A2 October 2021 Q4

4. A researcher is investigating the relationship between elevation, \(x\) metres, and annual mean temperature, \(t\,{}^\circ\mathrm{C}\).

From a random sample of 20 weather stations in Switzerland, the following results were obtained

\[\mathrm{S}_{xx} = 8\,820\,655 \qquad \mathrm{S}_{tt} = 444.7 \qquad \sum x = 28\,130 \qquad \sum t = 94.62\]

The product moment correlation coefficient for these data is found to be \(-0.959\)

(a) Interpret the value of this correlation coefficient. (1)

(b) Show that the equation of the regression line of \(t\) on \(x\) can be written as\[t = 14.3 - 0.00681x\] (4)

The random variable \(W\) represents the elevations of the weather stations in kilometres.

(c) Write down the equation of the regression line of \(t\) on \(w\) for these 20 weather stations in the form \(t = a + bw\) (1)

(d) Show that the residual sum of squares (RSS) for the model for \(t\) and \(x\) is 35.7 correct to one decimal place. (1)

One of the weather stations in the sample had a recorded elevation of 1100 metres and an annual mean temperature of \(1.4\,{}^\circ\mathrm{C}\)

(e)

(i) Calculate this weather station’s contribution to the residual sum of squares.
Give your answer as a percentage. (2)

(ii) Comment on the data for this weather station in light of your answer to part (e)(i). (1)

Scheme	Marks	AO
As elevation increases, temperature decreases.	B1	3.4
	(1)

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{S}_{xt} = -0.959\sqrt{8\,820\,655 \times 444.7}\ [= -60\,062.38727]\)	M1	2.1
\(b = \dfrac{\text{‘}-60\,062\ldots\text{’}}{8\,820\,655}\ [= -0.006809\ldots]\)	M1	1.1b
\(a = \dfrac{94.62}{20} - \text{‘}b\text{’}\dfrac{28\,130}{20}\ [= 14.308\ldots]\)	M1	1.1b
\(t = 14.3 - 0.00681x\) *	A1cso*	2.2a
	(4)

M1: Using pmcc to find \(\mathrm{S}_{xt}\)

M1: Setting up linear model by attempting to find \(b\)
Note: Allow M2 for \(b = r\sqrt{\dfrac{\mathrm{S}_{tt}}{\mathrm{S}_{xx}}}\)

M1: Setting up linear model by attempting to find \(a\)

A1cso*: Correct model \(t = 14.3 - 0.00681x\) with \(a\) = awrt 14.3 and \(b\) = awrt \(-0.00681\) dependent upon all previous M marks.

Scheme	Marks	AO
\(\left[w = \dfrac{x}{1000} \rightarrow\right] \quad t = 14.3 - 6.81w\)	B1	3.3
	(1)

Scheme	Marks	AO
\(444.7(1 - (-0.959)^2)\) or \(444.7 - \dfrac{(-60\,062\ldots)^2}{8\,820\,655}\) [ = 35.7*]	B1cso*	1.1b
	(1)

Mark scheme (e)

Scheme	Marks	AO
(i) (residual)² = \([1.4 - (14.3 - 0.00681(1100))]^2 \quad [= 29.2\ldots]\)	M1	3.4
\([29.2\ldots \div 35.7 \times 100\%]\) awrt 82%	A1	1.1b
	(2)
(ii) (As the point representing this data contributes to the majority of the RSS), the point is possibly an outlier and should be investigated.	B1	3.5a
	(1)
	(10 marks)

M1: Using the model to evaluate the squared residual

A1: awrt 82%

B1: Evaluating the result obtained from the model to suggest that this point may be an outlier