| Exam Board | WJEC |
|---|---|
| Module | Further Unit 2 (Further Unit 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate regression line then predict |
| Difficulty | Moderate -0.3 This is a straightforward regression question requiring recall of the formula that regression lines pass through (x̄, ȳ), substitution to find the equation, and standard interpretation. The algebra is simple, and part (b) asks for textbook reliability concerns (extrapolation/correlation strength). Slightly below average difficulty as it's mostly procedural with minimal problem-solving. |
| Spec | 5.09a Dependent/independent variables5.09c Calculate regression line5.09e Use regression: for estimation in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = 14\), \(\bar{y} = 8\) | B1 | Both |
| \(b = \frac{20 - 8}{19 - 14} = 2.4\) | M1 A1 | convincing |
| Equation of regression line is \(y - 20 = 2.4(x - 19)\) | M1 | oe |
| \(y = 2.4x - 25.6\) | A1 | convincing |
| 1st Alternative Method | ||
| \(y = mx + c\) | (M1) | oe |
| \(20 = 2.4 \times 19 + c\) | (M1) | For 1st equation |
| \(c = -25.6\) | (B1) | For sight of \(\frac{240}{30}\) and \(\frac{420}{30}\) |
| \(y = 2.4x - 25.6\) | (A1) | For correct equations |
| 2nd Alternative Method | ||
| \(y = bx + a\) | (M1) | For 1st equation |
| \(20 = 19b + a\) | (B1) | For sight of \(\frac{240}{30}\) and \(\frac{420}{30}\) |
| \(\frac{240}{30} = \frac{420}{30}b + a\) | (A1) | For correct equations |
| Solve simultaneously to get \(a = -25.6\) and \(b = 2.4\) | (M1) | |
| \(y = 2.4x - 25.6\) | (A1) | |
| When \(x = 26\): \(y = 2.4 \times 26 - 25.6 = 36.8\) | B1 | |
| (b) Comment on linearity e.g. Scatter diagram may not be linear. Correlation might be weak. | E1 | |
| Comment on range of \(x\) values. e.g. 26 may be beyond the range of observed \(x\) values. | E1 | |
| Total [8] |
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = 14$, $\bar{y} = 8$ | B1 | Both |
| $b = \frac{20 - 8}{19 - 14} = 2.4$ | M1 A1 | convincing |
| Equation of regression line is $y - 20 = 2.4(x - 19)$ | M1 | oe |
| $y = 2.4x - 25.6$ | A1 | convincing |
| **1st Alternative Method** | | |
| $y = mx + c$ | (M1) | oe |
| $20 = 2.4 \times 19 + c$ | (M1) | For 1st equation |
| $c = -25.6$ | (B1) | For sight of $\frac{240}{30}$ and $\frac{420}{30}$ |
| $y = 2.4x - 25.6$ | (A1) | For correct equations |
| **2nd Alternative Method** | | |
| $y = bx + a$ | (M1) | For 1st equation |
| $20 = 19b + a$ | (B1) | For sight of $\frac{240}{30}$ and $\frac{420}{30}$ |
| $\frac{240}{30} = \frac{420}{30}b + a$ | (A1) | For correct equations |
| Solve simultaneously to get $a = -25.6$ and $b = 2.4$ | (M1) | |
| $y = 2.4x - 25.6$ | (A1) | |
| When $x = 26$: $y = 2.4 \times 26 - 25.6 = 36.8$ | B1 | |
| **(b)** Comment on linearity e.g. Scatter diagram may not be linear. Correlation might be weak. | E1 | |
| Comment on range of $x$ values. e.g. 26 may be beyond the range of observed $x$ values. | E1 | |
| **Total [8]** | | |For a set of 30 pairs of observations of the variables $x$ and $y$, it is known that $\sum x = 420$ and $\sum y = 240$. The least squares regression line of $y$ on $x$ passes through the point with coordinates $(19, 20)$.
\begin{enumerate}[label=(\alph*)]
\item Show that the equation of the regression line of $y$ on $x$ is $y = 2 \cdot 4x - 25 \cdot 6$ and use it to predict the value of $y$ when $x = 26$. [6]
\item State two reasons why your prediction in part (a) may not be reliable. [2]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 2 2023 Q2 [8]}}