AS June 2018 Q1
1. The scores achieved on a maths test, \(m\), and the scores achieved on a physics test, \(p\), by 16 students are summarised below.
\[\sum m = 392 \qquad \sum p = 254 \qquad \sum p^2 = 4748 \qquad \mathrm{S}_{mm} = 1846 \qquad \mathrm{S}_{mp} = 1115\]Figure 1 shows a plot of the residuals.

For the person who scored 30 marks on the maths test,
The data for the person who scored 20 on the maths test is removed from the data set.
The product moment correlation coefficient between \(m\) and \(p\) is now recalculated for the remaining 15 students.
Give a reason for your answer. (1)
| Scheme | Marks | AO |
|---|---|---|
| \([S_{pp} = 4748 - \tfrac{254^2}{16} = 715.75]\) | ||
| \(r = \dfrac{1115}{\sqrt{1846 \times (\text{“}715.75\text{”})}}\) | M1 | 1.1b |
| \(r = 0.970014\ldots\) awrt 0.970 | A1 | 1.1b |
| (2) |
Notes
M1 for a complete correct method for finding \(r\)
A1 for awrt 0.970 (allow 0.97 from correct working)
| Scheme | Marks | AO |
|---|---|---|
| \(b = \dfrac{1115}{1846}\ [= 0.6040\ldots.]\) | M1 | 3.3 |
| \(a = \dfrac{254}{16} - \text{“}b\text{”}\dfrac{392}{16}\ [= 1.076\ldots]\) | M1 | 1.1b |
| \(p = 1.08 + 0.604m\) | A1 | 1.1b |
| (3) |
Notes
1st M1 for use of a correct model i.e. a correct expression for \(b\)
2nd M1 for use of a correct model i.e. a correct (ft) expression for \(a\)
A1 for correct model \(p = 1.08 + 0.604m\) with awrt 1.08 and awrt 0.604
No fractions and must be in terms of \(p\) and \(m\)
| Scheme | Marks | AO |
|---|---|---|
| \(\text{RSS} = \text{“}715.75\text{”} - \dfrac{1115^2}{1846}\) or \(\text{RSS} = \text{“}715.75\text{”}\left(1 - \text{“}0.970\text{”}^2\right)\) | M1 | 1.1b |
| \(\text{RSS} = 42.28033\ldots\) awrt 42.3 | A1 | 1.1b |
| (2) |
Notes
M1 for a correct expression for RSS
A1 for awrt 42.3
| Scheme | Marks | AO |
|---|---|---|
| \(p = 1.08 + 0.604(30) + \text{residual}\) | M1 | 3.4 |
| \(p =\) 18 | A1ft | 1.1b |
| (2) |
Notes
M1 for substitution of \(m = 30\) into the regression equation and adding the residual
A1ft for 18
| Scheme | Marks | AO |
|---|---|---|
| |residual| is large/may be an outlier | B1 | 3.5b |
| (1) |
Notes
B1 for identifying this point’s residual is far from 0/it may be an outlier/anomaly/ does not fit the trend
| Scheme | Marks | AO |
|---|---|---|
| New \(r\) should be closer to 1 than part (a) since the remaining points are likely to be closer to the new regression line. | B1 | 2.2b |
| (1) | ||
| (11 marks) |
Notes
B1 for closer to 1 than part (a) / increase oe and correct supporting reason about the relative strength of correlation (condone outlier removed)