| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 17 |
| 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 application of standard regression formulas with all summary statistics provided. Parts (a)-(c) are routine calculations, (d) requires basic understanding of extrapolation, and (e) is simple equation solving. Easier than average A-level as it's purely procedural with no conceptual challenges. |
| Spec | 2.02c Scatter diagrams and regression lines5.09b Least squares regression: concepts5.09c Calculate regression line5.09e Use regression: for estimation in context |
| \(h\) (hours) | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 |
| \(n\) | 116 | 114 | 109 | 101 | 94 | 94 | 86 | 81 | 80 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Scatter diagram with points plotted correctly (showing declining trend from approximately (16, 120) to (24, 80)) | B4 | |
| (b) \(S_{nn} = 17204 - \frac{180 \times 875}{9} = -296\) and \(S_{hh} = 3660 - \frac{180^2}{9} = 60\) then \(b = \frac{-296}{60} = -4.9333\) and \(a = \frac{875}{9} - [-4.9333 \times \frac{180}{9}] = 195.888\) so \(h = 195.9 - 4.93h\) | M1 M1 M1 A1 A1 | |
| (c) No. of clinches decreases by 4.93 per hour awake | B1 | |
| (d) e.g. ability likely to be roughly constant during normal waking hours; only decreases when awake for longer than usual | B2 | |
| (e) \(195.9 - 4.93h = 213.4 - 5.87h\) gives \(0.94h = 17.5\); \(h = 18.6\) hours | M1 M1 A1 | Total 17 marks |
**(a)** Scatter diagram with points plotted correctly (showing declining trend from approximately (16, 120) to (24, 80)) | B4 |
**(b)** $S_{nn} = 17204 - \frac{180 \times 875}{9} = -296$ and $S_{hh} = 3660 - \frac{180^2}{9} = 60$ then $b = \frac{-296}{60} = -4.9333$ and $a = \frac{875}{9} - [-4.9333 \times \frac{180}{9}] = 195.888$ so $h = 195.9 - 4.93h$ | M1 M1 M1 A1 A1 |
**(c)** No. of clinches decreases by 4.93 per hour awake | B1 |
**(d)** e.g. ability likely to be roughly constant during normal waking hours; only decreases when awake for longer than usual | B2 |
**(e)** $195.9 - 4.93h = 213.4 - 5.87h$ gives $0.94h = 17.5$; $h = 18.6$ hours | M1 M1 A1 | Total 17 marks
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**Total: 75 marks**7. A doctor wished to investigate the effects of staying awake for long periods on a person's ability to complete simple tasks. She recorded the number of times, $n$, that a subject could clinch his or her fist in 30 seconds after being awake for $h$ hours.
The results for one subject were as follows.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | }
\hline
$h$ (hours) & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 \\
\hline
$n$ & 116 & 114 & 109 & 101 & 94 & 94 & 86 & 81 & 80 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Plot a scatter diagram of $n$ against $h$ for these results.
You may use
$$\Sigma h = 180 , \quad \Sigma n = 875 , \quad \Sigma h ^ { 2 } = 3660 , \quad \Sigma h n = 17204 .$$
\item Obtain the equation of the regression line of $n$ on $h$ in the form $n = a + b h$.
\item Give a practical interpretation of the constant b.
\item Explain why this regression line would be unlikely to be appropriate for values of $h$ between 0 and 16 .\\
(2 marks)\\
Another subject underwent the same tests giving rise to a regression line of $n = 213.4 - 5.87$ h
\item After how many hours of being awake together would you expect these two subjects to be able to clench their fists the same number of times in 30 seconds?
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q7 [17]}}