AS October 2020 Q4

EdexcelAS paperCurrent spec14 marksLinear Regression

4. Some students are investigating the strength of wire by suspending a weight at the end of the wire. They measure the diameter of the wire, \(d\) mm, and the weight, \(w\) grams, when the wire fails. Their results are given in the following table.

	These 14 points are plotted below
\(d\)	0.5	0.6	0.7	0.8	0.9	1.1	1.3	1.6	2
\(w\)	1.2	1.7	2.3	3.0	3.8	5.6	7.7	11.6	18

	These 14 points are plotted below					Not yet plotted
\(d\)	2.4	2.8	3.3	3.5	3.9	4.5	4.6	4.8	5.4
\(w\)	25.9	34.9	47.4	52.7	63.9	81	83.6	89.9	109.4

The first 14 points are plotted on the axes on page 13.

Axes for a scatter diagram: w from 0 to 110 against d from 0 to 6, with the first 14 points plotted as crosses, rising steeply in a curve from (0.5, 1.2) to (3.9, 63.9)

(a) On the axes below, complete the scatter diagram for these data. (1)

(b) Use your calculator to write down the equation of the regression line of \(w\) on \(d\). (2)

(c) With reference to the scatter diagram, comment on the appropriateness of using this linear regression model to make predictions for \(w\) for different values of \(d\) between 0.5 and 5.4 (1)

The product moment correlation coefficient for these data is \(r = 0.987\) (to 3 significant figures).

(d) Calculate the residual sum of squares (RSS) for this model. (2)

Robert, one of the students, suggests that the model could be improved and intends to find the equation of the line of regression of \(w\) on \(u\), where \(u = d^2\)
He finds the following statistics

\[\mathrm{S}_{wu} = 5721.625 \qquad \mathrm{S}_{uu} = 1482.619 \qquad \sum u = 157.57\]

(e) By considering the physical nature of the problem, give a reason to support Robert’s suggestion. (1)

(f) Find the equation of the regression line of \(w\) on \(u\). (3)

(g) Find the residual sum of squares (RSS) for Robert’s model. (2)

(h) State, giving a reason based on these calculations, which of these models better describes these data. (1)

(i) Hence estimate the weight at which a piece of wire with diameter 3 mm will fail. (1)

Scheme	Marks	AO
Use overlay. All correct	B1	1.1b
	(1)

Mark scheme (b)

Scheme	Marks	AO
Need to choose model of the form: \(w = a + bd\) and have one of \(a\) or \(b\) correct to 2 sf	M1	3.3
\(w = 21.5d - 17.7\)	A1	1.1b
	(2)

Notes

M1 for selecting the appropriate model and one coefficient correct to 2sf

A1 for \(b\) = awrt 21.5 and \(a\) = awrt – 17.7

Mark scheme (c)

Scheme	Marks	AO
Not appropriate because eg the line is plotted and not close to the points or two lines with different gradients or overestimates values in the middle and underestimates the others or the points are more curved	B1	3.5a
	(1)

Notes

B1 for comment suggesting not very good with a suitable reason.

Mark scheme (d)

Scheme	Marks	AO
\(\left\{\mathrm{S}_{ww} = \sum w^2 - \dfrac{\left(\sum w\right)^2}{18} = 45178.68 - \dfrac{643.6^2}{18}\right\} = 22166.404..\)	M1	1.1b
\(\text{RSS} = \mathrm{S}_{ww}\left(1 - r^2\right) = 22166.404\ldots \times (1 - 0.987^2)\) = awrt 570 (g²)	A1	1.1b
	(2)

Notes

M1 for calculation of \(\mathrm{S}_{ww}\) or any other terms needed for their calculation

A1 for RSS = 570.3299… i.e. awrt 570

Scheme	Marks	AO
Thicker wire should be stronger and strength is proportional to area (i.e. \(d^2\))	B1	2.4
	(1)

Mark scheme (f)

Scheme	Marks	AO
\(w = cu + f\) where \(c = \dfrac{5721.625}{1482.619} = 3.85913\ldots\)	M1	3.3
\(f \left\{= \bar{w} - c\bar{u}\right\} = \dfrac{\text{“}643.6\text{”}}{18} - \text{“}3.8591...\text{”} \times \dfrac{157.57}{18}\) \(\{= 1.973\ldots\}\)	M1	1.1b
\(w = 1.97 + 3.86u\)	A1	1.1b
	(3)

Notes

1^st M1 for using correct expression for gradient

2^nd M1 for correct expression for intercept

A1 for correct line with coefficients awrt 3 sf

Mark scheme (g)

Scheme	Marks	AO
\(\text{RSS} = \mathrm{S}_{ww} \times (1 - r^2)\) or \(\mathrm{S}_{ww} - \dfrac{(\mathrm{S}_{wu})^2}{\mathrm{S}_{uu}}\), \(= 85.8824\ldots\) awrt 85.9 (g²)	M1, A1	1.1b (x2)
	(2)

Notes

M1 for a correct expression (ft their \(\mathrm{S}_{ww}\)) [NB \(r\) = awrt 0.998]

Scheme	Marks	AO
Use Robert’s model: \(w\ \{= 3.859 \times 3^2 + 1.973\}\) = awrt 36.7	B1	3.4
	(1)
	(14 marks)