| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate y on x from raw data table |
| Difficulty | Moderate -0.8 This is a straightforward application of standard linear regression formulas with all summations provided. Students simply substitute into the formulae for gradient and intercept, then interpret results. The calculations are routine and require no problem-solving insight beyond textbook procedures. |
| Spec | 2.02c Scatter diagrams and regression lines5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line5.09d Linear coding: effect on regression5.09e Use regression: for estimation in context |
| \(l\) (metres) | 0.5 | 0.8 | 1.0 | 1.5 | 2 | 4 | 6 |
| \(P ( \pounds )\) | 2.50 | 3.40 | 4.00 | 5.20 | 6.00 | 10.50 | 15.00 |
| Answer | Marks | Guidance |
|---|---|---|
| Scatter diagram with correctly plotted points | B3 | |
| \(S_p = 159.77 - \frac{15.8 \times 46.6}{7} = 54.5871\) | M1 | |
| \(S_{ll} = 60.14 - \frac{15.8^2}{7} = 24.4771\) | M1 | |
| \(b = \frac{54.5871}{24.4771} = 2.2301\) | M1 A1 | |
| \(a = \frac{46.6}{7} - (2.2301 \times \frac{15.8}{7}) = 1.6234\) | M1 A1 | |
| \(P = 1.62 + 2.23l\) | A1 | |
| increase in price in £ per extra metre of tubing | B1 | |
| \(1.62 + (2.23 \times 5.2) = £13.22\) | M1 A1 | |
| e.g. machine may only be able to produce tubes up to a certain length so longer ones would be very difficult and expensive to make | B2 | (15) |
| Scatter diagram with correctly plotted points | B3 | |
| $S_p = 159.77 - \frac{15.8 \times 46.6}{7} = 54.5871$ | M1 | |
| $S_{ll} = 60.14 - \frac{15.8^2}{7} = 24.4771$ | M1 | |
| $b = \frac{54.5871}{24.4771} = 2.2301$ | M1 A1 | |
| $a = \frac{46.6}{7} - (2.2301 \times \frac{15.8}{7}) = 1.6234$ | M1 A1 | |
| $P = 1.62 + 2.23l$ | A1 | |
| increase in price in £ per extra metre of tubing | B1 | |
| $1.62 + (2.23 \times 5.2) = £13.22$ | M1 A1 | |
| e.g. machine may only be able to produce tubes up to a certain length so longer ones would be very difficult and expensive to make | B2 | (15) |7. Pipes-R-us manufacture a special lightweight aluminium tubing.
The price $\pounds P$, for each length, $l$ metres, that the company sells is shown in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
$l$ (metres) & 0.5 & 0.8 & 1.0 & 1.5 & 2 & 4 & 6 \\
\hline
$P ( \pounds )$ & 2.50 & 3.40 & 4.00 & 5.20 & 6.00 & 10.50 & 15.00 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Represent these data on a scatter diagram.
You may use
$$\Sigma l = 15.8 , \quad \Sigma P = 46.6 , \quad \Sigma l ^ { 2 } = 60.14 , \quad \Sigma l P = 159.77$$
\item Find the equation of the regression line of $P$ on $l$ in the form $P = a + b l$.
\item Give a practical interpretation of the constant b.
In response to customer demand Pipes- $R$-us decide to start selling tubes cut to specific lengths. Initially the company decides to use the regression line found in part (b) as a pricing formula for this new service.
\item Calculate the price that Pipes- $R$-us should charge for 5.2 metres of the tubing.
\item Suggest a reason why Pipes- $R$-us might not offer prices based on the regression line for any length of tubing.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q7 [15]}}