## Question 10(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum x=25$, $\sum y=25+q$, $\sum xy=135+4q$, $\sum x^2=141$, $\sum y^2=159+q^2$ | B1 | Find required values |
| $S_{xy}=135+4q-25(25+q)/5=[10-q]$ (AEF); $[S_{xx}=141-25^2/5=16]$; $S_{yy}=(159+q^2)-(25+q)^2/5\ [=34-10q+4q^2/5]$ | M1 A1 | ($S_{xy}$, $S_{xx}$, $S_{yy}$ may be scaled by the same constant) |
| $40-4q=170-50q+4q^2$, $2q^2-23q+65=0$, $(2q-13)(q-5)=0$, $q=5$ | M1 A1 | Equate gradient $\frac{5}{4}$ in line of $x$ on $y$ to $S_{xy}/S_{yy}$ and solve quadratic to find integer value of $q$ |
| | **5** | |

## Question 10(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $c = 25/5 - (5/4)(25+q)/5 = -(5+q)/4$ | M1 A1 | Find $c$ from $\bar{x} - \frac{5}{4}\,\bar{y}$ |
| $= -5/2$ or $-2.5$ | A1 | |
| | **3** | |

## Question 10(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = S_{xy}/\sqrt{(S_{xx}S_{yy})} = 5/\sqrt{(16\times 4)}$ | M1 A1 | Find correlation coefficient $r$ |
| $= 5/8$ or $0.625$ | A1 | |
| | **3** | |

## Question 11E(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}mu_P^2 = \frac{1}{2}m(21ag/2) + mga\quad [u_P^2 = \frac{25}{2}ag]$ | M1 A1 | $u_P$ is speed of $P$ at lowest point, $v_Q$ is speed of $Q$ immediately after collision. Apply conservation of energy at lowest point (A0 if no $m$) |
| $\frac{1}{2}\cdot4mv_Q^2 = 4mag$ | M1 | Find speed $v_Q$ at lowest point by conservation of energy (A0 if no $m$) |
| $v_Q = \sqrt{2ag}$ or $1.41\sqrt{ag}$ or $4.47\sqrt{a}$ | A1 | |
| $mu_P = [\pm]\,mv_P + 4mv_Q$ | M1 | Find $v_P$ using conservation of momentum ($m$ may be omitted) |
| $v_P = [\pm](-5/\sqrt{2}+4\sqrt{2})\sqrt{ag}$ | A1 | |
| $|v_P| = \frac{3}{\sqrt{2}}\sqrt{ag}$ or $2.12\sqrt{ag}$ or $0.671\sqrt{a}$ (AEF) | A1 | Hence find speed of $P$ |
| | **7** | |

## Question 11E(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $V_P$ is speed of $P$ when it loses contact: $\frac{1}{2}mV_P^2 = \frac{1}{2}mv_P^2 - mga(1+\cos\alpha)$; $[V_P^2 = \frac{9}{2}ag - 2ga(1+\cos\alpha) = (\frac{5}{2}-2\cos\alpha)ag]$ | M1 A1 | Apply conservation of energy at $D$ (A0 if no $m$) |
| $[R_D=]\,mV_P^2/a - mg\cos\alpha = 0\quad [V_P^2 = ag\cos\alpha]$ | M1 A1 | Apply $F=ma$ radially at $D$ with reaction $= 0$ |
| $(\frac{5}{2}-2\cos\alpha)ag = ag\cos\alpha$, $\cos\alpha = 5/6$ or $0.833$ | A1 | Combine to find $\cos\alpha$ |
| | **5** | |

## Question 11O(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| $t_{s_A} / \sqrt{8} = \frac{1}{2}(16.7 - 13.5) [= 1.6]$ | M1 | Relate $s_A$ to semi-width of confidence interval |
| $t_{7, 0.975} = 2.365$ (to 3 s.f.) | A1 | State or use correct tabular $t$ value |
| $[s_A = \sqrt{8} \times 1.6 / 2.365 = 1.9135]$, $s_A^2 = 3.66[16]$ | A1 | Hence find unbiased estimate of $A$'s population variance |
| | **3** | |

## Question 11O(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu_A = \mu_B$, $H_1: \mu_A > \mu_B$ (AEF) | B1 | State hypotheses (B0 for $\bar{x}$ ...) |
| $[\bar{x}_A = 15.1]$, $\bar{x}_B = 85.2/6 = 14.2$ | B1 | Find sample mean for $B$ |
| $s_B^2 = (1221.06 - 85.2^2/6)/5 = 561/250$ or $2.244$ or $1.498^2$ (all to 3 s.f.) | M1 | Estimate or imply population variance for $B$ (allow biased here: $1.87$ or $1.367^2$) |
| $s^2 = (7s_A^2 + 5s_B^2)/12 = 3.0709$ or $1.752^2$ | M1 A1 | Estimate (pooled) common variance ($s_B^2$ not needed explicitly) |
| $t_{12, 0.95} = 1.782$ | B1 | State or use correct tabular $t$ value |
| $[-]\, t = (\bar{x}_A - \bar{x}_B) / (s\sqrt{1/8 + 1/6}) = 0.951$, $t < 1.78$ so [accept $H_0$] | M1 A1 | Find value of $t$ (or can compare $\bar{x}_A - \bar{x}_B = 0.9$ with $1.69$); Correct conclusion |
| mean mass of $B$ not less than mean mass of $A$ (AEF) | B1 | |
| | **9** | |

**SC1:** Implicitly taking $s_A^2$, $s_B^2$ as unequal population variances (may also earn first B1 B1 M1):
$$z = (\bar{x}_A - \bar{x}_B) / \sqrt{(s_A^2/8 + s_B^2/6)} = 0.9/\sqrt{(0.8317)} = 0.987$$
$z < 1.645$ so

**DepSC1:** mean mass of $B$ not less than mean mass of $A$ (AEF); Comparison with $z_{0.95}$ and conclusion (FT on $z$); (can earn at most 5/9)