| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 14 |
| 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 S1 linear regression question requiring standard application of formulas with given summations. Students substitute into least squares formulas (which are provided in the formula booklet) to find a and b. The context interpretation parts are routine for this topic. Easier than average as it's a direct application with no complications or problem-solving required. |
| Spec | 2.02c Scatter diagrams and regression lines5.09a 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 |
| \(m ( \mathrm {~g} )\) | 50 | 100 | 200 | 300 | 400 | 500 | 600 | 700 |
| \(l ( \mathrm {~cm} )\) | 7.8 | 10.7 | 16.5 | 22.1 | 28.0 | 33.9 | 35.2 | 35.6 |
| Answer | Marks |
|---|---|
| (a) Scatter diagram with 8 points plotted: approximately at (100, 8), (150, 10), (200, 18), (350, 21), (400, 27), (500, 31), (650, 34), (700, 35) | B4 |
| (b) e.g. the first six values lie roughly along a straight line but this changes for the two values above 500 g | B2 |
| (c) \(S_{ml} = 39540 - \frac{1550 \times 119}{6} = 8798.33\); \(S_{mm} = 552500 - \frac{1550^2}{6} = 152083\); \(b = \frac{8798.33}{152083} = 0.05785\); \(a = \frac{119}{6} - (0.05785 \times \frac{1550}{6}) = 4.888\); \(\therefore a = 4.89, b = 0.0579\) | M1 M1 M1 A1 M1 A1 |
| (d) a is the length of the spring with no mass suspended from it; b is the extra extension for each additional gram suspended from spring | B1 B1 |
| (14) |
| Answer | Marks | Guidance |
|---|---|---|
| Total | (75) |
**(a)** Scatter diagram with 8 points plotted: approximately at (100, 8), (150, 10), (200, 18), (350, 21), (400, 27), (500, 31), (650, 34), (700, 35) | B4 |
**(b)** e.g. the first six values lie roughly along a straight line but this changes for the two values above 500 g | B2 |
**(c)** $S_{ml} = 39540 - \frac{1550 \times 119}{6} = 8798.33$; $S_{mm} = 552500 - \frac{1550^2}{6} = 152083$; $b = \frac{8798.33}{152083} = 0.05785$; $a = \frac{119}{6} - (0.05785 \times \frac{1550}{6}) = 4.888$; $\therefore a = 4.89, b = 0.0579$ | M1 M1 M1 A1 M1 A1 |
**(d)** a is the length of the spring with no mass suspended from it; b is the extra extension for each additional gram suspended from spring | B1 B1 |
| | | (14) |
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**Total** | | (75) |6. A physics student recorded the length, $l \mathrm {~cm}$, of a spring when different masses, $m$ grams, were suspended from it giving the following results.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | }
\hline
$m ( \mathrm {~g} )$ & 50 & 100 & 200 & 300 & 400 & 500 & 600 & 700 \\
\hline
$l ( \mathrm {~cm} )$ & 7.8 & 10.7 & 16.5 & 22.1 & 28.0 & 33.9 & 35.2 & 35.6 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Represent these data on a scatter diagram with $l$ on the vertical axis.
The student decides to find the equation of a regression line of the form $l = a + b m$ using only the data for $m \leq 500 \mathrm {~g}$.
\item Give a reason to support the fitting of such a regression line and explain why the student is excluding two of his values.\\
(2 marks)\\
You may use
$$\Sigma m = 1550 , \quad \Sigma l = 119 , \quad \Sigma m ^ { 2 } = 552500 , \quad \Sigma l ^ { 2 } = 2869.2 , \quad \Sigma m l = 39540 .$$
\item Find the values of $a$ and $b$.
\item Explain the significance of the values of $a$ and $b$ in this situation.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q6 [14]}}