| Exam Board | Edexcel |
|---|---|
| Module | FS2 AS (Further Statistics 2 AS) |
| Year | 2020 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Linearize non-linear relationships |
| Difficulty | Standard +0.3 This is a standard Further Statistics linearization question with routine calculations: plotting points, finding regression equations using given summary statistics, calculating RSS from correlation coefficient, and transforming variables (w on d²). All steps are algorithmic with no novel insight required, though it's slightly above average difficulty due to being Further Maths content and requiring multiple statistical techniques. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.09a Dependent/independent variables5.09c Calculate regression line |
| \cline { 2 - 13 } \multicolumn{1}{l|}{} | These 14 points are plotted on page 13 | Not yet plotted | ||||||||||||||||
| \(d\) | 0.5 | 0.6 | 0.7 | 0.8 | 0.9 | 1.1 | 1.3 | 1.6 | 2 | 2.4 | 2.8 | 3.3 | 3.5 | 3.9 | \(\mathbf { 4 . 5 }\) | \(\mathbf { 4 . 6 }\) | \(\mathbf { 4 . 8 }\) | \(\mathbf { 5 . 4 }\) |
| \(w\) | 1.2 | 1.7 | 2.3 | 3.0 | 3.8 | 5.6 | 7.7 | 11.6 | 18 | 25.9 | 34.9 | 47.4 | 52.7 | 63.9 | \(\mathbf { 8 1 }\) | \(\mathbf { 8 3 . 6 }\) | \(\mathbf { 8 9 . 9 }\) | \(\mathbf { 1 0 9 . 4 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Scatter diagram – use overlay. All correct | B1 | For fully correct scatter diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Need to choose model of the form \(w = a + bd\) and have one of \(a\) or \(b\) correct to 2 sf | M1 | For selecting appropriate model and one coefficient correct to 2sf |
| \(\underline{w = 21.5d - 17.7}\) | A1 | \(b\) = awrt 21.5 and \(a\) = awrt \(-17.7\) |
| Answer | Marks | Guidance |
|---|---|---|
| Not appropriate because e.g. 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 | For comment suggesting not very good with a suitable reason |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{ww} = \sum w^2 - \dfrac{(\sum w)^2}{18} = 45178.68 - \dfrac{643.6^2}{18} = 22166.404\ldots\) | M1 | For calculation of \(S_{ww}\) or any other terms needed |
| \(\text{RSS} = S_{ww}(1-r^2) = 22166.404\ldots \times (1 - 0.987^2) =\) awrt \(\mathbf{570}\) \((g^2)\) | A1 | For RSS = 570.3299… i.e. awrt 570 |
| Answer | Marks | Guidance |
|---|---|---|
| Thicker wire should be stronger and strength is proportional to area (i.e. \(d^2\)) | B1 | For comment realising strength is proportional to \(d^2\) (area) |
| Answer | Marks | Guidance |
|---|---|---|
| \(w = cu + f\) where \(c = \dfrac{5721.625}{1482.619} = 3.85913\ldots\) | M1 | For using correct expression for gradient |
| \(f = \bar{w} - c\bar{u} = \dfrac{643.6}{18} - 3.8591\ldots \times \dfrac{157.57}{18} = 1.973\ldots\) | M1 | For correct expression for intercept |
| \(\underline{w = 1.97 + 3.86u}\) | A1 | For correct line with coefficients awrt 3 sf |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{RSS} = S_{ww} \times (1-r^2)\) or \(S_{ww} - \dfrac{(S_{wu})^2}{S_{uu}} = 85.8824\ldots\) awrt \(\mathbf{85.9}\) \((g^2)\) | M1, A1 | For correct expression (ft their \(S_{ww}\)) [NB \(r\) = awrt 0.998] |
| Answer | Marks | Guidance |
|---|---|---|
| Robert's model is better since RSS is reduced | B1 | For comment about reduced RSS (RSS needs to be lower but needn't be correct) |
| Answer | Marks |
|---|---|
| Use Robert's model: \(w = \{3.859 \times 3^2 + 1.973\} =\) awrt \(\mathbf{36.7}\) | B1 |
# Question 4(a):
Scatter diagram – use overlay. All correct | B1 | For fully correct scatter diagram
---
# Question 4(b):
Need to choose model of the form $w = a + bd$ and have one of $a$ or $b$ correct to 2 sf | M1 | For selecting appropriate model and one coefficient correct to 2sf
