| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate y on x from raw data table |
| Difficulty | Moderate -0.3 This is a standard S1 linear regression question with coding that follows a routine template. Students are given summary statistics, must apply standard formulas for Sxx and Sxy, find the regression line, make predictions, and decode. The coding is straightforward (subtracting a constant), all steps are algorithmic, and the reliability comment in part (f) is a standard textbook response about extrapolation. Slightly easier than average due to given summations and formulaic nature. |
| Spec | 5.09b Least squares regression: concepts5.09c Calculate regression line5.09e Use regression: for estimation in context |
| \(t\) | \(l\) |
| 20.4 | 2461.12 |
| 27.3 | 2461.41 |
| 32.1 | 2461.73 |
| 39.0 | 2461.88 |
| 42.9 | 2462.03 |
| 49.7 | 2462.37 |
| 58.3 | 2462.69 |
| 67.4 | 2463.05 |
| Answer | Marks |
|---|---|
| \(\sum x = \sum t = 337.1\), \(\sum y = 16.28\) | B1, B1 |
| Can be implied | |
| \(S_{xy} = 757.467 - \frac{337.1 \times 16.28}{8} = 71.4685\) | M1A1 |
| either method, awrt 71.5 | |
| \(S_{xx} = 15965.01 - \frac{337.1^2}{8} = 1760.45875\) | A1 |
| awrt 1760 |
| Answer | Marks |
|---|---|
| \(b = \frac{71.4685}{1760.45875} = 0.04059652\) | M1 |
| / correct way up, awrt 0.0406 | |
| \(a = \frac{16.28}{8} - b \times \frac{337.1}{8} = 0.324364\) | M1A1 |
| using correct formula, awrt 0.324 | |
| \(y = 0.324 + 0.0406x\) | A1 |
| 3 sf or better but award for copying from above |
| Answer | Marks |
|---|---|
| At \(t = 40\), \(x = 40\), \(y = 1.948\), \(l = 2461.948\) | M1A1A1 |
| sub x=40, awrt 1.95, awrt 2461.95 |
| Answer | Marks |
|---|---|
| \(l - 2460 = 0.324 + 0.0406t\) | M1 |
| \(l = 2460.324 + 0.0406t\) | |
| LHS required | |
| awrt 2460.32, f.t. their 0.0406, \(l\) and \(t\) |
| Answer | Marks |
|---|---|
| At \(t = 90\), \(l = 2463.978\) | B1 |
| awrt 2464 |
| Answer | Marks |
|---|---|
| \(90°C\) outside range of data; unlikely to be reliable | B1 |
**Part (a):**
| $\sum x = \sum t = 337.1$, $\sum y = 16.28$ | B1, B1 |
| Can be implied | |
| $S_{xy} = 757.467 - \frac{337.1 \times 16.28}{8} = 71.4685$ | M1A1 |
| either method, awrt 71.5 | |
| $S_{xx} = 15965.01 - \frac{337.1^2}{8} = 1760.45875$ | A1 |
| awrt 1760 | |
**Part (b):**
| $b = \frac{71.4685}{1760.45875} = 0.04059652$ | M1 |
| / correct way up, awrt 0.0406 | |
| $a = \frac{16.28}{8} - b \times \frac{337.1}{8} = 0.324364$ | M1A1 |
| using correct formula, awrt 0.324 | |
| $y = 0.324 + 0.0406x$ | A1 |
| 3 sf or better but award for copying from above | |
**Part (c):**
| At $t = 40$, $x = 40$, $y = 1.948$, $l = 2461.948$ | M1A1A1 |
| sub x=40, awrt 1.95, awrt 2461.95 | |
**Part (d):**
| $l - 2460 = 0.324 + 0.0406t$ | M1 |
| $l = 2460.324 + 0.0406t$ | |
| LHS required | |
| awrt 2460.32, f.t. their 0.0406, $l$ and $t$ | |
**Part (e):**
| At $t = 90$, $l = 2463.978$ | B1 |
| awrt 2464 | |
**Part (f):**
| $90°C$ outside range of data; unlikely to be reliable | B1 |
**Total 20 marks**
---\begin{enumerate}
\item A metallurgist measured the length, $l \mathrm {~mm}$, of a copper rod at various temperatures, $t ^ { \circ } \mathrm { C }$, and recorded the following results.
\end{enumerate}
\begin{center}
\begin{tabular}{|l|l|}
\hline
$t$ & $l$ \\
\hline
20.4 & 2461.12 \\
\hline
27.3 & 2461.41 \\
\hline
32.1 & 2461.73 \\
\hline
39.0 & 2461.88 \\
\hline
42.9 & 2462.03 \\
\hline
49.7 & 2462.37 \\
\hline
58.3 & 2462.69 \\
\hline
67.4 & 2463.05 \\
\hline
\end{tabular}
\end{center}
The results were then coded such that $x = t$ and $y = l - 2460.00$.\\
(a) Calculate $S _ { x y }$ and $S _ { x x }$.\\
(You may use $\Sigma x ^ { 2 } = 15965.01$ and $\Sigma x y = 757.467$ )\\
(b) Find the equation of the regression line of $y$ on $x$ in the form $y = a + b x$.\\
(c) Estimate the length of the rod at $40 ^ { \circ } \mathrm { C }$.\\
(d) Find the equation of the regression line of $l$ on $t$.\\
(e) Estimate the length of the rod at $90 ^ { \circ } \mathrm { C }$.\\
(f) Comment on the reliability of your estimate in part (e).
\hfill \mbox{\textit{Edexcel S1 2006 Q3 [18]}}