A2 June 2024 Q1
1. Two students are experimenting with some water in a plastic bottle. The bottle is filled with water and a hole is put in the bottom of the bottle. The students record the time, \(t\) seconds, it takes for the water level to fall to each of 10 given values of the height, \(h\) cm, above the hole.
Student \(A\) models the data with an equation of the form \(t = a + b\sqrt{h}\)
The data is coded using \(v = t - 40\) and \(w = \sqrt{h}\) and the following information is obtained.
\[\sum v = 626 \qquad \sum v^2 = 64\,678 \qquad \sum w = 22.47 \qquad \mathrm{S}_{ww} = 4.52 \qquad \mathrm{S}_{vw} = -338.83\]The time it takes the water level to fall to a height of 9 cm above the hole is 47 seconds.
Give your answer to 2 decimal places. (2)
Given that the residual sum of squares (RSS) for the model of \(t\) on \(\sqrt{h}\) is the same as the RSS for the model of \(v\) on \(w\),
Student \(B\) models the data with an equation of the form \(t = c + dh\)
The regression line of \(t\) on \(h\) is calculated and the residual sum of squares (RSS) is found to be 980 to 3 significant figures.
| Scheme | Marks | AO |
|---|---|---|
| \(b = \dfrac{-338.83}{4.52}\ \ [= -74.96\ldots]\) | M1 | 3.3 |
| \(a = \dfrac{626}{10} - \text{“}b\text{”}\dfrac{22.47}{10}\ \ [= 231.04\ldots]\) | M1 | 1.1b |
| \(t - 40 = \text{“}231.04\ldots\text{”} + \left(\text{“}-74.96\ldots\text{”}\right)\sqrt{h}\) | dM1 | 3.1a |
| \(t = 271.04\ldots - 74.96\ldots\sqrt{h}\) | A1 | 1.1b |
| (4) |
Notes
M1: For use of a correct model ie a correct expression for \(b\)
M1: For use of a correct model ie a correct expression (ft) for \(a\)
dM1: dep on both previous method marks for proceeding from an equation of the form \(v = \text{“}a\text{”} + \text{“}b\text{”}w\) to a correct un-simplified model in terms of \(h\) and \(t\) ft their \(a\) and \(b\)
A1: For a correct model \(t = 271.04\ldots - 74.96\ldots\sqrt{h}\) with awrt 271 and awrt 75 (corrected from the printed mark scheme: the note prints 74.95…; \(b = -74.96\ldots\))
| Scheme | Marks | AO |
|---|---|---|
| Residual \(= 47 - \left(\text{“}271.04\ldots\text{”} - \text{“}74.96\ldots\text{”} \times \sqrt{9}\right)\) | M1 | 3.4 |
| \(= 0.8466\ldots\) | A1 | 1.1b |
| (2) |
Notes
M1: For a correct method to find the residual. States \(h = 9\) and a correct expression to find the residual for their model of the form \(t = a + b\sqrt{h}\) or 9 substituted into a correct expression. Must be subtracting the correct way round. Alternatively, may code the data to \(v\) and \(w\) and attempt \(47 - \left(\text{“}231.04\ldots\text{”} - \text{“}74.96\ldots\text{”} \times 3\right)\)
A1: awrt 0.85 Allow answers in range awrt 0.84 to awrt 0.86 if working shown following a correct model in (a). If no working is shown then look for awrt 0.847
| Scheme | Marks | AO |
|---|---|---|
| \(\text{RSS} = \left[64\,678 - \dfrac{626^2}{10}\right] - \dfrac{(-338.83)^2}{4.52} \quad \left(= 25\,490.4 - \dfrac{(-338.83)^2}{4.52}\right)\) | M1 | 1.1b |
| \(= 90.89\ldots\ [\mathrm{s}^2]\) | A1 | 1.1b |
| (2) |
Notes
M1: For a correct expression for RSS. 25490.4 may be seen as \(\dfrac{127\,452}{5}\)
May also use \(\text{RSS} = \mathrm{S}_{vv}(1 - r^2) = 25\,490.4\left(1 - \dfrac{(-338.83)^2}{4.52 \times 25\,490.4}\right)\)
A1: awrt 90.9
| Scheme | Marks | AO |
|---|---|---|
| Student \(A\)’s model as the sum of squares of the residuals is lower | B1 | 2.4 |
| (1) | ||
| (9 marks) |
Notes
B1: Explaining a reason for their conclusion that A is a more suitable model provided their positive RSS found in (c) is less than 980. e.g. RSS is smaller so model \(A\)
Condone references to the model being more accurate oe. Must be a comparison with \(B\) or implied so do not accept statements such as “\(A\) because it has a small RSS”