| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Bivariate data |
| Type | Estimate correlation from scatter diagram |
| Difficulty | Easy -1.3 Part (a) requires visual estimation of correlation from scatter diagrams, which is a basic skill involving pattern recognition rather than calculation. Part (b) involves standard calculation of PMCC using given data and a routine interpretation—this is textbook S1 material with no problem-solving or novel insight required. The question is easier than average A-level maths due to its purely procedural nature. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08b Linear coding: effect on pmcc |
| \(\boldsymbol { x }\) | 22.5 | 22.7 | 22.6 | 22.4 | 22.5 | 22.8 | 22.6 | 22.7 | 22.8 | 22.4 | 22.9 | 22.6 |
| \(\boldsymbol { y }\) | 160.3 | 159.4 | 157.8 | 158.0 | 157.3 | 159.8 | 158.3 | 159.6 | 161.3 | 156.4 | 162.5 | 161.2 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | \(r \approx 0.7\) (any value in range \(0.6\) to \(0.9\)) | B1 |
| (ii) | \(r \approx 0\) (any value in range \(-0.2\) to \(0.2\)) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n}\) | M1 | Correct formula structure |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{xx} = 6162.03 - \frac{271.9^2}{12} = 0.6492\) | A1 | |
| \(S_{yy} = 304826.51 - \frac{1912.3^2}{12} = 35.3392\) | A1 | |
| \(S_{xy} = 43288.91 - \frac{271.9 \times 1912.3}{12} = 3.8942\) | A1 | |
| \(r = \frac{S_{xy}}{\sqrt{S_{xx} \cdot S_{yy}}} = \frac{3.8942}{\sqrt{0.6492 \times 35.3392}}\) | M1 | Correct formula |
| \(r = 0.813\) | A1 | awrt \(0.813\) |
| Answer | Marks | Guidance |
|---|---|---|
| There is a moderate/fairly strong positive correlation between circumference and weight of cricket balls | B1 | Must be in context |
| This suggests that as circumference increases, weight tends to increase | B1 | Correct interpretation in context |
# Question 1:
## Part (a)
**(i)** | $r \approx 0.7$ (any value in range $0.6$ to $0.9$) | B1 | Positive correlation, moderately strong
**(ii)** | $r \approx 0$ (any value in range $-0.2$ to $0.2$) | B1 | Little or no correlation
## Part (b)(i)
| $S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n}$ | M1 | Correct formula structure |
$\sum x = 271.9$, $\sum y = 1912.3$, $\sum x^2 = 6162.03$, $\sum y^2 = 304826.51$, $\sum xy = 43288.91$
| $S_{xx} = 6162.03 - \frac{271.9^2}{12} = 0.6492$ | A1 | |
| $S_{yy} = 304826.51 - \frac{1912.3^2}{12} = 35.3392$ | A1 | |
| $S_{xy} = 43288.91 - \frac{271.9 \times 1912.3}{12} = 3.8942$ | A1 | |
| $r = \frac{S_{xy}}{\sqrt{S_{xx} \cdot S_{yy}}} = \frac{3.8942}{\sqrt{0.6492 \times 35.3392}}$ | M1 | Correct formula |
| $r = 0.813$ | A1 | awrt $0.813$ |
## Part (b)(ii)
| There is a moderate/fairly strong positive correlation between circumference and weight of cricket balls | B1 | Must be in context |
| This suggests that as circumference increases, weight tends to increase | B1 | Correct interpretation in context |
---1
\begin{enumerate}[label=(\alph*)]
\item Estimate, without undertaking any calculations, the value of the product moment correlation coefficient between the variables $x$ and $y$ for each of the two scatter diagrams.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{(i)}
\includegraphics[alt={},max width=\textwidth]{156f9453-ebc6-4406-b5bc-08d1918ebc62-02_487_652_733_356}
\end{center}
\end{figure}
\includegraphics[max width=\textwidth, alt={}, center]{156f9453-ebc6-4406-b5bc-08d1918ebc62-02_576_714_733_1153}
\item The table gives the circumference, $x$ centimetres, and the weight, $y$ grams, of each of 12 new cricket balls.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | c | c | c | }
\hline
$\boldsymbol { x }$ & 22.5 & 22.7 & 22.6 & 22.4 & 22.5 & 22.8 & 22.6 & 22.7 & 22.8 & 22.4 & 22.9 & 22.6 \\
\hline
$\boldsymbol { y }$ & 160.3 & 159.4 & 157.8 & 158.0 & 157.3 & 159.8 & 158.3 & 159.6 & 161.3 & 156.4 & 162.5 & 161.2 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\item Calculate the value of the product moment correlation coefficient between $x$ and $y$.
\item Assuming that the 12 balls may be considered to be a random sample, interpret your value in context.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2011 Q1 [7]}}