| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Convert regression equation between coded and original |
| Difficulty | Moderate -0.3 This is a straightforward application of standard regression formulas with coding transformations. Part (a) requires calculating b = Sxy/Sxx and a = ȳ - bx̄ using given summary statistics. Part (b) involves routine algebraic substitution to convert between coded and original variables. Part (c) is direct substitution. All steps are mechanical with no conceptual challenges beyond understanding the coding relationship, making this slightly easier than average. |
| Spec | 5.09b Least squares regression: concepts5.09c Calculate regression line5.09d Linear coding: effect on regression5.09e Use regression: for estimation in context |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{xy} = 8880 - \frac{130 \times 48}{8} = (8100)\) | B1 | May be implied |
| Answer | Marks | Guidance |
|---|---|---|
| \(b = \frac{s_{xy}}{s_{xx}} = \frac{8100}{20487.5} = 0.395363...\) | M1 A1 | Allow use of their \(S_{xy}\) for M; awrt 0.395 |
| \(a = \frac{48}{8} - (0.395363...)\frac{130}{8} = -0.424649...\) | M1 A1 | Allow use of their \(b\) for M; awrt \(-0.425\) |
| \(y = -0.425 + 0.395x\) | B1\(\sqrt{}\) | 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| \(f - 100 = -0.424649... + 0.395...(m - 250)\) | M1 A1\(\sqrt{}\) | Subst \(f - 100\) & \(m - 250\) |
| \(f = 0.735 + 0.395m\) | A1 | 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| \(m = 235 \Rightarrow f = 93.64489....\) | B1 | awrt 93.6/93.7 |
# Question 3:
## Part (a)
$S_{xy} = 8880 - \frac{130 \times 48}{8} = (8100)$ | B1 | May be implied
$S_{xx} = 20487.5$
$b = \frac{s_{xy}}{s_{xx}} = \frac{8100}{20487.5} = 0.395363...$ | M1 A1 | Allow use of their $S_{xy}$ for M; awrt 0.395
$a = \frac{48}{8} - (0.395363...)\frac{130}{8} = -0.424649...$ | M1 A1 | Allow use of their $b$ for M; awrt $-0.425$
$y = -0.425 + 0.395x$ | B1$\sqrt{}$ | 3 s.f.
Special case answer only B0 M0 B1 M0 B1 B1 (fully correct 3sf) ($\equiv$ to B0 M0 A1 M0 A1 B1 on the epen)
## Part (b)
$f - 100 = -0.424649... + 0.395...(m - 250)$ | M1 A1$\sqrt{}$ | Subst $f - 100$ & $m - 250$
$f = 0.735 + 0.395m$ | A1 | 3 s.f.
## Part (c)
$m = 235 \Rightarrow f = 93.64489....$ | B1 | awrt 93.6/93.7
---\begin{enumerate}
\item A long distance lorry driver recorded the distance travelled, $m$ miles, and the amount of fuel used, $f$ litres, each day. Summarised below are data from the driver's records for a random sample of 8 days.
\end{enumerate}
The data are coded such that $x = m - 250$ and $y = f - 100$.
$$\Sigma x = 130 \quad \Sigma y = 48 \quad \Sigma x y = 8880 \quad \mathrm {~S} _ { x x } = 20487.5$$
(a) Find the equation of the regression line of $y$ on $x$ in the form $y = a + b x$.\\
(b) Hence find the equation of the regression line of $f$ on $m$.\\
(c) Predict the amount of fuel used on a journey of 235 miles.\\
\hfill \mbox{\textit{Edexcel S1 2005 Q3 [10]}}