| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Identify response/explanatory variables |
| Difficulty | Moderate -0.8 This is a straightforward application of standard linear regression formulas with all summations provided. Students simply substitute into memorized formulas for regression line, correlation coefficient, and perform a routine hypothesis test—no problem-solving or interpretation challenges beyond basic recall. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation5.09b Least squares regression: concepts5.09c Calculate regression line |
| \(x\) | 12.2 | 10.4 | 5.2 | 6.3 | 11.8 | 10.0 | 14.2 | 2.3 |
| \(y\) | 15 | 9 | 10 | 7 | 8 | 11 | 12 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(b = \frac{(761.3 - 72.4 \times 78/8)}{(769.9 - 72.4^2/8)}\) | M1 | Calculate gradient \(b\) in \(y - \bar{y} = b(x - \bar{x})\) |
| \(= \frac{55.4}{114.68}\ [or\ \frac{6.925}{14.335}]\) | ||
| \(= \frac{1385}{2867}\) or \(0.483[1]\) | A1 | |
| \(y - 9.75 = 0.483(x - 9.05)\) | M1 A1 | Find regression line |
| *Or* \(y = 5.38 + 0.483x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r = \frac{(761.3 - 72.4 \times 78/8)}{\sqrt{\{(769.9 - 72.4^2/8)(820 - 78^2/8)\}}}\) | M1 | Find correlation coefficient \(r\) |
| \(= \frac{55.4}{\sqrt{(114.68 \times 59.5)}}\ [or\ \frac{6.925}{\sqrt{(14.335 \times 7.4375)}}]\) | A1 | |
| \(= 0.671\) | *A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \rho = 0,\quad H_1: \rho > 0\) | B1 | State both hypotheses |
| \(r_{8,\,5\%} = 0.621\) | *B1 | State/use correct tabular one-tail \(r\) value |
| Reject \(H_0\) if \( | r | >\) tabular value |
| There is positive correlation | A1 | Correct conclusion (AEF, dep *A1, *B1) |
| Subtotal: 4 | Total: [11] |
## Question 8(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $b = \frac{(761.3 - 72.4 \times 78/8)}{(769.9 - 72.4^2/8)}$ | M1 | Calculate gradient $b$ in $y - \bar{y} = b(x - \bar{x})$ |
| $= \frac{55.4}{114.68}\ [or\ \frac{6.925}{14.335}]$ | | |
| $= \frac{1385}{2867}$ or $0.483[1]$ | A1 | |
| $y - 9.75 = 0.483(x - 9.05)$ | M1 A1 | Find regression line |
| *Or* $y = 5.38 + 0.483x$ | | |
**Subtotal: 4**
### Question 8(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r = \frac{(761.3 - 72.4 \times 78/8)}{\sqrt{\{(769.9 - 72.4^2/8)(820 - 78^2/8)\}}}$ | M1 | Find correlation coefficient $r$ |
| $= \frac{55.4}{\sqrt{(114.68 \times 59.5)}}\ [or\ \frac{6.925}{\sqrt{(14.335 \times 7.4375)}}]$ | A1 | |
| $= 0.671$ | *A1 | |
**Subtotal: 3**
### Question 8(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \rho = 0,\quad H_1: \rho > 0$ | B1 | State both hypotheses |
| $r_{8,\,5\%} = 0.621$ | *B1 | State/use correct tabular one-tail $r$ value |
| Reject $H_0$ if $|r| >$ tabular value | M1 | Valid method for reaching conclusion |
| There is positive correlation | A1 | Correct conclusion (AEF, dep *A1, *B1) |
**Subtotal: 4 | Total: [11]**
---8 The yield of a particular crop on a farm is thought to depend principally on the amount of sunshine during the growing season. For a random sample of 8 years, the average yield, $y$ kilograms per square metre, and the average amount of sunshine per day, $x$ hours, are recorded. The results are given in the following table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | }
\hline
$x$ & 12.2 & 10.4 & 5.2 & 6.3 & 11.8 & 10.0 & 14.2 & 2.3 \\
\hline
$y$ & 15 & 9 & 10 & 7 & 8 & 11 & 12 & 6 \\
\hline
\end{tabular}
\end{center}
$$\left[ \Sigma x = 72.4 , \Sigma x ^ { 2 } = 769.9 , \Sigma y = 78 , \Sigma y ^ { 2 } = 820 , \Sigma x y = 761.3 . \right]$$
(i) Find the equation of the regression line of $y$ on $x$.\\
(ii) Find the product moment correlation coefficient.\\
(iii) Test, at the $5 \%$ significance level, whether there is positive correlation between the average yield and the average amount of sunshine per day.
\hfill \mbox{\textit{CAIE FP2 2012 Q8 [11]}}