| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| 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 routine application of formulas with all summations provided. Parts (i)-(ii) are direct calculations, (iii) tests knowledge that regression lines intersect at (x̄, ȳ), and (iv) requires stating to use x-on-y regression. Slightly easier than average as it's computational with no interpretation challenges or unusual contexts. |
| Spec | 5.09c Calculate regression line5.09e Use regression: for estimation in context |
| 5.1 | 5.8 | 6.5 | 7.1 | 7.6 | 8.4 | 9.5 | 10.5 | ||
| 6.2 | 6.1 | 5.9 | 5.6 | 5.3 | 5.4 | 5.3 | 5.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{xx} = 481.13 - \frac{60.5^2}{8}\) or 23.59875 or 23.6 or \(\frac{18879}{800}\) | M1 | Correct sub in any correct \(S_{xx}\) or \(S_{xy}\) formula or correct value of either \(S\); Alternative: \(44.9 = 8a + 60.5b\) |
| \(S_{xy} = 334.65 - \frac{60.5 \times 44.9}{8}\) or \(-4.90625\) or \(-4.91\) or \(-\frac{157}{32}\) | M1 | \(334.65 = 60.5a + 481.13b\) |
| \(b = \frac{334.65 - \frac{60.5\times44.9}{8}}{481.13 - \frac{60.5^2}{8}}\) | M1 | Correct sub in both \(S\)s and in a correct \(b\) formula; hence \(a = 7.18\) or \(b = -0.208\) |
| or \(-0.20790\) or \(-0.208\) or \(-\frac{3925}{18879}\) | M1 | |
| \(y - \frac{44.9}{8} = {-0.20790}(x - \frac{60.5}{8})\) | M1 | or \(a = \frac{44.9}{8} - {-0.20790} \times \frac{60.5}{8}\) |
| \(y = -0.208x + 7.18\) (or \(+7.19\)) (3 sf) | A1 | or \(y = -\frac{3925}{18879}x + 7.18/9\); Must include "\(y =\)"; Allow \(y = -0.21x + 7.2\) (awrt 2 sf); no working, correct ans M1M1M1M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(-0.208 \times 9.2 + 7.18\) | M1 | ft their equation from (i); but no ft from \(x\) on \(y\) line |
| \(= 5.27\) or \(5.28\) (km/l) (3 sf) | A1ft |
| Answer | Marks | Guidance |
|---|---|---|
| \((7.56,\ 5.61)\) (3 sf) or \(\left(\frac{121}{16}, \frac{449}{80}\right)\) | B1 | Ignore calc'n of reg line, if done; NOT \(\left(\frac{60.5}{8}, \frac{44.9}{8}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Use reg line of \(x\) on \(y\) (either equation or line) | M1 | Must specify or imply \(x\) on \(y\), otherwise M0A0; NOT "Use either \(x\) on \(y\) or \(y\) on \(x\)"; If calc \(x\) on \(y\) reg line (allow errors) M1 |
| Sub \(y = 5.8\) or fuel \(= 5.8\) or km/l \(= 5.8\) | A1 | NOT "and read off \(y\) coord"; Subst 5.8 into their \(x\) on \(y\) line A1; Ignore all else |
# Question 4:
## Part (i)
$S_{xx} = 481.13 - \frac{60.5^2}{8}$ or 23.59875 or 23.6 or $\frac{18879}{800}$ | M1 | Correct sub in any correct $S_{xx}$ or $S_{xy}$ formula or correct value of either $S$; Alternative: $44.9 = 8a + 60.5b$
$S_{xy} = 334.65 - \frac{60.5 \times 44.9}{8}$ or $-4.90625$ or $-4.91$ or $-\frac{157}{32}$ | M1 | $334.65 = 60.5a + 481.13b$
$b = \frac{334.65 - \frac{60.5\times44.9}{8}}{481.13 - \frac{60.5^2}{8}}$ | M1 | Correct sub in both $S$s and in a correct $b$ formula; hence $a = 7.18$ or $b = -0.208$
or $-0.20790$ or $-0.208$ or $-\frac{3925}{18879}$ | M1 |
$y - \frac{44.9}{8} = {-0.20790}(x - \frac{60.5}{8})$ | M1 | or $a = \frac{44.9}{8} - {-0.20790} \times \frac{60.5}{8}$
$y = -0.208x + 7.18$ (or $+7.19$) (3 sf) | A1 | or $y = -\frac{3925}{18879}x + 7.18/9$; Must include "$y =$"; Allow $y = -0.21x + 7.2$ (awrt 2 sf); no working, correct ans M1M1M1M1A1
**[4]**
## Part (ii)
$-0.208 \times 9.2 + 7.18$ | M1 | ft their equation from (i); but no ft from $x$ on $y$ line
$= 5.27$ or $5.28$ (km/l) (3 sf) | A1ft |
**[2]**
## Part (iii)
$(7.56,\ 5.61)$ (3 sf) or $\left(\frac{121}{16}, \frac{449}{80}\right)$ | B1 | Ignore calc'n of reg line, if done; NOT $\left(\frac{60.5}{8}, \frac{44.9}{8}\right)$
**[1]**
## Part (iv)
Use reg line of $x$ on $y$ (either equation or line) | M1 | Must specify or imply $x$ on $y$, otherwise M0A0; NOT "Use either $x$ on $y$ or $y$ on $x$"; If calc $x$ on $y$ reg line (allow errors) M1
Sub $y = 5.8$ or fuel $= 5.8$ or km/l $= 5.8$ | A1 | NOT "and read off $y$ coord"; Subst 5.8 into their $x$ on $y$ line A1; Ignore all else
**[2]**
---4 The table shows the load a lorry was carrying, $x$ tonnes, and the fuel economy, $y \mathrm {~km}$ per litre, for 8 different journeys. You should assume that neither variable is controlled.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
\begin{tabular}{ l }
Load \\
$( x$ tonnes $)$ \\
\end{tabular} & 5.1 & 5.8 & 6.5 & 7.1 & 7.6 & 8.4 & 9.5 & 10.5 \\
\hline
\begin{tabular}{ l }
Fuel economy \\
$( y \mathrm {~km}$ per litre $)$ \\
\end{tabular} & 6.2 & 6.1 & 5.9 & 5.6 & 5.3 & 5.4 & 5.3 & 5.1 \\
\hline
\end{tabular}
\end{center}
$$n = 8 \quad \sum x = 60.5 \quad \sum y = 44.9 \quad \sum x ^ { 2 } = 481.13 \quad \sum y ^ { 2 } = 253.17 \quad \sum x y = 334.65$$
(i) Calculate the equation of the regression line of $y$ on $x$.\\
(ii) Estimate the fuel economy for a load of 9.2 tonnes.\\
(iii) An analyst calculated the equation of the regression line of $x$ on $y$. Without calculating this equation, state the coordinates of the point where the two regression lines intersect.\\
(iv) Describe briefly the method required to estimate the load when the fuel economy is 5.8 km per litre.
\hfill \mbox{\textit{OCR S1 2015 Q4 [9]}}