$\underline{w = 21.5d - 17.7}$ | A1 | $b$ = awrt 21.5 and $a$ = awrt $-17.7$
---
# Question 4(c):
Not appropriate because e.g. 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 | For comment suggesting not very good with a suitable reason
---
# Question 4(d):
$S_{ww} = \sum w^2 - \dfrac{(\sum w)^2}{18} = 45178.68 - \dfrac{643.6^2}{18} = 22166.404\ldots$ | M1 | For calculation of $S_{ww}$ or any other terms needed
$\text{RSS} = S_{ww}(1-r^2) = 22166.404\ldots \times (1 - 0.987^2) =$ awrt $\mathbf{570}$ $(g^2)$ | A1 | For RSS = 570.3299… i.e. awrt 570
---
# Question 4(e):
Thicker wire should be stronger and strength is proportional to area (i.e. $d^2$) | B1 | For comment realising strength is proportional to $d^2$ (area)
---
# Question 4(f):
$w = cu + f$ where $c = \dfrac{5721.625}{1482.619} = 3.85913\ldots$ | M1 | For using correct expression for gradient
$f = \bar{w} - c\bar{u} = \dfrac{643.6}{18} - 3.8591\ldots \times \dfrac{157.57}{18} = 1.973\ldots$ | M1 | For correct expression for intercept
$\underline{w = 1.97 + 3.86u}$ | A1 | For correct line with coefficients awrt 3 sf
---
# Question 4(g):
$\text{RSS} = S_{ww} \times (1-r^2)$ or $S_{ww} - \dfrac{(S_{wu})^2}{S_{uu}} = 85.8824\ldots$ awrt $\mathbf{85.9}$ $(g^2)$ | M1, A1 | For correct expression (ft their $S_{ww}$) [NB $r$ = awrt 0.998]
---
# Question 4(h):
Robert's model is better since RSS is reduced | B1 | For comment about reduced RSS (RSS needs to be lower but needn't be correct)
---
# Question 4(i):
Use Robert's model: $w = \{3.859 \times 3^2 + 1.973\} =$ awrt $\mathbf{36.7}$ | B1 |\begin{enumerate}
\item 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, $d \mathrm {~mm}$, and the weight, $w$ grams, when the wire fails. Their results are given in the following table.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | c | c | c | c | c | c | c | c | c | }
\cline { 2 - 13 }
\multicolumn{1}{l|}{} & \multicolumn{12}{|c|}{These 14 points are plotted on page 13} & \multicolumn{5}{|c|}{Not yet plotted} & \\
\hline
$d$ & 0.5 & 0.6 & 0.7 & 0.8 & 0.9 & 1.1 & 1.3 & 1.6 & 2 & 2.4 & 2.8 & 3.3 & 3.5 & 3.9 & $\mathbf { 4 . 5 }$ & $\mathbf { 4 . 6 }$ & $\mathbf { 4 . 8 }$ & $\mathbf { 5 . 4 }$ \\
\hline
$w$ & 1.2 & 1.7 & 2.3 & 3.0 & 3.8 & 5.6 & 7.7 & 11.6 & 18 & 25.9 & 34.9 & 47.4 & 52.7 & 63.9 & $\mathbf { 8 1 }$ & $\mathbf { 8 3 . 6 }$ & $\mathbf { 8 9 . 9 }$ & $\mathbf { 1 0 9 . 4 }$ \\
\hline
\end{tabular}
\end{center}
The first 14 points are plotted on the axes on page 13.\\
(a) On the axes on page 13, complete the scatter diagram for these data.\\
(b) Use your calculator to write down the equation of the regression line of $w$ on $d$.\\
(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
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.
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
$$\mathrm { S } _ { w u } = 5721.625 \quad \mathrm {~S} _ { u u } = 1482.619 \quad \sum u = 157.57$$
(e) By considering the physical nature of the problem, give a reason to support Robert's suggestion.\\
(f) Find the equation of the regression line of $w$ on $u$.\\
(g) Find the residual sum of squares (RSS) for Robert's model.\\
(h) State, giving a reason based on these calculations, which of these models better describes these data.\\
(i) Hence estimate the weight at which a piece of wire with diameter 3 mm will fail.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Question 4 continued}
\includegraphics[alt={},max width=\textwidth]{fbd7b196-5372-4956-8d38-92f05c92a5f7-13_2315_1363_301_358}
\end{center}
\end{figure}
\hfill \mbox{\textit{Edexcel FS2 AS 2020 Q4 [14]}}