| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate PMCC from summary statistics |
| Difficulty | Standard +0.3 This is a straightforward application of standard linear regression formulas from Further Statistics 2. Parts (a)-(d) involve direct substitution into PMCC and regression line formulas with given summary statistics (S_ff, S_fw already calculated). Part (e) requires basic residual plot interpretation, and part (f) asks for a standard refinement suggestion. All steps are routine for FS2 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08b Linear coding: effect on pmcc5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line5.09d Linear coding: effect on regression5.09e Use regression: for estimation in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_{ww} = 13447 - \frac{303^2}{8} = 1970.875\) | ||
| \(r = \frac{269.5}{\sqrt{42 \times 1970.875}}\) | M1 | Complete correct method for finding \(r\) |
| \(r = 0.9367\ldots\) awrt \(0.937\) | A1 | for awrt 0.937 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| As the amount of fertiliser increases the yield increases | B1 | Correct contextual statement |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b = \frac{269.5}{42} \ [= 6.41666\ldots]\) | M1 | For use of a correct model ie a correct expression for \(b\) |
| \(a = \frac{303}{8} - 'b' \cdot \frac{28}{8} \ [= 15.41666\ldots]\) | M1 | For use of a correct model ie a correct expression (ft) for \(a\) |
| \(w = 15.4 + 6.42f\) | A1 | For a correct model \(w = 15.4 + 6.42f\) with awrt 15.4 and awrt 6.42 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3.21\) tonnes | B1ft | awrt 3.21, condone \(-3.21\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The residual plot is close to an 'n' shape or the residuals appear not to be randomly scattered | M1 | Explaining a reason for their conclusion eg there is a pattern/trend in the residuals. Do not accept residuals not close to zero |
| The model in part (c) is unlikely to be suitable | A1 | Concluding it is not valid oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Fit a curve rather than a line | B1 | A comment about not using a linear line eg use a quadratic model, logarithmic graph, exponential |
## Question 2:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_{ww} = 13447 - \frac{303^2}{8} = 1970.875$ | | |
| $r = \frac{269.5}{\sqrt{42 \times 1970.875}}$ | M1 | Complete correct method for finding $r$ |
| $r = 0.9367\ldots$ awrt $0.937$ | A1 | for awrt 0.937 |
**(2 marks)**
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| As the amount of **fertiliser** increases the **yield** increases | B1 | Correct contextual statement |
**(1 mark)**
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b = \frac{269.5}{42} \ [= 6.41666\ldots]$ | M1 | For use of a correct model ie a correct expression for $b$ |
| $a = \frac{303}{8} - 'b' \cdot \frac{28}{8} \ [= 15.41666\ldots]$ | M1 | For use of a correct model ie a correct expression (ft) for $a$ |
| $w = 15.4 + 6.42f$ | A1 | For a correct model $w = 15.4 + 6.42f$ with awrt 15.4 and awrt 6.42 |
**(3 marks)**
### Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3.21$ tonnes | B1ft | awrt 3.21, condone $-3.21$ |
**(1 mark)**
### Part (e)
| Answer/Working | Mark | Guidance |
|---|---|---|
| The **residual** plot is close to an 'n' shape or the **residuals** appear not to be **randomly** scattered | M1 | Explaining a reason for their conclusion eg there is **a pattern/trend** in the **residuals**. Do not accept residuals not close to zero |
| The model in part (c) is unlikely to be suitable | A1 | Concluding it is not valid oe |
**(2 marks)**
### Part (f)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Fit a curve rather than a line | B1 | A comment about not using a linear line eg use a quadratic model, logarithmic graph, exponential |
**(1 mark)**2 A large field of wheat is split into 8 plots of equal area. Each plot is treated with a different amount of fertiliser, $f$ grams $/ \mathrm { m } ^ { 2 }$. The yield of wheat, $w$ tonnes, from each plot is recorded. The results are summarised below.
$$\sum f = 28 \quad \sum w = 303 \quad \sum w ^ { 2 } = 13447 \quad \mathrm {~S} _ { f f } = 42 \quad \mathrm {~S} _ { f w } = 269.5$$
\begin{enumerate}[label=(\alph*)]
\item Calculate the product moment correlation coefficient between $f$ and $w$
\item Interpret the value of your product moment correlation coefficient.
\item Find the equation of the regression line of $w$ on $f$ in the form $w = a + b f$
\item Using your equation, estimate the decrease in yield when the amount of fertiliser decreases by 0.5 grams $/ \mathrm { m } ^ { 2 }$
The residuals of the data recorded are calculated and plotted on the graph below.\\
\includegraphics[max width=\textwidth, alt={}, center]{67df73d4-6ce4-45f7-8a69-aa94292ea814-04_1232_1294_1169_301}
\item With reference to this graph, comment on the suitability of the model you found in part (c).
\item Suggest how you might be able to refine your model.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FS2 2019 Q2 [10]}}