelation & Regression

3. A scientist is researching whether or not birds of prey exposed to pollutants lay eggs with thinner shells. He collects a random sample of egg shells from each of 6 different nests and tests for pollutant level, \(p\), and measures the thinning of the shell, \(t\). The results are shown in the table below.

\(p\)	3	8	30	25	15	12
\(t\)	1	3	9	10	5	6

[You may use \(\sum p^2 = 1967\) and \(\sum pt = 694\)]

(a) Draw a scatter diagram on the axes below to represent these data. (2)

Axes for scatter diagram: p from 0 to 35, t from 0 to 11

(b) Explain why a linear regression model may be appropriate to describe the relationship between \(p\) and \(t\). (1)

(d) Find the equation of the regression line of \(t\) on \(p\), giving your answer in the form \(t = a + bp\). (4)

(e) Plot the point \((\bar{p}, \bar{t})\) and draw the regression line on your scatter diagram. (2)

The scientist reviews similar studies and finds that pollutant levels above 16 are likely to result in the death of a chick soon after hatching.

(f) Estimate the minimum thinning of the shell that is likely to result in the death of a chick. (2)

Scheme	Marks
Use overlay	B1 B1
	(2)

Scheme	Marks
Points (appear to) lie close to a (straight) line or “strong /high correlation”	B1
	(1)

Mark scheme (c)

Scheme	Marks
\(\sum p = 93\) and \(\sum t = 34\) (may be seen in table)	M1
\(S_{pt} = 694 - \dfrac{\text{"}93\text{"}\times\text{"}34\text{"}}{6} = [167]\) or \(S_{pp} = 1967 - \dfrac{\text{"}93\text{"}^2}{6} = [525.5]\)	M1
\(S_{pt} = 167\); \(S_{pp} =\) awrt 526	A1; A1
	(4)

Notes

1st M1 for attempting \(\sum p\) and \(\sum t\). Allow \(80 \lt \sum p \lt 100\) and \(30 \lt \sum t \lt 40\)

2nd M1 for one correct expression for \(S_{pt}\) or \(S_{pp}\), f.t. their \(\sum p\) and \(\sum t\).

1st A1 for \(S_{pt}\)

2nd A1 for \(S_{pp}\)

Mark scheme (d)

Scheme	Marks
\(b = \left[\dfrac{S_{pt}}{S_{pp}} =\right] \dfrac{\text{"}167\text{"}}{\text{"}525.5\text{"}} = [0.31779\ldots]\) (check their answer if expression not seen)	B1ft
\(a = \dfrac{\text{"}34\text{"}}{6} - \text{"}0.31779\ldots\text{"}\times\dfrac{\text{"}93\text{"}}{6} = 5.666\ldots - 0.31779\ldots\times 15.5 =,\ 0.74088\ldots\) awrt 0.74	M1, A1
\(\boldsymbol{t = 0.741 + 0.318p}\) (Accept \(a = \frac{2336}{3153}\) and \(b = \frac{334}{1051}\) in their equation)	A1
	(4)

Notes

B1ft for correct expression for the gradient, f.t. their 167 and 525.5 from (c)

M1 for correct use of \(a = \bar{t} - b\bar{p}\) f.t. their values. Condone 5.6 for \(\bar{t}\)

1st A1 for awrt 0.74 NB use of 526 gives 0.745566… and gets A0

2nd A1 for a correct equation for \(t\) in terms of \(p\) with \(a\) and \(b\) awrt 3sf. An equn in \(y\) or \(x\) is A0

Mark scheme (e)

Scheme	Marks
\((\bar{p}, \bar{t}) = (15.5, 5.7)\) plotted on the graph (not wholly outside the circle)	B1
Correct line plotted as per overlay. For \(p = 5\); \(2 \lt t \lt 3\) and for \(p = 30\); \(10 \lt t \lt 11\) Their line must stretch roughly as far as the points and go through the \((\bar{p}, \bar{t})\) circle	B1
	(2)

Mark scheme (f)

Scheme	Marks
\(t = \text{"}0.741\text{"} + \text{"}0.318\text{"}\times 16\)	M1
\(= 5.825\ldots\) awrt 5.8	A1
	(2)
	(15 marks)

Notes

M1 for clear use of their line (equation or on graph) and \(p = 16\) to estimate \(t\). This may be an expression or lines marked on the diagram

A1 for awrt 5.8, even if their line is not fully correct. Accept “\(t \gt 5.8\)”(oe). Answer only 2/2