| Exam Board | Edexcel |
|---|---|
| Module | FS2 AS (Further Statistics 2 AS) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate y on x from summary statistics |
| Difficulty | Standard +0.3 This is a standard Further Statistics linear regression question requiring routine application of formulas (correlation coefficient, regression line, RSS) with given summary statistics. All calculations are straightforward substitutions with no conceptual challenges, though part (f) requires basic interpretation of residual patterns. Slightly above average difficulty due to being Further Maths content and multi-part structure, but remains a textbook exercise. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.09c Calculate regression line5.09e Use regression: for estimation in context |
| \(\mathbf { x }\) | - 0.4 | - 0.2 | 0.3 | 0.8 | 1.1 | 1.4 | 1.8 | 2.1 | 2.5 | 2.6 |
| R esidual | - 0.63 | - 0.32 | - 0.52 | - 0.73 | 0.74 | 2.22 | 1.84 | 0.32 | \(f\) | - 1.88 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r = \frac{284.4 - \frac{251(12)}{10}}{\sqrt{10.36 \times 40.9}}\) | M1 | For a complete correct method for finding \(r\) |
| \(r = -0.79671\ldots\) awrt \(\mathbf{-0.797}\) | A1 | For awrt \(-0.797\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(b = \frac{-16.4}{10.36}\) | M1 | For use of correct model i.e. correct expression for \(b\) (ft their \(S_{xy}\)) |
| \(a = \frac{251}{10} - b \cdot \frac{12}{10}\) | M1 | For use of correct model i.e. correct (ft) expression for \(a\) |
| \(y = 27.0 - 1.58x\) | A1 | A correct answer here can imply both method marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = [27.0 - 1.58(2)] = 23.84\) awrt 23.8 | B1ft | For awrt 23.8, evaluating their model from (b) with \(x = 2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(RSS = 40.9 - \frac{(-16.4)^2}{10.36}\) | M1 | For a correct expression for RSS |
| \(RSS = 14.938\ldots\) awrt 14.9 | A1 | For awrt 14.9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum\text{residuals} = 0 \rightarrow -0.63 + (-0.32) + \ldots + f + (-1.88) = 0\) | M1 | For use of \(\sum\text{residuals} = 0\); use of regression equation needs correct sign |
| \(f = \mathbf{-1.04}\) | A1 | For \(-1.04\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The residuals should be randomly scattered above and below zero so linear model may not be appropriate | B1 | For identifying residuals are not randomly scattered above and below zero and concluding linear regression model may not be appropriate |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r = \frac{284.4 - \frac{251(12)}{10}}{\sqrt{10.36 \times 40.9}}$ | M1 | For a complete correct method for finding $r$ |
| $r = -0.79671\ldots$ awrt $\mathbf{-0.797}$ | A1 | For awrt $-0.797$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $b = \frac{-16.4}{10.36}$ | M1 | For use of correct model i.e. correct expression for $b$ (ft their $S_{xy}$) |
| $a = \frac{251}{10} - b \cdot \frac{12}{10}$ | M1 | For use of correct model i.e. correct (ft) expression for $a$ |
| $y = 27.0 - 1.58x$ | A1 | A correct answer here can imply both method marks |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = [27.0 - 1.58(2)] = 23.84$ awrt **23.8** | B1ft | For awrt 23.8, evaluating their model from (b) with $x = 2$ |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $RSS = 40.9 - \frac{(-16.4)^2}{10.36}$ | M1 | For a correct expression for RSS |
| $RSS = 14.938\ldots$ awrt **14.9** | A1 | For awrt 14.9 |
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum\text{residuals} = 0 \rightarrow -0.63 + (-0.32) + \ldots + f + (-1.88) = 0$ | M1 | For use of $\sum\text{residuals} = 0$; use of regression equation needs correct sign |
| $f = \mathbf{-1.04}$ | A1 | For $-1.04$ |
### Part (f):
| Answer/Working | Marks | Guidance |
|---|---|---|
| The residuals should be randomly scattered above and below zero so linear model may not be appropriate | B1 | For identifying residuals are not randomly scattered above and below zero and concluding linear regression model may not be appropriate |
---\begin{enumerate}
\item A scientist wants to develop a model to describe the relationship between the average daily temperature, $\mathrm { x } ^ { \circ } \mathrm { C }$, and a household's daily energy consumption, ykWh , in winter.
\end{enumerate}
A random sample of the average temperature and energy consumption are taken from 10 winter days and are summarised below.
$$\begin{gathered}
\sum x = 12 \quad \sum x ^ { 2 } = 24.76 \quad \sum y = 251 \quad \sum y ^ { 2 } = 6341 \quad \sum x y = 284.8 \\
S _ { x x } = 10.36 \quad S _ { y y } = 40.9
\end{gathered}$$
(a) Find the product moment correlation coefficient between y and x .\\
(b) Find the equation of the regression line of $y$ on $x$ in the form $y = a + b x$\\
(c) Use your equation to estimate the daily energy consumption when the average daily temperature is $2 ^ { \circ } \mathrm { C }$\\
(d) Calculate the residual sum of squares (RSS).
The table shows the residual for each value of x .
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | c | }
\hline
$\mathbf { x }$ & - 0.4 & - 0.2 & 0.3 & 0.8 & 1.1 & 1.4 & 1.8 & 2.1 & 2.5 & 2.6 \\
\hline
R esidual & - 0.63 & - 0.32 & - 0.52 & - 0.73 & 0.74 & 2.22 & 1.84 & 0.32 & $f$ & - 1.88 \\
\hline
\end{tabular}
\end{center}
(e) Find the value of f.\\
(f) By considering the signs of the residuals, explain whether or not the linear regression model is a suitable model for these data.
\hfill \mbox{\textit{Edexcel FS2 AS Q3 [11]}}