| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate regression line then predict |
| Difficulty | Standard +0.3 This is a straightforward application of standard linear regression formulas from A-level statistics. All parts involve direct substitution into memorized formulas (means, regression line, correlation coefficient, hypothesis test) with no conceptual challenges or problem-solving required. While it's a multi-part question worth several marks, each step is routine calculation that any competent student would recognize immediately, making it slightly easier than the average A-level question. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation5.09c Calculate regression line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\bar{x} = 50.1\), \(\bar{y} = 51.5\) | B1 | 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Calculate gradient \(b'\) in \(x - \bar{x} = b'(y - \bar{y})\): \(b' = (40236 - 752 \times 773/15)/(45351 - 773^2/15)\) | M1 | |
| \(= 1482.9/5515.7 = 0.268\) | A1 | |
| Use regression line at \(y = 56\): \(x = 50.13 + 0.2689(56 - 51.53)\) | M1 | |
| \([x = 36.28 + 0.2689y] = 51\) | A1 | 4 marks total |
| OR gradient \(b\) in \(y - \bar{y} = b(x - \bar{x})\): \(b = (40236 - 752 \times 773/15)/(38814 - 752^2/15)\) | (M1) | |
| \(= 1482.9/1113.7 = 1.33\) | (A1) | |
| Use regression line at \(y = 56\): \(x = 50.13 + (56 - 51.53)/1.332\) | (M1) | |
| \([y = -15.22 + 1.332x] = 53\) | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(r = (40236 - 752 \times 773/15)/\sqrt{\{(38814 - 752^2/15)(45351 - 773^2/15)\}}\) | M1 | |
| \(= 1482.9/\sqrt{(1113.7 \times 5515.7)}\) | ||
| \(= 1483/(33.37 \times 74.27)\) | ||
| *or* \(98.86/\sqrt{(74.25 \times 367.7)}\) | ||
| \(= 98.86/(8.617 \times 19.18)\) | ||
| \(= 0.598\) | *A1 | 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \rho = 0\), \(H_1: \rho \neq 0\) | B1 | |
| \(r_{15,\,5\%} = 0.514\) (to 2 dp) | *B1 | Correct tabular 2-tail \(r\) value |
| Reject \(H_0\) if \( | r | >\) tabular value |
| There is a non-zero coefficient | A1 | 4 marks total; AEF, dep *A1, *B1 |
# Question 9:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{x} = 50.1$, $\bar{y} = 51.5$ | B1 | 1 mark total |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Calculate gradient $b'$ in $x - \bar{x} = b'(y - \bar{y})$: $b' = (40236 - 752 \times 773/15)/(45351 - 773^2/15)$ | M1 | |
| $= 1482.9/5515.7 = 0.268$ | A1 | |
| Use regression line at $y = 56$: $x = 50.13 + 0.2689(56 - 51.53)$ | M1 | |
| $[x = 36.28 + 0.2689y] = 51$ | A1 | 4 marks total |
| **OR** gradient $b$ in $y - \bar{y} = b(x - \bar{x})$: $b = (40236 - 752 \times 773/15)/(38814 - 752^2/15)$ | (M1) | |
| $= 1482.9/1113.7 = 1.33$ | (A1) | |
| Use regression line at $y = 56$: $x = 50.13 + (56 - 51.53)/1.332$ | (M1) | |
| $[y = -15.22 + 1.332x] = 53$ | (A1) | |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $r = (40236 - 752 \times 773/15)/\sqrt{\{(38814 - 752^2/15)(45351 - 773^2/15)\}}$ | M1 | |
| $= 1482.9/\sqrt{(1113.7 \times 5515.7)}$ | | |
| $= 1483/(33.37 \times 74.27)$ | | |
| *or* $98.86/\sqrt{(74.25 \times 367.7)}$ | | |
| $= 98.86/(8.617 \times 19.18)$ | | |
| $= 0.598$ | *A1 | 2 marks total |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \rho = 0$, $H_1: \rho \neq 0$ | B1 | |
| $r_{15,\,5\%} = 0.514$ (to 2 dp) | *B1 | Correct tabular 2-tail $r$ value |
| Reject $H_0$ if $|r| >$ tabular value | M1 | Valid method for reaching conclusion |
| There is a non-zero coefficient | A1 | 4 marks total; AEF, dep *A1, *B1 |
---9 The marks achieved by a random sample of 15 college students in a Physics examination ( $x$ ) and in a General Studies examination (y) are summarised as follows.
$$\Sigma x = 752 \quad \Sigma x ^ { 2 } = 38814 \quad \Sigma y = 773 \quad \Sigma y ^ { 2 } = 45351 \quad \Sigma x y = 40236$$
(i) Find the mean values, $\bar { x }$ and $\bar { y }$.\\
(ii) Another college student achieved a mark of 56 in the General Studies examination, but was unable to take the Physics examination. Use the equation of a suitable regression line to estimate the mark that the student would have obtained in the Physics examination.\\
(iii) Find the product moment correlation coefficient for the given data.\\
(iv) Stating your hypotheses, test at the $5 \%$ level of significance whether there is a non-zero product moment correlation coefficient between examination marks in Physics and in General Studies achieved by college students.
\hfill \mbox{\textit{CAIE FP2 2011 Q9 [11]}}