| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| 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 S1 linear regression question requiring standard calculations (finding regression line from data, interpolation, and interpretation). Part (a)(i) tests understanding that the intercept should be 15 from context, parts (a)(ii)-(b) involve routine formula application with calculator, and part (c) tests basic understanding of extrapolation limitations. The calculations are mechanical and the conceptual demands are minimal for A-level statistics. |
| Spec | 5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line5.09d Linear coding: effect on regression |
| \(\boldsymbol { x }\) | 25 | 50 | 75 | 100 | 125 | 150 | 175 | 200 | 250 | 300 |
| \(\boldsymbol { y }\) | 14.7 | 13.4 | 12.8 | 11.9 | 11.0 | 10.3 | 9.7 | 9.0 | 7.5 | 6.7 |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = 15\) | B1 | Accept values in range 14–16 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = \frac{25+50+...+300}{10} = 162.5\) | M1 | For attempt at \(\bar{x}\) and \(\bar{y}\) |
| \(\bar{y} = \frac{14.7+13.4+...+6.7}{10} = 10.63\) | ||
| \(S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 25^2+50^2+...+300^2 - \frac{1625^2}{10}\) | M1 | For correct formula for \(S_{xx}\) |
| \(S_{xx} = 349375\) | ||
| \(S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n}\) | M1 | For correct formula for \(S_{xy}\) |
| \(S_{xy} = -2775\) | ||
| \(b = \frac{S_{xy}}{S_{xx}} = \frac{-2775}{349375} = -0.00795\) | A1 | Accept \(-0.00795\) to \(-0.00794\) |
| \(a = \bar{y} - b\bar{x} = 10.63 - (-0.00795)(162.5) = 15.52\) | ||
| \(y = 15.52 - 0.00795x\) | Accept awrt \(a=15.5\), \(b=-0.00795\) |
| Answer | Marks | Guidance |
|---|---|---|
| For each additional 1 kPa of pressure applied, the depth of the seal decreases by approximately 0.00795 cm | B1 B1 | B1 for decrease/negative relationship; B1 for context (pressure and depth) with numerical value |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 15.52 - 0.00795 \times 225\) | M1 | Substituting \(x = 225\) |
| \(y = 13.73\) cm | A1 | Accept awrt 13.7 |
| Answer | Marks | Guidance |
|---|---|---|
| 225 kPa is outside the range of the data (25 to 300), so this is extrapolation | B1 | Must mention extrapolation or data range |
| Answer | Marks | Guidance |
|---|---|---|
| At 525 kPa the equation gives a negative/very small depth, which is not physically possible as the seal must have a positive depth / the seal cannot be compressed below 0 cm | B1 | Accept equivalent contextual reason about physical impossibility |
## Question 6:
**(a)(i)**
| $a = 15$ | B1 | Accept values in range 14–16 |
**(a)(ii)**
| $\bar{x} = \frac{25+50+...+300}{10} = 162.5$ | M1 | For attempt at $\bar{x}$ and $\bar{y}$ |
| $\bar{y} = \frac{14.7+13.4+...+6.7}{10} = 10.63$ | | |
| $S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 25^2+50^2+...+300^2 - \frac{1625^2}{10}$ | M1 | For correct formula for $S_{xx}$ |
| $S_{xx} = 349375$ | | |
| $S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n}$ | M1 | For correct formula for $S_{xy}$ |
| $S_{xy} = -2775$ | | |
| $b = \frac{S_{xy}}{S_{xx}} = \frac{-2775}{349375} = -0.00795$ | A1 | Accept $-0.00795$ to $-0.00794$ |
| $a = \bar{y} - b\bar{x} = 10.63 - (-0.00795)(162.5) = 15.52$ | | |
| $y = 15.52 - 0.00795x$ | | Accept awrt $a=15.5$, $b=-0.00795$ |
**(a)(iii)**
| For each additional 1 kPa of pressure applied, the depth of the seal decreases by approximately 0.00795 cm | B1 B1 | B1 for decrease/negative relationship; B1 for context (pressure and depth) with numerical value |
**(b)**
| $y = 15.52 - 0.00795 \times 225$ | M1 | Substituting $x = 225$ |
| $y = 13.73$ cm | A1 | Accept awrt 13.7 |
**(c)(i)**
| 225 kPa is outside the range of the data (25 to 300), so this is extrapolation | B1 | Must mention extrapolation or data range |
**(c)(ii)**
| At 525 kPa the equation gives a negative/very small depth, which is not physically possible as the seal must have a positive depth / the seal cannot be compressed below 0 cm | B1 | Accept equivalent contextual reason about physical impossibility |6 A rubber seal is fitted to the bottom of a flood barrier. When no pressure is applied, the depth of the seal is 15 cm . When pressure is applied, a watertight seal is created between the flood barrier and the ground.
The table shows the pressure, $x$ kilopascals ( kPa ), applied to the seal and the resultant depth, $y$ centimetres, of the seal.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | c | }
\hline
$\boldsymbol { x }$ & 25 & 50 & 75 & 100 & 125 & 150 & 175 & 200 & 250 & 300 \\
\hline
$\boldsymbol { y }$ & 14.7 & 13.4 & 12.8 & 11.9 & 11.0 & 10.3 & 9.7 & 9.0 & 7.5 & 6.7 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item State the value that you would expect for $a$ in the equation of the least squares regression line, $y = a + b x$.
\item Calculate the equation of the least squares regression line, $y = a + b x$.
\item Interpret, in context, your value for $b$.
\end{enumerate}\item Calculate an estimate of the depth of the seal when it is subjected to a pressure of 225 kPa .
\item \begin{enumerate}[label=(\roman*)]
\item Give a statistical reason as to why your equation is unlikely to give a realistic estimate of the depth of the seal if it were to be subjected to a pressure of 400 kPa .
\item Give a reason based on the context of this question as to why your equation will not give a realistic estimate of the depth of the seal if it were to be subjected to a pressure of 525 kPa .\\[0pt]
[3 marks]
\begin{center}
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\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{8aeacd54-a5a1-4f2d-b936-2faf635ffce7-21_2484_1707_221_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{8aeacd54-a5a1-4f2d-b936-2faf635ffce7-23_2484_1707_221_153}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2014 Q6 [12]}}