| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Bivariate data |
| Type | Calculate r from summary statistics |
| Difficulty | Moderate -0.3 Part (a) is a standard S1 calculation requiring the PMCC formula with given summary statistics—routine but multi-step arithmetic. Parts (b) and (c) test interpretation of correlation vs causation, which is conceptual but covered in standard S1 teaching. The question is slightly easier than average due to straightforward formula application and predictable interpretation points. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08c Pearson: measure of straight-line fit |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{kk} = 10131 - \frac{703^2}{50} = 246.82\) | M1 | |
| \(S_{HH} = 1338.5 - \frac{217^2}{50} = 396.72\) | M1 | |
| \(S_{kH} = 3253.5 - \frac{703 \times 217}{50} = 202.48\) | M1 | |
| \(r = \frac{202.48}{\sqrt{246.82 \times 396.72}} = 0.6471\) | M1 A1 | |
| \(r\) is strongly +ve showing that older pupils tend to do more homework | B1 | |
| e.g. younger pupils are set less homework and also do little or no paid work etc. apparent correlation is because both correlate with age | B2 | |
| e.g. collect data from one year group, likely to have similar amounts of homework set, so will see effect of paid work | B2 | (10) |
$S_{kk} = 10131 - \frac{703^2}{50} = 246.82$ | M1 |
$S_{HH} = 1338.5 - \frac{217^2}{50} = 396.72$ | M1 |
$S_{kH} = 3253.5 - \frac{703 \times 217}{50} = 202.48$ | M1 |
$r = \frac{202.48}{\sqrt{246.82 \times 396.72}} = 0.6471$ | M1 A1 |
$r$ is strongly +ve showing that older pupils tend to do more homework | B1 |
e.g. younger pupils are set less homework and also do little or no paid work etc. apparent correlation is because both correlate with age | B2 |
e.g. collect data from one year group, likely to have similar amounts of homework set, so will see effect of paid work | B2 | (10)2. A statistics student gave a questionnaire to a random sample of 50 pupils at his school. The sample included pupils aged from 11 to 18 years old.
The student summarised the data on age in completed years, $A$, and the number of hours spent doing homework in the previous week, $H$, giving the following:
$$\Sigma A = 703 , \quad \Sigma H = 217 , \quad \Sigma A ^ { 2 } = 10131 , \quad \Sigma H ^ { 2 } = 1338.5 , \quad \Sigma A H = 3253.5$$
\begin{enumerate}[label=(\alph*)]
\item Calculate the product moment correlation coefficient for these data and explain what is shown by your result.\\
(6 marks)\\
The student also asked each pupil how many hours of paid work they had done in the previous week. He then calculated the product moment correlation coefficient for the data on hours doing homework and hours doing paid work, giving a value of $r = 0.5213$
The student concluded that paid work did not interfere with homework as pupils doing more paid work also tended to do more homework.
\item Explain why this conclusion may not be valid.
\item Explain briefly how the student could more effectively investigate the effect of paid work on homework.\\
(2 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q2 [10]}}