S1 June 2013 Q1
1. A meteorologist believes that there is a relationship between the height above sea level, \(h\) m, and the air temperature, \(t\) °C. Data is collected at the same time from 9 different places on the same mountain. The data is summarised in the table below.
| \(h\) | 1400 | 1100 | 260 | 840 | 900 | 550 | 1230 | 100 | 770 |
|---|---|---|---|---|---|---|---|---|---|
| \(t\) | 3 | 10 | 20 | 9 | 10 | 13 | 5 | 24 | 16 |
[You may assume that \(\sum h = 7150\), \(\sum t = 110\), \(\sum h^2 = 7\,171\,500\), \(\sum t^2 = 1716\), \(\sum th = 64\,980\) and \(\mathrm{S}_{tt} = 371.56\)]
| Scheme | Marks |
|---|---|
| \((\mathrm{S}_{th}) = 64980 - \dfrac{7150\times 110}{9} = -22408.9\ldots\) \(-\underline{22\,400}\) | M1 A1 |
| \((\mathrm{S}_{hh}) = 7171500 - \dfrac{7150^2}{9} = 1491222.2\ldots\) \(\underline{1\,490\,000}\) | A1 |
| (3) |
Notes
M1 for at least one correct expression (condone transcription error)
1st A1 for \(\mathrm{S}_{hh} =\) awrt 1 490 000 or \(\mathrm{S}_{th} =\) awrt \(-22\,400\) (Condone \(S_{xx}\) or \(S_{xy} = \ldots\) or even \(S_{yy} = \ldots\))
2nd A1 for \(\mathrm{S}_{th} = -22\,400\) and \(\mathrm{S}_{hh} = 1\,490\,000\) only. [This mark is assessing correct rounding]
(Allow no labels but mis-labelling \(\mathrm{S}_{th}\) as \(\mathrm{S}_{hh}\) etc loses the final A1)
| Scheme | Marks |
|---|---|
| \(r = \dfrac{-22408.9}{\sqrt{1491222\times 371.56}} = -0.95200068\ldots\) awrt \(-\underline{\mathbf{0.952}}\) | M1A1 |
| (2) |
Notes
M1 for attempt at correct formula. Allow minor transcription errors of 2 or 3 digits. Must have their \(\mathrm{S}_{hh}\), \(\mathrm{S}_{th}\) and given \(\mathrm{S}_{tt}\) (3sf or better) in the correct places. Condone missing “−”
Award M1A0 for awrt \(-0.95\) with no expression seen. M0 for \(\dfrac{64980}{\sqrt{7171500\times 1716}}\) (corrected from the printed mark scheme, which has 7.864 in place of 1716)
| Scheme | Marks |
|---|---|
| Yes as \(r\) is close to \(-1\) (if \(-1 \lt r \lt -0.5\)) or Yes as \(r\) is close to 1 (if \(1 \gt r \gt 0.5\)) [ If \(-0.5 \leqslant r \leqslant 0.5\) allow “no since \(r\) is close to 0”] [ If \(|r| \gt 1\) award B0] | B1ft |
| (1) |
Notes
B1ft must comment on supporting and state: high/strong/clear (negative or positive) correlation
“points lie close to a straight line” is B0 since there is no evidence of this.
| Scheme | Marks |
|---|---|
| \(b = \dfrac{-22408.9}{1491222.2} = -0.015027\ldots\) (allow \(\frac{-56}{3725}\)) awrt \(-0.015\) | M1 A1 |
| \(a = \dfrac{110}{9} - \text{"their } b\text{"}\times\dfrac{7150}{9} = (12.2 - {-0.015}\times 794.4) = 24.1604\ldots\) so \(\boldsymbol{t = 24.2 - 0.015h}\) | M1, A1 |
| (4) |
Notes
1st M1 for a correct expression for \(b\). Follow through their \(\mathrm{S}_{hh}\) & \(\mathrm{S}_{th}\). Condone missing “−”
1st A1 for awrt \(-0.015\) or allow exact fraction from rounded values.
2nd M1 for a correct method for \(a\). Follow through their value of \(b\)
2nd A1 for a correct equation for \(t\) and \(h\) with \(a\) = awrt 24.2 and \(b\) = awrt \(-0.015\). No fractions
| Scheme | Marks |
|---|---|
| 0.015 is the drop in temp, (in °C), for every 1(m) increase in height above sea level. | B1 |
| (1) |
Notes
B1 Must mention \(h\) (or height) and \(t\) (or temperature) and their (1 sf) value of \(b\) in a correct comment
| Scheme | Marks |
|---|---|
| Change = (“\(24.2 - 0.015\)”\(\times 500\)) − (“\(24.2 - 0.015\)”\(\times 1000\)) or \(500\times\)“0.015” | M1 |
| \(= \pm 7.5\) (awrt \(\pm 7.5\)) (only ft a value < 100) | A1ft |
| (2) | |
| (13 marks) |
Notes
M1 for a correct expression seen based on their equation. Allow transcription error of 1 digit.
If answer is \(500\times\) their \(b\) to 2sf and < 100 (M1A1), If answer is \(500\times\) their \(b\) to 2sf and \(\geqslant 100\) (M1A0)