S1 June 2017 Q1
1. A clothes shop manager records the weekly sales figures, £\(s\), and the average weekly temperature, \(t\) °C, for 6 weeks during the summer. The sales figures were coded so that
\[w = \frac{s}{1000}\]The data are summarised as follows
\[\mathrm{S}_{ww} = 50 \quad \sum wt = 784 \quad \sum t^2 = 2435 \quad \sum t = 119 \quad \sum w = 42\]The manager of the clothes shop believes that a linear regression model may be appropriate to describe these data.
| Scheme | Marks |
|---|---|
| \([\mathrm{S}_{wt}] = 784 - \dfrac{119 \times 42}{6} =,\ \ -49\) | M1 A1 |
| \([\mathrm{S}_{tt}] = 2435 - \dfrac{119^2}{6} =,\ \ 74.8\dot{3}\) or \(74\tfrac{5}{6}\) or \(\dfrac{449}{6}\) (accept awrt 74.8) | A1 |
| (3) |
Notes
M1 for a correct expression for \(\mathrm{S}_{wt}\) or \(\mathrm{S}_{tt}\) (May be implied by either correct answer)
1st A1 for \([\mathrm{S}_{wt}] = -49\) 2nd A1 for \([\mathrm{S}_{tt}]\) = awrt 74.8
SC If both values correct but clearly mislabelled award M1A0A1
| Scheme | Marks |
|---|---|
| \(\mathrm{S}_{ss} = 5 \times 10^7\) or 50 000 000 (o.e.) | B1 |
| \(\mathrm{S}_{st} = \)\(-\)49 000 | B1ft |
| (2) |
Notes
2nd B1ft for multiplying their \(\mathrm{S}_{wt}\) by 1000
| Scheme | Marks |
|---|---|
| \(r = \dfrac{\text{"}-49\text{"}}{\sqrt{50 \times \text{"}74.8\dot{3}\text{"}}}\) or \(\dfrac{\text{"}-49\,000\text{"}}{\sqrt{\text{"}5 \times 10^7\text{"} \times \text{"}74.8\dot{3}\text{"}}}\) \(=,\ -0.80105\ldots\) = awrt \(-\)0.801 | M1, A1 |
| (2) |
Notes
M1 for a correct expression using their values provided \(\mathrm{S}_{tt}\) and \(\mathrm{S}_{ss}\) both > 0
A1 for awrt – 0.801 (Correct ans. only M1A1, – 0.80 with no working M1A0)
| Scheme | Marks |
|---|---|
| \(r\) is close to – 1 or \(|r|\) is close to 1 or “strong” (o.e.) [negative] correlation … so “yes” or does support the belief | B1ft |
| (1) |
Notes
B1ft for a correct comment that uses their value of \(r\) as support, provided \(0.5 \leqslant |r| \leqslant 1\)
For \(|r| \lt 0.5\) comment must be “does not support”, because “weak” (o.e.) correlation.
NB “points lie close to a straight line” is B0 unless supported by mention of their value of \(r\)
| Scheme | Marks |
|---|---|
| \(b = \dfrac{\text{"}-49\text{"}}{\text{"}74.8\dot{3}\text{"}} = [-0.6547\ldots],\ \ a = \dfrac{42}{6} - b \times \dfrac{119}{6} = [19.9866\ldots]\) or \(a = 7 - b \times 19.8\dot{3}\) | M1, M1 |
| So \(w = 20.0 - 0.655t\) | A1 |
| (3) |
Notes
1st M1 for a correct expression for \(b\) or awrt \(-0.66\) or \(-0.65\) Ft their answers from (a)
2nd M1 for a correct expression for \(a\) ft their value for \(b\)
A1 for a correct equation in \(w\) and \(t\) only with \(a = 20\) or awrt 20.0 and \(b\) = awrt – 0.655 (No fractions)
If their \(a\) and \(b\) are given to more than 3 sf, accept answers in (f) to 3sf or better.
| Scheme | Marks |
|---|---|
| \(s = 20\,000 - 655t\) or \(c = 20\,000\) and \(d = -655\) | B1ft B1ft |
| (2) |
Notes
1st B1 ft for correct \(c\) or “their 20.0”\(\times 1000\) 2nd B1ft for correct \(d\) or their “– 0.655”\(\times 1000\)
Values can be in an \(s\), \(t\) eq’n or \(c =\), \(d =\) (Their \(a\) and \(b\) needn’t be to 3 sf and ft their letter for \(t\))
| Scheme | Marks |
|---|---|
| Decrease in sales of [£] 655 ( ignore any minus sign) | B1ft |
| (1) | |
| (14 marks) |
Notes
B1ft for stating clearly both decrease (o.e.) and [£] 655. Ft their \(d\) and allow “increase” if \(d \gt 0\)