| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Bivariate data |
| Type | Draw scatter diagram from data |
| Difficulty | Moderate -0.3 This is a standard S1 correlation and regression question requiring routine application of given formulas with provided summations. While it involves multiple parts and careful arithmetic with negative numbers, it requires no problem-solving insight—students simply substitute values into standard PMCC and regression line formulas, then interpret the result. Slightly easier than average due to all summations being provided. |
| Spec | 2.02c Scatter diagrams and regression lines2.02d Informal interpretation of correlation5.08a Pearson correlation: calculate pmcc5.09a Dependent/independent variables5.09c Calculate regression line |
| \(T\) (°C) | 4 | \(-3\) | \(-2\) | \(-6\) | 0 | 3 | 7 | \(-1\) | 3 | 2 |
| \(A\) (\%) | 8.5 | 14.1 | 17.0 | 20.3 | 17.9 | 15.5 | 12.4 | 12.8 | 13.7 | 11.6 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) [Scatter diagram with line of best fit: \(A = 14.8 - 0.605T\)] | B4 | |
| (b) \(S_{TT} = 137 - \frac{7^2}{10} = 132.1\) | M1 | |
| \(S_{AA} = 2172.66 - \frac{143.8^2}{10} = 104.816\) | M1 | |
| \(S_{TA} = 20.7 - \frac{7 \times 143.8}{10} = 79.96\) | M1 | |
| \(r = \frac{-79.96}{\sqrt{132.1 \times 104.816}} = -0.6795\) | M1 A1 | |
| e.g. fairly strong –ve correlation so belief seems reasonable | B1 | |
| (c) \(q = \frac{-79.96}{132.1} = -0.60530\) | M1 A1 | |
| \(p = \frac{143.8}{10} - (-0.60530 \times \frac{7}{10}) = 14.804\) | M1 A1 | |
| \(A = 14.8 - 0.605T\) | ||
| (d) Line on graph above | B2 | (16 marks total) |
**(a)** [Scatter diagram with line of best fit: $A = 14.8 - 0.605T$] | B4 |
**(b)** $S_{TT} = 137 - \frac{7^2}{10} = 132.1$ | M1 |
$S_{AA} = 2172.66 - \frac{143.8^2}{10} = 104.816$ | M1 |
$S_{TA} = 20.7 - \frac{7 \times 143.8}{10} = 79.96$ | M1 |
$r = \frac{-79.96}{\sqrt{132.1 \times 104.816}} = -0.6795$ | M1 A1 |
e.g. fairly strong –ve correlation so belief seems reasonable | B1 |
**(c)** $q = \frac{-79.96}{132.1} = -0.60530$ | M1 A1 |
$p = \frac{143.8}{10} - (-0.60530 \times \frac{7}{10}) = 14.804$ | M1 A1 |
$A = 14.8 - 0.605T$ | |
**(d)** Line on graph above | B2 | **(16 marks total)**
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**Total: (75 marks)**The Principal of a school believes that more students are absent on days when the temperature is lower. Over a two-week period in December she records the percentage of students who are absent, $A\%$, and the temperature, $T°$C, at 9 am each morning giving these results.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
$T$ (°C) & 4 & $-3$ & $-2$ & $-6$ & 0 & 3 & 7 & $-1$ & 3 & 2 \\
\hline
$A$ (\%) & 8.5 & 14.1 & 17.0 & 20.3 & 17.9 & 15.5 & 12.4 & 12.8 & 13.7 & 11.6 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Represent these data on a scatter diagram. [4 marks]
\end{enumerate}
You may use
$$\Sigma T = 7, \quad \Sigma A = 143.8, \quad \Sigma T^2 = 137, \quad \Sigma A^2 = 2172.66, \quad \Sigma TA = 20.7$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Calculate the product moment correlation coefficient for these data and comment on the Principal's hypothesis. [6 marks]
\item Find an equation of the regression line of $A$ on $T$ in the form $A = p + qT$. [4 marks]
\item Draw the regression line on your scatter diagram. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q6 [16]}}