## Question 10:

### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(\sum x)^2 = 6(844.20 - 6 \times 36.66) = 3775.44$ | M1 A1 | Find $\sum x$ and $\sum y$ (M1 for either) |
| $\sum x = 61.2$ or $\bar{x} = 10.2$ | | **SC**: Allow M1 if 5 used instead of 6 (max 4/11), giving $62.97$, $50.97$ and $r = 0.961[5]$ |
| $(\sum y)^2 = 6(481.5 - 6 \times 9.69) = 2540.16$ | A1 | |
| $\sum y = 50.4$ or $\bar{y} = 8.4$ | | |
| $S_{xy} = 625.59 - 61.2 \times 50.4/6 = 111.51$ | | Find correlation coefficient $r$ |
| $S_{xx} = 844.20 - 61.2^2/6$ or $6 \times 36.66 = 219.96$ | M1 A1 | |
| $S_{yy} = 481.50 - 50.4^2/6$ or $6 \times 9.69 = 58.14$ | | |
| $r = S_{xy}/\sqrt{(S_{xx}\,S_{yy})} = 0.986$ | | Total: [5] |

# Question (ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $b_1 = S_{xy} / S_{xx} = 0.507$ | B1 | Calculate gradients in both lines |
| $b_2 = S_{xy} / S_{yy}$ or $r^2/b_1 = 1.918$ | | |
| $y = 50.4/6 + 0.507(x - 61.2/6) = 0.507x + 3.23$ | M1 A1 | Find one regression line, e.g. $y$ on $x$ |
| $x = 61.2/6 + 1.918(y - 50.4/6) = 1.92y - 5.91$ | A1 | Find other regression line, e.g. $x$ on $y$ |
| | [4] | SC: Use PA $-1$ if results only correct to 2 s.f. |

# Question (iii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 6.36$ (allow $6.37$ or $6.38$) | B1 | Find $x$ when $y = 6.4$ |
| Reliable since $r \approx 1$ [$\sqrt{}$ on $r$] *or* $6.4$ within range of $y$ | B1 | Valid comment on reliability |
| | [2] | |

# Question 11A:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $W \times a\sin\Phi + 2W \times (3a/2)\sin\Phi = T \times AE$ | M1 A1 | Take moments for rod about $A$ (AEF) |
| $W \times a\sin 2\theta + 2W \times (3a/2)\sin 2\theta = T \times 2a\sin\theta$ | A1 | Substituting $\Phi = 2\theta$ and $AE = 2a\sin\theta$ |
| $T = 2W\sin 2\theta / \sin\theta$ $[= 4W\cos\theta]$ | A1 | Hence find tension $T$ in terms of $\theta$ |
| $T = 3W(BC - 3a/2)/(3a/2) = W(8\cos\theta - 3)$ | M1 A1 A1 | Find $T$ in terms of $\theta$ using Hooke's Law |
| $4\cos\theta = 8\cos\theta - 3$ $\cos\theta = \frac{3}{4}$ **A.G.** | M1 A1 | Equate to find $\cos\theta$ |
| $T = 3W$ | B1 | Hence find $T$ (allow assumption of $\cos\theta = \frac{3}{4}$) |
| | [10] | |
| $X = T\sin\theta \left[= 3W\sqrt{7}/4\right]$ | B1 | Find horizontal force $X$ at $A$ (ignore sign) |
| $Y = 3W - T\cos\theta \left[= 3W/4\right]$ | | Find vertical force $Y$ at $A$ (ignore sign) |
| $\sqrt{X^2 + Y^2} = \frac{3}{\sqrt{2}}W$ *or* $3\sqrt{2}/2\ W$ *or* $2.12\ W$ | M1 A1 | Find magnitude of force at $A$ |
| | [4] | |

# Question 11B:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\Sigma x = 12\bar{x} = 12 \times \frac{1}{2}(25.17 + 26.83) = 12 \times 26 = 312$ | M1 A1 M1 | Find $\Sigma x$ via sample mean $\bar{x}$ |
| $t\ s_A / \sqrt{12} = \frac{1}{2}(26.83 - 25.17)$, $s_A = 0.83\sqrt{12}/2.201 = 1.306$ or $s_A^2 = 1.706$ | A1 | Find estimated s.d. $s_A$ or $s_A^2$; with $t = t_{11,\ 0.975} = 2.201$ (to 3 s.f.) |
| $s_A^2 = \{\Sigma x^2 - (\Sigma x)^2/12\}/11$; $\Sigma x^2 = 11s_A^2 + (\Sigma x)^2/12 = 8130[.8]$ | M1 A1 | Find $\Sigma x^2$ from $s_A$ or $s_A^2$ (M0 for $s_A^2 = \{\ldots\}/12$) |
| | [6] | |
| $H_0: \mu_A = \mu_B$, $H_1: \mu_A > \mu_B$ | B1 | State hypotheses |
| $s_B^2 = (4507.62 - 177^2/7)/6 = [5.341]$ | M1 | Estimate $B$'s population variance (to 3 d.p.); allow biased: $4.578$ |
| $s^2 = (11s_A^2 + 6s_B^2)/17 = (18.77 + 32.05)/17 = 2.989$ or $1.729^2$ (to 3 s.f.) | M1 A1 | Find pooled estimate of common variance |
| $t = (26 - 177/7)/s\sqrt{1/12 + 1/7} = (26 - 25.29)/s\sqrt{1/12 + 1/7} = 0.7143/0.8222 = 0.869$ | M1 A1 *B1 | Calculate value of $t$ (or $-t$) (to 3 s.f.) |
| $t_{17,\ 0.9} = 1.333$ | | State or use correct tabular $t$ value (to 3 s.f.) |
| $t <$ tabular value, so Petra's belief not supported *or* wing span of $A$ not greater | B1$\checkmark$ | Consistent conclusion (AEF, $\sqrt{}$ on $t$, dep *B1) |
| | [8] | |