| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Convert regression equation between coded and original |
| Difficulty | Moderate -0.8 This is a straightforward application of standard regression formulas with coded variables. Part (a) requires direct substitution into b = S_pv/S_vv and a = p̄ - bv̄, while part (b) involves converting x=85 to v using the given coding, then substituting. All steps are routine with no conceptual challenges or problem-solving required. |
| Spec | 5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line5.09d Linear coding: effect on regression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(b = \frac{1.688}{5.753} = 0.293\) | M1A1 | M1 for correct expression for \(b\). \(\frac{1.688}{1.168}\) is M0. A1 for awrt 0.29 |
| \(a = 3.22 - 4.42 \times 0.293 = 1.9231...\) | M1 | For use of \(a = \bar{p} - b\bar{v}\), follow through their value of \(b\) (or even just the letter \(b\)) |
| \(p = 1.92 + 0.293v\) | A1 | For complete equation with \(a\) = awrt 1.92 and \(b\) = awrt 0.293. \(y\) or \(p = 1.92 + 0.293x\) is A0. Correct answer with no working is 4/4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v = \frac{85-5}{10} = 8\) | M1 | For attempt to find value of \(v\) when \(x = 85\) (at least 2 correct terms in \(\pm\frac{85-5}{10}\)), or attempt to find equation for \(p\) in terms of \(x\) using \(x=85\). Attempt at equation of \(p\) in \(x\) requires \(p = 1.92 + 0.293\frac{(x-5)}{10}\) |
| \(p = 1.92 + 0.293 \times 8 = 4.3\) (awrt 4.3) | A1 | For awrt 4.3 (award when first seen, apply ISW). N.B. \(p = 1.92 + 0.293 \times 85\) is M0A0 |
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $b = \frac{1.688}{5.753} = 0.293$ | M1A1 | M1 for correct expression for $b$. $\frac{1.688}{1.168}$ is M0. A1 for awrt 0.29 |
| $a = 3.22 - 4.42 \times 0.293 = 1.9231...$ | M1 | For use of $a = \bar{p} - b\bar{v}$, follow through their value of $b$ (or even just the letter $b$) |
| $p = 1.92 + 0.293v$ | A1 | For complete equation with $a$ = awrt 1.92 and $b$ = awrt 0.293. $y$ or $p = 1.92 + 0.293x$ is A0. Correct answer with no working is 4/4 |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v = \frac{85-5}{10} = 8$ | M1 | For attempt to find value of $v$ when $x = 85$ (at least 2 correct terms in $\pm\frac{85-5}{10}$), or attempt to find equation for $p$ in terms of $x$ using $x=85$. Attempt at equation of $p$ in $x$ requires $p = 1.92 + 0.293\frac{(x-5)}{10}$ |
| $p = 1.92 + 0.293 \times 8 = 4.3$ (awrt 4.3) | A1 | For awrt 4.3 (award when first seen, apply ISW). N.B. $p = 1.92 + 0.293 \times 85$ is M0A0 |
---\begin{enumerate}
\item A farmer collected data on the annual rainfall, $x \mathrm {~cm}$, and the annual yield of peas, $p$ tonnes per acre.
\end{enumerate}
The data for annual rainfall was coded using $v = \frac { x - 5 } { 10 }$ and the following statistics were found.
$$S _ { v v } = 5.753 \quad S _ { p v } = 1.688 \quad S _ { p p } = 1.168 \quad \bar { p } = 3.22 \quad \bar { v } = 4.42$$
(a) Find the equation of the regression line of $p$ on $v$ in the form $p = a + b v$.\\
(b) Using your regression line estimate the annual yield of peas per acre when the annual rainfall is 85 cm .\\
\hfill \mbox{\textit{Edexcel S1 2011 Q4 [6]}}