| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Convert regression equation between coded and original |
| Difficulty | Moderate -0.3 This is a straightforward application of standard linear regression formulas with coded data. Students must calculate the regression line using given summations, decode back to original variables, interpret the slope, and make a prediction. While it requires careful algebraic manipulation and understanding of coding transformations, all steps follow routine procedures taught in S1 with no novel problem-solving required. The 10 marks for part (a) reflect computational length rather than conceptual difficulty. |
| Spec | 5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line5.09d Linear coding: effect on regression5.09e Use regression: for estimation in context |
| Answer | Marks | Guidance |
|---|---|---|
| \(b = \frac{15 \times 484 - 143 \times 391}{15 \times 2413 - (143)^2} = -3.0899\) | M1 A1 | AWRT −3.09 |
| \(a = \frac{391}{15} - (-3.0899)\left(\frac{143}{15}\right) = 55.5237\) | M1 A1 | AWRT 55.5 |
| \(y = 55.52 - 3.09x\) | B1 ft | |
| \(h - 100 = 55.52 - 3.09(s - 20) \Rightarrow h = 217.32 - 3.09s\) | M1 A1 ft | AWRT 217; 3.09 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| For every extra revolution/minute the life of the drill is reduced by 3 hours. | B1 B1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(s = 30 \Rightarrow h = 124.6\) | AWRT 125 | M1 A1 ft |
## (a)
$b = \frac{15 \times 484 - 143 \times 391}{15 \times 2413 - (143)^2} = -3.0899$ | M1 A1 | AWRT −3.09
$a = \frac{391}{15} - (-3.0899)\left(\frac{143}{15}\right) = 55.5237$ | M1 A1 | AWRT 55.5
$y = 55.52 - 3.09x$ | B1 ft |
$h - 100 = 55.52 - 3.09(s - 20) \Rightarrow h = 217.32 - 3.09s$ | M1 A1 ft | AWRT 217; 3.09 A1 | (10 marks)
## (b)
For every extra revolution/minute the life of the drill is reduced by 3 hours. | B1 B1 | (2 marks)
## (c)
$s = 30 \Rightarrow h = 124.6$ | AWRT 125 | M1 A1 ft | (2 marks)
**(14 marks)**
---A drilling machine can run at various speeds, but in general the higher the speed the sooner the drill needs to be replaced. Over several months, 15 pairs of observations relating to speed, $s$ revolutions per minute, and life of drill, $h$ hours, are collected.
For convenience the data are coded so that $x = s - 20$ and $y = h - 100$ and the following summations obtained.
$\Sigma x = 143; \Sigma y = 391; \Sigma x^2 = 2413; \Sigma y^2 = 22441; \Sigma xy = 484$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the regression line of $h$ on $s$. [10]
\item Interpret the slope of your regression line. [2]
\end{enumerate}
Estimate the life of a drill revolving at 30 revolutions per minute.
[2]
\hfill \mbox{\textit{Edexcel S1 Q4 [14]}}