| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate x on y regression line |
| Difficulty | Moderate -0.3 This is a standard S1 regression question requiring calculation of regression line coefficients, correlation coefficient, and interpretation. All formulas are routine (Sxx, Syy, Sxy), and the question follows a typical textbook structure with no novel problem-solving required. The calculations are straightforward given the summary statistics, making it slightly easier than average for A-level but still requiring careful arithmetic and understanding of when regression is appropriate. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line |
| Car | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) |
| \(x\) | 0.95 | 1.20 | 1.37 | 1.76 | 2.25 | 2.50 | 2.875 |
| \(y\) | 21.3 | 17.2 | 15.5 | 19.1 | 14.7 | 11.4 | 9.0 |
| Answer | Marks |
|---|---|
| (a) \(S_{xy} = 109.177\), \(S_{yy} = -16.298\) | B1 B1 |
| \(\bar{x} - \frac{12905}{109177} = \frac{16298}{109177}(\bar{y} - \frac{1082}{7})\) | M1 A1 |
| \(x - 1.84357 = -0.14928(y - 15.4571)\) or \(x = -0.149y + 4.15\) | M1 A1 |
| (b) \(S_{xx} = 3.1038\), \(r = -0.885\) | M1 A1 A1 B1 |
| Quite good negative correlation | |
| (c) \(y = 12\) gives \(x \simeq 2.36\) | M1 A1 |
| Not necessarily accurate — \(n\) is small, which reduces significance of strong correlation | B1 |
| (d) When \(y\) is close to 0, \(x\) tends to 4·15, suggesting that a 4·15 litre car would travel no km on any amount of fuel — meaningless | B1 |
| 15 marks total |
(a) $S_{xy} = 109.177$, $S_{yy} = -16.298$ | B1 B1 |
$\bar{x} - \frac{12905}{109177} = \frac{16298}{109177}(\bar{y} - \frac{1082}{7})$ | M1 A1 |
$x - 1.84357 = -0.14928(y - 15.4571)$ or $x = -0.149y + 4.15$ | M1 A1 |
(b) $S_{xx} = 3.1038$, $r = -0.885$ | M1 A1 A1 B1 |
Quite good negative correlation |
(c) $y = 12$ gives $x \simeq 2.36$ | M1 A1 |
Not necessarily accurate — $n$ is small, which reduces significance of strong correlation | B1 |
(d) When $y$ is close to 0, $x$ tends to 4·15, suggesting that a 4·15 litre car would travel no km on any amount of fuel — meaningless | B1 |
| 15 marks total |The following data was collected for seven cars, showing their engine size, $x$ litres, and their fuel consumption, $y$ km per litre, on a long journey.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
Car & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ \\
\hline
$x$ & 0.95 & 1.20 & 1.37 & 1.76 & 2.25 & 2.50 & 2.875 \\
\hline
$y$ & 21.3 & 17.2 & 15.5 & 19.1 & 14.7 & 11.4 & 9.0 \\
\hline
\end{tabular}
\end{center}
$\sum x = 12.905$, $\sum x^2 = 26.8951$, $\sum y = 108.2$, $\sum y^2 = 1781.64$, $\sum xy = 183.176$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the equation of the regression line of $x$ on $y$, expressing your answer in the form $x = ay + b$. [6 marks]
\item Calculate the product moment correlation coefficient between $y$ and $x$ and give a brief interpretation of its value. [4 marks]
\item Use the equation of the regression line to estimate the value of $x$ when $y = 12$. State, with a reason, how accurate you would expect this estimate to be. [3 marks]
\item Comment on the use of the line to find values of $x$ as $y$ gets very small. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q7 [15]}}