| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Hypothesis test for zero correlation |
| Difficulty | Standard +0.3 This is a standard A-level statistics question requiring routine application of correlation coefficient formula, a straightforward hypothesis test using tables, and regression line calculation. All steps are procedural with no conceptual challenges beyond remembering formulas, making it slightly easier than average. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation5.09c Calculate regression line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r = (92.01 - 18.7 \times 34.7/12)/\sqrt{\{(106.43 - 18.7^2/12)(133.43 - 34.7^2/12)\}}\) | M1 | Find correlation coefficient \(r\) |
| \(= 37.94/\sqrt{(77.29 \times 33.09)}\) | ||
| \(= 37.94/(8.791 \times 5.752)\) | ||
| or \(3.161/\sqrt{(6.441 \times 2.757)}\) | ||
| \(= 3.161/(2.538 \times 1.661)\) | A1 | |
| \(= 0.750\) [allow \(0.75\)] | *A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \rho = 0\), \(H_1: \rho \neq 0\) | B1 | State both hypotheses |
| \(r_{12,\,1\%} = 0.708\) (to 2 dp) | *B1 | Use correct tabular 2-tail \(r\) value |
| Reject \(H_0\) if \( | r | >\) tabular value |
| There is non-zero correlation | A1 | Correct conclusion (AEF, dep *A1, *B1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| *EITHER:* \(b = (92.01 - 18.7 \times 34.7/12)/(106.43 - 18.7^2/12)\) | Calculate gradient \(b\) in \(y - \bar{y} = b(x - \bar{x})\) | |
| \(= 37.94/77.29 = 0.491\) | B1 | |
| \(y = 34.7/12 + 0.491(x - 18.7/12)\) | M1 | Use regression line for \(y\) at \(x=2\) |
| \([y = 0.491x + 2.13]\ = 2.13 + 0.491x = 3.11 \pm 0.01\) | A1 | |
| *OR:* \(b' = (92.01 - 18.7 \times 34.7/12)/(133.43 - 34.7^2/12)\) | Calculate gradient \(b'\) in \(x - \bar{x} = b'(y - \bar{y})\) | |
| \(= 37.94/33.09 = 1.15\) | (B1) | |
| \(y = 34.7/12 + (x - 18.7/12)/1.15\) | (M1) | Use regression line for \(y\) at \(x=2\) |
| \([x = 1.15y - 1.76]\ = 1.53 + x/1.15 = 3.28 \pm 0.01\) | (A1) |
## Question 10(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r = (92.01 - 18.7 \times 34.7/12)/\sqrt{\{(106.43 - 18.7^2/12)(133.43 - 34.7^2/12)\}}$ | M1 | Find correlation coefficient $r$ |
| $= 37.94/\sqrt{(77.29 \times 33.09)}$ | | |
| $= 37.94/(8.791 \times 5.752)$ | | |
| or $3.161/\sqrt{(6.441 \times 2.757)}$ | | |
| $= 3.161/(2.538 \times 1.661)$ | A1 | |
| $= 0.750$ [allow $0.75$] | *A1 | |
**Total part: [3]**
## Question 10(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \rho = 0$, $H_1: \rho \neq 0$ | B1 | State both hypotheses |
| $r_{12,\,1\%} = 0.708$ (to 2 dp) | *B1 | Use correct tabular 2-tail $r$ value |
| Reject $H_0$ if $|r| >$ tabular value | M1 | Valid method for reaching conclusion |
| There is non-zero correlation | A1 | Correct conclusion (AEF, dep *A1, *B1) |
**Total part: [4]**
## Question 10(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| *EITHER:* $b = (92.01 - 18.7 \times 34.7/12)/(106.43 - 18.7^2/12)$ | | Calculate gradient $b$ in $y - \bar{y} = b(x - \bar{x})$ |
| $= 37.94/77.29 = 0.491$ | B1 | |
| $y = 34.7/12 + 0.491(x - 18.7/12)$ | M1 | Use regression line for $y$ at $x=2$ |
| $[y = 0.491x + 2.13]\ = 2.13 + 0.491x = 3.11 \pm 0.01$ | A1 | |
| *OR:* $b' = (92.01 - 18.7 \times 34.7/12)/(133.43 - 34.7^2/12)$ | | Calculate gradient $b'$ in $x - \bar{x} = b'(y - \bar{y})$ |
| $= 37.94/33.09 = 1.15$ | (B1) | |
| $y = 34.7/12 + (x - 18.7/12)/1.15$ | (M1) | Use regression line for $y$ at $x=2$ |
| $[x = 1.15y - 1.76]\ = 1.53 + x/1.15 = 3.28 \pm 0.01$ | (A1) | |
**Total part: [3], Question Total: [10]**10 The mid-day temperature, $x ^ { \circ } \mathrm { C }$, and the amount of sunshine, $y$ hours, were recorded at a winter holiday resort on each of 12 days, chosen at random during the winter season. The results are summarised as follows.
$$\Sigma x = 18.7 \quad \Sigma x ^ { 2 } = 106.43 \quad \Sigma y = 34.7 \quad \Sigma y ^ { 2 } = 133.43 \quad \Sigma x y = 92.01$$
(i) Find the product moment correlation coefficient for the data.\\
(ii) Stating your hypotheses, test at the $1 \%$ significance level whether there is a non-zero correlation between mid-day temperature and amount of sunshine.\\
(iii) Use the equation of a suitable regression line to estimate the number of hours of sunshine on a day when the mid-day temperature is $2 ^ { \circ } \mathrm { C }$.
\hfill \mbox{\textit{CAIE FP2 2011 Q10 [10]}}