| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 16 |
| 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 S1 regression question requiring standard calculations with given summary statistics (Σx², Σy², Σxy). Students follow a routine algorithm to find a and b, then interpret results. The only mild challenge is part (f) on extrapolation, but this is a standard textbook concept. Significantly easier than average A-level maths questions. |
| 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 |
| \(x ( \mathrm { mph } )\) | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 |
| \(y \left( { } ^ { \circ } \mathrm { C } \right)\) | 53 | 55 | 63 | 65 | 78 | 83 | 91 | 101 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Scales & labels correct | B1 | |
| Points plotted correctly | B2, 1, 0 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Points lie reasonably close to a straight line | B1 (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(b = \frac{8 \times 20615 - 260 \times 589}{8 \times 9500 - (260)^2} = \frac{11780}{8400} = 1.40238\ldots\) | M1 A1 | Accept awrt 1.40 |
| \(a = \frac{589}{8} - (1.40238\ldots)\left(\frac{260}{8}\right) = 28.0476175\ldots\) | M1 A1 (4) | Accept awrt 28.0 |
| \(\therefore y = 28.0 + 1.40x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a \Rightarrow\) surrounding air temperature when tyre is stationary | B1 | |
| \(b \Rightarrow\) for every extra mph, temperature rises by \(1.40\ ^\circ\text{C}\) | B1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = 28.0 + 1.40 \times 50 = 98\) | B1 | |
| Regression line is only a line of best fit and does not necessarily pass through all points | B1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 12 mph – reasonable to use line; 12 is just below lowest \(x\)-value | B1; B1 | |
| 85 mph – not reasonable to use line; 85 is well outside range of values | B1; B1 | (4) |
## Question 6:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Scales & labels correct | B1 | |
| Points plotted correctly | B2, 1, 0 (3) | |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Points lie reasonably close to a straight line | B1 (1) | |
### Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $b = \frac{8 \times 20615 - 260 \times 589}{8 \times 9500 - (260)^2} = \frac{11780}{8400} = 1.40238\ldots$ | M1 A1 | Accept awrt 1.40 |
| $a = \frac{589}{8} - (1.40238\ldots)\left(\frac{260}{8}\right) = 28.0476175\ldots$ | M1 A1 (4) | Accept awrt 28.0 |
| $\therefore y = 28.0 + 1.40x$ | | |
### Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a \Rightarrow$ surrounding air temperature when tyre is stationary | B1 | |
| $b \Rightarrow$ for every extra mph, temperature rises by $1.40\ ^\circ\text{C}$ | B1 (2) | |
## Question (e):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 28.0 + 1.40 \times 50 = 98$ | B1 | |
| Regression line is only a line of best fit and does not necessarily pass through all points | B1 | (2) |
## Question (f):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 12 mph – reasonable to use line; 12 is just below lowest $x$-value | B1; B1 | |
| 85 mph – not reasonable to use line; 85 is well outside range of values | B1; B1 | (4) |
**Total: (16)**6. To test the heating of tyre material, tyres are run on a test rig at chosen speeds under given conditions of load, pressure and surrounding temperature. The following table gives values of $x$, the test rig speed in miles per hour (mph), and the temperature, $y ^ { \circ } \mathrm { C }$, generated in the shoulder of the tyre for a particular tyre material.
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | l | l | c | }
\hline
$x ( \mathrm { mph } )$ & 15 & 20 & 25 & 30 & 35 & 40 & 45 & 50 \\
\hline
$y \left( { } ^ { \circ } \mathrm { C } \right)$ & 53 & 55 & 63 & 65 & 78 & 83 & 91 & 101 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Draw a scatter diagram to represent these data.
\item Give a reason to support the fitting of a regression line of the form $y = a + b x$ through these points.
\item Find the values of $a$ and $b$.\\
(You may use $\Sigma x ^ { 2 } = 9500 , \Sigma y ^ { 2 } = 45483 , \Sigma x y = 20615$ )
\item Give an interpretation for each of $a$ and $b$.
\item Use your line to estimate the temperature at 50 mph and explain why this estimate differs from the value given in the table.
A tyre specialist wants to estimate the temperature of this tyre material at 12 mph and 85 mph .
\item Explain briefly whether or not you would recommend the specialist to use this regression equation to obtain these estimates.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q6 [16]}}