| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 21 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Linearize non-linear relationships |
| Difficulty | Standard +0.3 This is a straightforward S1 regression question with clear scaffolding. Part (a) is trivial algebra (dividing by t), part (b) is routine plotting, part (c) applies standard regression formulas with given summations, and parts (d)-(e) involve direct substitution and interpretation. The linearization technique is explicitly guided, requiring no independent insight. Slightly easier than average due to heavy scaffolding and provision of all necessary summations. |
| Spec | 2.02c Scatter diagrams and regression lines5.08a Pearson correlation: calculate pmcc5.09a Dependent/independent variables5.09b Least squares regression: concepts |
| \(t\) | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| \(h\) | 68 | 126 | 174 | 216 | 240 | 252 | 266 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(h = t(p - qt)\) | \(B1\) | |
| \(\frac{h}{t} = p - qt\) | ||
| (b) | \(t\) | 1 |
| \(\frac{h}{t}\) | 68 | 64 |
| Scatter graph drawn | ||
| (c) \(\sum t = 28\), \(\sum t^2 = 140\), \(\sum \left(\frac{h}{t}\right) = \sum h = 1342\) | \(B1\) \(B1\) | |
| \(\frac{h}{t} = \frac{371}{7} = \frac{7(1342) - 28(371)}{7(140) - 28^2}(t - \frac{28}{7})\) | \(M1\) \(A1\) \(A1\) | |
| \(\frac{h}{t} = -5.07t + 73.3\) | \(M1\) \(A1\) \(A1\) | |
| \(p = 73.3\), \(q = 5.07\) | ||
| (d) \(t = 10\): \(h/10 = 22.6\) | \(M1\) \(A1\) | |
| \(h = 226\) | ||
| Coming down again | \(M1\) \(A1\) | |
| (e) \(r = \frac{-994}{\sqrt{7(20385) - 371^2}} = -0.999\) | \(M1\) \(M1\) \(A1\) | |
| Shows that the formula is a very good fit to the data and confirms that \(\frac{h}{t}\) decreases as \(t\) increases. | \(A1\) |
**(a)** $h = t(p - qt)$ | $B1$ |
$\frac{h}{t} = p - qt$ | |
**(b)** | $t$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | | |
| $\frac{h}{t}$ | 68 | 64 | 58 | 54 | 48 | 42 | 38 | | $M1$ $A1$ |
Scatter graph drawn | | | | | | | | $B3$ |
**(c)** $\sum t = 28$, $\sum t^2 = 140$, $\sum \left(\frac{h}{t}\right) = \sum h = 1342$ | $B1$ $B1$ |
$\frac{h}{t} = \frac{371}{7} = \frac{7(1342) - 28(371)}{7(140) - 28^2}(t - \frac{28}{7})$ | $M1$ $A1$ $A1$ |
$\frac{h}{t} = -5.07t + 73.3$ | $M1$ $A1$ $A1$ |
$p = 73.3$, $q = 5.07$ | |
**(d)** $t = 10$: $h/10 = 22.6$ | $M1$ $A1$ |
$h = 226$ | |
Coming down again | $M1$ $A1$ |
**(e)** $r = \frac{-994}{\sqrt{7(20385) - 371^2}} = -0.999$ | $M1$ $M1$ $A1$ |
Shows that the formula is a very good fit to the data and confirms that $\frac{h}{t}$ decreases as $t$ increases. | $A1$ |
**Total: 21 marks**A missile was fired vertically upwards and its height above ground level, $h$ metres, was found at various times $t$ seconds after it was released. The results are given in the following table:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$t$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
$h$ & 68 & 126 & 174 & 216 & 240 & 252 & 266 \\
\hline
\end{tabular}
It is thought that this data can be fitted to the formula $h = pt - qt^2$.
\begin{enumerate}[label=(\alph*)]
\item Show that this equation can be written as $\frac{h}{t} = p - qt$. [1 mark]
\item Plot a scatter diagram of $\frac{h}{t}$ against $t$. [5 marks]
\end{enumerate}
Given that $\sum h = 1342$, $\sum \frac{h}{t} = 371$ and $\sum \frac{h^2}{t^2} = 20385$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find the equation of the regression line of $\frac{h}{t}$ on $t$ and hence write down the values of $p$ and $q$. [8 marks]
\item Use your equation to find the value of $h$ when $t = 10$. Comment on the implication of your answer. [3 marks]
\item Find the product-moment correlation coefficient between $\frac{h}{t}$ and $t$ and state the significance of its value. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q6 [21]}}