S1 June 2018 Q6
6. A group of climbers collected information about the height above sea level, \(h\) metres, and the air temperature, \(t\) °C, at the same time at 8 different points on the same mountain.
The data are summarised by
\[\sum h = 6370 \qquad \sum t = 61 \qquad \sum th = 31\,070 \qquad \sum t^2 = 693\]The product moment correlation coefficient for these data is \(-0.985\)
One of the climbers has just stopped for a short break before climbing the next 150 metres.
| Scheme | Marks |
|---|---|
| \((\mathrm{S}_{th}) = 31070 - \dfrac{61 \times 6370}{8}\) or \(31070 - 48571.25\) ; \((\mathrm{S}_{tt}) = 693 - \dfrac{61^2}{8}\) or \(693 - 465.125\) | M1; M1 |
| \((\mathrm{S}_{th}) = -17\,501.25\) and \((\mathrm{S}_{tt}) = 227.875\) (*) | A1cso |
| (3) |
Notes
1st M1 for a correct expression for \(\mathrm{S}_{th}\) 2nd M1 for a correct expression for \(\mathrm{S}_{tt}\)
Allow 1 slip e.g. 6730; \(61 \times 6370\) or 388570; \(61^2\) or 3721. Consistent use of \(n \neq 8\) M0M1
A1cso for both answers correct and both Ms scored.
| Scheme | Marks |
|---|---|
| \(|r|\) close to 1 or \(r\) is close to \(-1\) therefore it does support the linear model | B1 |
| (1) |
Notes
B1 for correct and relevant comment about the value of \(r\) and saying it does support or “yes”
Allow “ it is..”“strong” or “near perfect” correlation BUT B0 for “perfect”or “highly negative”
| Scheme | Marks |
|---|---|
| \(\left[r = \dfrac{\mathrm{S}_{yx}}{\sqrt{\mathrm{S}_{yy} \times \mathrm{S}_{xx}}}\right]\) so \(r = \dfrac{\mathrm{S}_{th}}{\sqrt{\mathrm{S}_{tt} \times \mathrm{S}_{hh}}}\) or \(r^2 = \dfrac{(\mathrm{S}_{th})^2}{\mathrm{S}_{tt} \times \mathrm{S}_{hh}}\) or \(\mathrm{S}_{hh} = \dfrac{(\mathrm{S}_{th})^2}{r^2 \times \mathrm{S}_{tt}}\) or substitute 1 value | M1 |
| e.g. \(\pm 0.985 = \dfrac{\pm 17501.25}{\sqrt{227.875 \times \mathrm{S}_{hh}}}\) or \(\mathrm{S}_{hh} = \dfrac{(\pm 17501.25)^2}{(\pm 0.985)^2 \times 227.875}\) o.e. ,(= 1 385 380.258 ) | A1, |
| = awrt 1 390 000 | A1 |
| \(b = \dfrac{-17501.25}{1385380.258} = -0.0126328\ldots\), = awrt \(-0.013\) | M1, A1 |
| [NB \(\bar{t} = 7.625\), \(\bar{h} = 796.25\) ] \(a = \dfrac{61}{8} - \text{"}-0.0126\ldots\text{"} \times \dfrac{6370}{8}\) [= 17.6838…] | M1 |
| So \(t = 17.7 - 0.0126h\) | A1 |
| (7) |
Notes
In (c) condone \(x\) for \(h\) and \(y\) for \(t\) except in 4th A1
1st M1 for the sight of the formula for \(r\) and an attempt to do something useful with it
1st A1 for a correct numerical expr’n in \(\mathrm{S}_{hh}\) or \(\sqrt{\mathrm{S}_{hh}}\) Accept with 3sf values (ignore – signs)
2nd A1 for awrt 1 390 000 (3sf gives 1 384 422.948 but scores 1st A1 and 2nd A0)
2nd M1 for a correct expression for \(b\) seen (ft their values to 3sf) Use of \(\mathrm{S}_{tt} \to -76.8\) is M0
3rd A1 for awrt – 0.013 (candidates using 3sf for \(\mathrm{S}_{hh}\) should therefore get this)
Beware \(\frac{\mathrm{S}_{tt}}{\mathrm{S}_{th}} = \frac{227.875}{-17501.25} = -0.0130\ldots\) but is 2nd M0 3rd A0 Ans only of – 0.0126..is M1A1A1M1A1
3rd M1 for a correct use of \(\bar{t}\) and \(\bar{h}\) to find \(a\) ft their \(b\) (allow letter \(b\) or even \(b = -0.985\) )
4th A1 for a correct equation with \(a\) = awrt 17.7 and \(b\) = awrt – 0.0126 [No \(y\) and \(x\)]
| Scheme | Marks |
|---|---|
| \(a\) is an estimate of the temperature at sea level is (17.7 °C) | B1 |
| (1) |
Notes
B1 for stating or implying that it is the temperature (value not needed) at sea level
| Scheme | Marks |
|---|---|
| \((\mp)\,150 \times b\) (o.e. e.g. [17.7 – 0.0126\(h\)] – [17.7 – 0.0126(\(h\) + 150)] ) | M1 |
| = 1.89 awrt 2 (°C) | A1 |
| (2) | |
| (14 marks) |
Notes
M1 for a correct expression equivalent to \((\mp)\) \(150b\). Can use letter \(b\) or ft their value(s).
A1 for awrt 2 (°C not required) Allow \(\pm\) can give if “\(a\)” incorrect or “\(b\)” from M0A0 in (c)
Common wrong answer of 11520 can score M1A0 even if no working seen.