| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2003 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate y on x from raw data table |
| Difficulty | Moderate -0.3 This is a standard S1 linear regression question with coding that follows a routine template. All summary statistics are provided, requiring only substitution into formulas for the regression line, then reverse-coding. The interpretation and prediction parts are straightforward applications. While it has multiple parts (5 marks total based on typical S1 marking), each step is mechanical with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.09a 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 |
| Year | Number of new car sale, \(c\) | Cost of advertising (£000), \(p\) |
| 1990 | 4240 | 120 |
| 1991 | 4380 | 126 |
| 1992 | 4420 | 132 |
| 1993 | 4440 | 134 |
| 1994 | 4430 | 137 |
| 1995 | 4520 | 144 |
| 1996 | 4590 | 148 |
| 1997 | 4660 | 150 |
| 1998 | 4700 | 153 |
| 1999 | 4790 | 158 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x\): 20 26 32 34 37 44 48 50 53 58 | B1 | |
| \(y\): 24 38 42 44 43 52 59 66 70 79 | B1 | |
| Change in cost of advertising influences number of new car sales | B1 | |
| Graph: scale and labels | B1 | |
| Points all correct | B2 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_{xy} = 22611 - \frac{402 \times 517}{10} = 1827.6\) | M1A1 | |
| \(S_{xx} = 17538 - \frac{402^2}{10} = 1377.6\) | A1 | |
| \(b = \frac{S_{xy}}{S_{xx}} = \frac{1827.6}{1377.6} = 1.3267...\) | M1A1 | |
| \(a = \frac{517}{10} - (1.326655...)\times\frac{402}{10} = -1.63153...\) | B1 | |
| \(\therefore y = -1.63 + 1.33x\) | B1ft | (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{c-4000}{10} = -1.63 + 1.33(p-100)\) | M1A1ft | |
| \(c = 2653.7 + 13.3p\) | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| No. sold if no money spent on advertising | B1 | |
| \(p=0\) is well outside valid range — meaningless | B1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2 \times 13.3 = 27\) extra cars sold | B1 | |
| Only valid in range of data for 1990s | B1 | (2 marks) |
# Question 6:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x$: 20 26 32 34 37 44 48 50 53 58 | B1 | |
| $y$: 24 38 42 44 43 52 59 66 70 79 | B1 | |
| Change in cost of advertising influences number of new car sales | B1 | |
| Graph: scale and labels | B1 | |
| Points all correct | B2 | **(5 marks)** |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{xy} = 22611 - \frac{402 \times 517}{10} = 1827.6$ | M1A1 | |
| $S_{xx} = 17538 - \frac{402^2}{10} = 1377.6$ | A1 | |
| $b = \frac{S_{xy}}{S_{xx}} = \frac{1827.6}{1377.6} = 1.3267...$ | M1A1 | |
| $a = \frac{517}{10} - (1.326655...)\times\frac{402}{10} = -1.63153...$ | B1 | |
| $\therefore y = -1.63 + 1.33x$ | B1ft | **(7 marks)** |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{c-4000}{10} = -1.63 + 1.33(p-100)$ | M1A1ft | |
| $c = 2653.7 + 13.3p$ | A1 | **(3 marks)** |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| No. sold if no money spent on advertising | B1 | |
| $p=0$ is well outside valid range — meaningless | B1 | **(2 marks)** |
## Part (e)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2 \times 13.3 = 27$ extra cars sold | B1 | |
| Only valid in range of data for 1990s | B1 | **(2 marks)** |6. The chief executive of Rex cars wants to investigate the relationship between the number of new car sales and the amount of money spent on advertising. She collects data from company records on the number of new car sales, $c$, and the cost of advertising each year, $p$ (£000). The data are shown in the table below.
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Year & Number of new car sale, $c$ & Cost of advertising (£000), $p$ \\
\hline
1990 & 4240 & 120 \\
\hline
1991 & 4380 & 126 \\
\hline
1992 & 4420 & 132 \\
\hline
1993 & 4440 & 134 \\
\hline
1994 & 4430 & 137 \\
\hline
1995 & 4520 & 144 \\
\hline
1996 & 4590 & 148 \\
\hline
1997 & 4660 & 150 \\
\hline
1998 & 4700 & 153 \\
\hline
1999 & 4790 & 158 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Using the coding $x = ( p - 100 )$ and $y = \frac { 1 } { 10 } ( c - 4000 )$, draw a scatter diagram to represent these data. Explain why $x$ is the explanatory variable.
\item Find the equation of the least squares regression line of $y$ on $x$.
$$\text { [Use } \left. \Sigma x = 402 , \Sigma y = 517 , \Sigma x ^ { 2 } = 17538 \text { and } \Sigma x y = 22611 . \right]$$
\item Deduce the equation of the least squares regression line of $c$ on $p$ in the form $c = a + b p$.
\item Interpret the value of $a$.
\item Predict the number of extra new cars sales for an increase of $\pounds 2000$ in advertising budget. Comment on the validity of your answer.\\
(2)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2003 Q6 [19]}}