| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| 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 application of standard linear regression formulas from S1. Students need to calculate summary statistics (Σx, Σy, Σx², Σxy) from the given data table and substitute into the regression line formula y = a + bx. The interpretation and context questions are routine. This is easier than average because it's a textbook calculation with no conceptual challenges or problem-solving required. |
| Spec | 2.02c Scatter diagrams and regression lines5.09a Dependent/independent variables5.09c Calculate regression line5.09e Use regression: for estimation in context |
| \(\boldsymbol { x }\) | 86 | 60 | 65 | 46 | 71 | 93 | 56 | 81 | 75 | 57 |
| \(\boldsymbol { y }\) | 940 | 790 | 620 | 530 | 770 | 1050 | 690 | 780 | 860 | 550 |
| Answer | Marks | Guidance |
|---|---|---|
| Plot the 4 remaining points: (56, 690), (81, 780), (75, 860), (57, 550) | B1 | All 4 points plotted correctly |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = \frac{86+60+65+46+71+93+56+81+75+57}{10} = \frac{690}{10} = 69\) | M1 | Attempt to find \(\bar{x}\) and \(\bar{y}\) |
| \(\bar{y} = \frac{940+790+620+530+770+1050+690+780+860+550}{10} = \frac{7580}{10} = 758\) | A1 | Both means correct |
| \(S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 50064 - \frac{690^2}{10} = 50064 - 47610 = 2454\) | M1 | Attempt \(S_{xx}\) |
| \(S_{xy} = \sum xy - \frac{\sum x \sum y}{n} = 541400 - \frac{690 \times 7580}{10} = 541400 - 523020 = 18380\) | M1 | Attempt \(S_{xy}\) |
| \(b = \frac{S_{xy}}{S_{xx}} = \frac{18380}{2454} \approx 7.49\) | A1 | Correct value of \(b\) |
| \(a = \bar{y} - b\bar{x} = 758 - 7.49 \times 69 \approx 241\) | A1 | Correct equation: \(y = 241 + 7.49x\) |
| Answer | Marks | Guidance |
|---|---|---|
| For each additional customer, the takings increase by approximately £7.49 | B1 B1 | B1 for "increase" linked to customers, B1 for £7.49 in context |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 0\) means no customers, which is not a realistic/possible scenario for a butcher's shop | B1 | Accept equivalent explanation that 0 customers is outside the range of the data |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 241 + 7.49 \times 50 = 241 + 374.5 = 615.5 \approx £620\) | B1 | Answer £620 (to nearest £10) |
# Question 5:
## Part (a)
| Plot the 4 remaining points: (56, 690), (81, 780), (75, 860), (57, 550) | B1 | All 4 points plotted correctly |
## Part (b)(i)
| $\bar{x} = \frac{86+60+65+46+71+93+56+81+75+57}{10} = \frac{690}{10} = 69$ | M1 | Attempt to find $\bar{x}$ and $\bar{y}$ |
| $\bar{y} = \frac{940+790+620+530+770+1050+690+780+860+550}{10} = \frac{7580}{10} = 758$ | A1 | Both means correct |
| $S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 50064 - \frac{690^2}{10} = 50064 - 47610 = 2454$ | M1 | Attempt $S_{xx}$ |
| $S_{xy} = \sum xy - \frac{\sum x \sum y}{n} = 541400 - \frac{690 \times 7580}{10} = 541400 - 523020 = 18380$ | M1 | Attempt $S_{xy}$ |
| $b = \frac{S_{xy}}{S_{xx}} = \frac{18380}{2454} \approx 7.49$ | A1 | Correct value of $b$ |
| $a = \bar{y} - b\bar{x} = 758 - 7.49 \times 69 \approx 241$ | A1 | Correct equation: $y = 241 + 7.49x$ |
## Part (b)(ii)
| For each additional customer, the takings increase by approximately £7.49 | B1 B1 | B1 for "increase" linked to customers, B1 for £7.49 in context |
## Part (b)(iii)
| $x = 0$ means no customers, which is not a realistic/possible scenario for a butcher's shop | B1 | Accept equivalent explanation that 0 customers is outside the range of the data |
## Part (c)
| $y = 241 + 7.49 \times 50 = 241 + 374.5 = 615.5 \approx £620$ | B1 | Answer £620 (to nearest £10) |
I can see these are answer space pages (blank lined pages for students to write their answers) from what appears to be an AQA Statistics exam paper (P/Jun15/SS1B).
These pages do not contain any mark scheme content - they are simply blank answer spaces for Questions 5 and 6. The actual question text for Question 6 is visible, but no mark scheme or worked solutions are shown on these pages.
To access the mark scheme for this paper, I would recommend checking the **AQA website** directly at aqa.org.uk, where past paper mark schemes are freely available for download.
However, based on the **question content** visible (Question 6), I can outline the expected working:
---5 The table shows the number of customers, $x$, and the takings, $\pounds y$, recorded to the nearest $\pounds 10$, at a local butcher's shop on each of 10 randomly selected weekdays.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | c | }
\hline
$\boldsymbol { x }$ & 86 & 60 & 65 & 46 & 71 & 93 & 56 & 81 & 75 & 57 \\
\hline
$\boldsymbol { y }$ & 940 & 790 & 620 & 530 & 770 & 1050 & 690 & 780 & 860 & 550 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item The first 6 pairs of data values in this table are plotted on the scatter diagram shown on the opposite page.
Plot the final 4 pairs of data values on the scatter diagram.
\item \begin{enumerate}[label=(\roman*)]
\item Calculate the equation of the least squares regression line in the form $y = a + b x$ and draw your line on the scatter diagram.
\item Interpret your value for $b$ in the context of the question.
\item State why your value for $a$ has no practical interpretation.
\end{enumerate}\item Estimate, to the nearest $\pounds 10$, the shop's takings when the number of customers is 50 .\\[0pt]
[1 mark]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{4c679380-894f-4d36-aec8-296b662058e2-14_1255_1705_1448_155}
\end{center}
Butcher's shop
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Answer space for question 5}
\includegraphics[alt={},max width=\textwidth]{4c679380-894f-4d36-aec8-296b662058e2-15_2335_1760_372_100}
\end{center}
\end{figure}
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2015 Q5 [11]}}