| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Convert regression equation between coded and original |
| Difficulty | Standard +0.3 This is a standard S1 regression question requiring systematic application of coding formulas and algebraic manipulation. Part (a) involves routine calculations with coded variables (finding means, Sxy, Sxx, then transforming back), while part (b) is straightforward substitution and solving. The multi-step nature and 12 total marks make it slightly above average difficulty, but all techniques are standard textbook procedures with no novel insight required. |
| Spec | 5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line5.09e Use regression: for estimation in context |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | M1 | \(S_{yy} = 3871 - \frac{57 \times 2222}{20} = -2461.7\) |
| M1 | \(S_{xx} = 401 - \frac{57^2}{20} = 238.55\) | |
| M1 A1 | \(b = \frac{-2461.7}{238.55} = -10.3194\) | |
| M1 A1 | \(a = \frac{2222}{20} - (-10.3194 \times \frac{57}{20}) = 140.5104\) | |
| A1 | \(y = 140.5104 - 10.3194x\) | |
| M1 | \(P - 300 = 140.5104 - 10.3194(T - 20)\) | |
| A1 | \(P = 646.9 - 10.3T\) | |
| (b) | M1 | \(460 = 646.9 - 10.3T\) |
| M1 A1, (12) | \(T = \frac{646.9 - 460}{10.3} = 18.1 \therefore 18°\mathrm{C}\) (nearest degree) |
(a) | M1 | $S_{yy} = 3871 - \frac{57 \times 2222}{20} = -2461.7$
| M1 | $S_{xx} = 401 - \frac{57^2}{20} = 238.55$
| M1 A1 | $b = \frac{-2461.7}{238.55} = -10.3194$
| M1 A1 | $a = \frac{2222}{20} - (-10.3194 \times \frac{57}{20}) = 140.5104$
| A1 | $y = 140.5104 - 10.3194x$
| M1 | $P - 300 = 140.5104 - 10.3194(T - 20)$
| A1 | $P = 646.9 - 10.3T$
(b) | M1 | $460 = 646.9 - 10.3T$
| M1 A1, (12) | $T = \frac{646.9 - 460}{10.3} = 18.1 \therefore 18°\mathrm{C}$ (nearest degree)
---The owner of a mobile burger-bar believes that hot weather reduces his sales.
To investigate the effect on his business he collected data on his daily sales, $£P$, and the maximum temperature, $T$°C, on each of 20 days. He then coded the data, using $x = T - 20$ and $y = P - 300$, and calculated the summary statistics given below.
$$\Sigma x = 57, \quad \Sigma y = 2222, \quad \Sigma x^2 = 401, \quad \Sigma y^2 = 305576, \quad \Sigma xy = 3871.$$
\begin{enumerate}[label=(\alph*)]
\item Find an equation of the regression line of $P$ on $T$. [9 marks]
\end{enumerate}
The owner of the bar doesn't believe it is profitable for him to run the bar if he takes less than £460 in a day.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item According to your regression line at what maximum daily temperature, to the nearest degree Celsius, does it become unprofitable for him to run the bar? [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q4 [12]}}