| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | January |
| Marks | 10 |
| 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 memorized formulas for S_xx, S_xy, then b and a, followed by basic interpretation and substitution. No problem-solving or conceptual insight required beyond routine calculation. |
| Spec | 5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line5.09e Use regression: for estimation in context |
| 2 | 2.5 | 3 | 4 | 4.5 | 4.5 | 5 | 3 | 6 | 6.5 | ||
| 22 | 34 | 33 | 37 | 40 | 45 | 49 | 30 | 58 | 58 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S_{xy} = 1818.5 - \frac{41 \times 406}{10} = 153.9\) | M1, A1 | AWRT 154; can be seen in (b) |
| \(S_{xx} = 188 - \frac{41^2}{10} = 19.9\) | A1 | Can be seen in (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(b = \frac{153.9}{19.9} = 7.733668...\) | M1, A1 | AWRT 7.73 |
| \(a = 40.6 - b \times 4.1 = 8.89796...\) | M1 | Correct formula for \(a\); ft their \(b\) |
| \(y = 8.89 + 7.73x\) | A1 | Accept \(a = 8.89\), \(b = 7.73\) even if not written as final equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| A typical car will travel 7700 miles every year | B1ft | For their \(b \times 1000\) to at least 2 sf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = 5\), \(y = 8.89 + 7.73 \times 5 = 47.5\) | M1 | Substituting \(x = 5\) into final answer to (b) |
| Mileage predicted is AWRT 48000 | A1 | Accept "48 thousands" |
## Question 4:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_{xy} = 1818.5 - \frac{41 \times 406}{10} = 153.9$ | M1, A1 | AWRT 154; can be seen in (b) |
| $S_{xx} = 188 - \frac{41^2}{10} = 19.9$ | A1 | Can be seen in (b) |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $b = \frac{153.9}{19.9} = 7.733668...$ | M1, A1 | AWRT 7.73 |
| $a = 40.6 - b \times 4.1 = 8.89796...$ | M1 | Correct formula for $a$; ft their $b$ |
| $y = 8.89 + 7.73x$ | A1 | Accept $a = 8.89$, $b = 7.73$ even if not written as final equation |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| A typical car will travel 7700 miles every year | B1ft | For their $b \times 1000$ to at least 2 sf |
### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 5$, $y = 8.89 + 7.73 \times 5 = 47.5$ | M1 | Substituting $x = 5$ into final answer to (b) |
| Mileage predicted is AWRT 48000 | A1 | Accept "48 thousands" |
---4. A second hand car dealer has 10 cars for sale. She decides to investigate the link between the age of the cars, $x$ years, and the mileage, $y$ thousand miles. The data collected from the cars are shown in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | c | }
\hline
\begin{tabular}{ c }
Age, $x$ \\
(years) \\
\end{tabular} & 2 & 2.5 & 3 & 4 & 4.5 & 4.5 & 5 & 3 & 6 & 6.5 \\
\hline
\begin{tabular}{ c }
Mileage, $y$ \\
(thousands) \\
\end{tabular} & 22 & 34 & 33 & 37 & 40 & 45 & 49 & 30 & 58 & 58 \\
\hline
\end{tabular}
\end{center}
[You may assume that $\sum x = 41 , \sum y = 406 , \sum x ^ { 2 } = 188 , \sum x y = 1818.5$ ]
\begin{enumerate}[label=(\alph*)]
\item Find $S _ { x x }$ and $S _ { x y }$.
\item Find the equation of the least squares regression line in the form $y = a + b x$. Give the values of $a$ and $b$ to 2 decimal places.
\item Give a practical interpretation of the slope $b$.
\item Using your answer to part (b), find the mileage predicted by the regression line for a 5 year old car.\\
$\_\_\_\_$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2008 Q4 [10]}}