| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate y on x from raw data table |
| Difficulty | Standard +0.3 This is a straightforward application of standard regression formulas with all summations provided. Students must choose y-on-x (since x is controlled), substitute into the formula for regression coefficients, then solve for x when y=0. The only mild challenge is the interpretation/reliability comment, but this is routine for A-level statistics questions. |
| Spec | 5.09a Dependent/independent variables5.09c Calculate regression line5.09e Use regression: for estimation in context |
| \(x\) | 0.10 | 0.15 | 0.20 | 0.25 | 0.30 | 0.35 | 0.40 | 0.45 | 0.50 | 0.55 |
| \(y\) | 48 | 40 | 44 | 35 | 39 | 24 | 10 | 13 | 9 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y\) depends on \(x\) so choose \(y\) on \(x\) | B1 | State choice of line with reason (A.E.F.) |
| \(b = (66.1 - 3.25\times268/10)\ /\ (1.2625 - 3.25^2/10)\) | Find coefficient \(b\) in regression line for \(y\) | |
| \(= -21/0.20625 = -101.8\) *or* \(-102\) | M1 A1 | |
| \(y = b(x - 0.325) + 26.8\) | Find equation of regression line | |
| \(= 59.9 - 102x\) | M1 A1 | |
| SR: M1 A1 for finding \(x\) on \(y\): \(x = 0.563 - 0.00088y\) | (M1 A1) | Part mark: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(59.9/102 = 0.587\) *or* \(0.588\) *or* \(0.59\) | M1 A1 | Find \(x\) when \(y = 0\) |
| SR: If using eqn of \(x\) on \(y\): \(0.563\) | (B1) | |
| OK since point just outside range | Valid comment on reliability | |
| *or* OK as \(r \approx -1\) *or* \( | r | \approx 1\) (A.E.F.) |
## Question 7:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y$ depends on $x$ so choose $y$ on $x$ | B1 | State choice of line with reason (A.E.F.) |
| $b = (66.1 - 3.25\times268/10)\ /\ (1.2625 - 3.25^2/10)$ | | Find coefficient $b$ in regression line for $y$ |
| $= -21/0.20625 = -101.8$ *or* $-102$ | M1 A1 | |
| $y = b(x - 0.325) + 26.8$ | | Find equation of regression line |
| $= 59.9 - 102x$ | M1 A1 | |
| **SR:** M1 A1 for finding $x$ on $y$: $x = 0.563 - 0.00088y$ | (M1 A1) | Part mark: 5 |
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $59.9/102 = 0.587$ *or* $0.588$ *or* $0.59$ | M1 A1 | Find $x$ when $y = 0$ |
| **SR:** If using eqn of $x$ on $y$: $0.563$ | (B1) | |
| OK since point just outside range | | Valid comment on reliability |
| *or* OK as $r \approx -1$ *or* $|r| \approx 1$ (A.E.F.) | B1 | Part mark: 3. Total: **[8]** |
---7 An experiment was carried out to determine how much weedkiller to apply per $100 \mathrm {~m} ^ { 2 }$ in a large field. Ten $100 \mathrm {~m} ^ { 2 }$ areas of the field were randomly chosen and sprayed with predetermined volumes of the weedkiller. The volume of the weedkiller is denoted by $x$ litres and the number of weeds that survived is denoted by $y$. The results are given in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | c | }
\hline
$x$ & 0.10 & 0.15 & 0.20 & 0.25 & 0.30 & 0.35 & 0.40 & 0.45 & 0.50 & 0.55 \\
\hline
$y$ & 48 & 40 & 44 & 35 & 39 & 24 & 10 & 13 & 9 & 6 \\
\hline
\end{tabular}
\end{center}
$$\left[ \Sigma x = 3.25 , \Sigma x ^ { 2 } = 1.2625 , \Sigma y = 268 , \Sigma y ^ { 2 } = 9548 , \Sigma x y = 66.10 . \right]$$
It is given that the product moment correlation coefficient for the data is - 0.951 , correct to 3 decimal places.\\
(i) Calculate the equation of a suitable regression line, giving a reason for your choice of line.\\
(ii) Estimate the best volume of weedkiller to apply, and comment on the reliability of your estimate.
\hfill \mbox{\textit{CAIE FP2 2009 Q7 [8]}}