## Question 9:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Calculate sample mean: $\bar{d} = 2623/5 = 524.6$ | M1 | |
| Estimate population variance (1st sample): $s_1^2 = (1376081 - 2623^2/5)/4 = [13.8$ or $3.715^2]$ | M1 | allow biased: $11.04$ or $3.323^2$ |
| Find confidence interval (allow $z$ in place of $t$): $524.6 \pm t\sqrt{(13.8/5)}$ | M1 | inconsistent use of 4 or 5 loses M1 |
| Correct tabular value: $t_{4,\,0.975} = 2.776$ (2 d.p.) | A1 | |
| Evaluate CI to 3 s.f.: $524.6 \pm 4.6[1]$ or $[520.0,\;529.2]$ | A1 | needs correct $s$, $t$ |

**Subtotal: 5 marks**

| Answer/Working | Marks | Guidance |
|---|---|---|
| State hypotheses: $H_0: \mu_b = \mu_a$, $H_1: \mu_b \neq \mu_a$ | B1 | |
| Estimate pop. variance (2nd sample): $s_2^2 = (2720780 - 5216^2/10)/9 = [572/45$ or $12.711$ or $3.565^2]$ | M1 | allow biased: $11.44$ or $3.382^2$ |
| Pooled variance estimate: $s^2 = (4 \times 13.8 + 9 \times 12.71)/13 = 848/65$ or $13.05$ | M1 A1 | |
| Calculate $t$ (to 2 dp): $t = (524.6 - 521.6)/(s\sqrt{(5^{-1} + 10^{-1})}) = 1.52$ | M1 *A1 | |
| Compare tabular $t$ value: $t_{13,\,0.95} = 1.77[1]$ | *B1 | |
| Conclusion: No difference in means | B1 | AEF, dep *A1, *B1 |

**Subtotal: 8 marks | Total: 13 marks**

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## Question 10a:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Take moments about $P$ for system [rod]: $F_A h = Wa\cos\theta$ | M1 A1 | |
| Take moments about $B$ for rod: $F_A h = R_A\cdot 2a\cos\theta$ | M1 A1 | |
| Eliminate $F_A$ to give $R_A$: $R_A = \frac{1}{2}W + (Wa\sin\theta)/h$ | M1 | |
| Find inequality for $\mu$: $\mu \geq F_A/R_A$, $\mu \geq 2a\cos\theta/(h + 2a\sin\theta)$ | M1 A1 | A.G. |

**Subtotal: 7 marks**

## Question 10a(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Use $kW = T$ to express in terms of $F_A$ or $R_A$: $kW = F_A/\sin\theta$ or $(W - R_A)/\cos\theta$ | M1 | |
| Substitute for $F_A$ or $R_A$: $k = (a/h)\cot\theta$ or $(\frac{1}{2} - (a/h)\sin\theta)/\cos\theta$ | M1 A1 | |
| Substitute for $h$ and $\theta$: $k = \sqrt{5}/6$ or $0.373$ | A1 | |

**Subtotal: 4 marks**

## Question 10a(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Find horizontal component $N_P$: $N_P = T\sin\theta$ or $kW\sin\theta$ | M1 | |
| Substitute for $k$ and $\theta$: $(\sqrt{5}/6)(2/3)W = \sqrt{5}W/9$ or $0.248W$ | M1 A1 | |

**Subtotal: 3 marks | Total: 14 marks**

# Question 10b:

## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use regression line or 1st normal equation: $\Sigma y/5 = 2.5\Sigma x/5 - 1.5$ | B1 | |
| Use data to substitute for $\Sigma x$ and $\Sigma y$: $11 + p + q = 2.5 \times 15 - 5 \times 1.5$ | M1 | |
| $p + q = 37.5 - 7.5 - 11 = 19$ **A.G.** | A1 | |
| **Total: 3 marks** | | |

## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use formula for $b$ or 2nd normal equation: $2.5 = (32+2p+6q-15\times30/5)/(61-15^2/5)$ | M2 A1 | |
| *or* $32+2p+6q = 2.5\times61 - 1.5\times15$ | A1 | |
| (A.E.F.) $p + 3q = 49$ (*or* $3q - p = 41$) | | |
| Solve any two simultaneous equations for $p$, $q$: $p = 4,\ q = 15$ | M1 A1 | |
| **Total: 6 marks** | | |

## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = (32+2p+6q-15\times30/5)/\sqrt{\{(61-15^2/5)(49+p^2+q^2-(11+p+q)^2/5)\}}$ | M1 | |
| $= (130 - 15\times30/5)/\sqrt{\{(61-15^2/5)(290-30^2/5)\}}$ | A1 | |
| *or* $2.5\sqrt{\{(61-15^2/5)/(49+p^2+q^2-(11+p+q)^2/5)\}}$ | (M1) | |
| $= 2.5\sqrt{\{(61-15^2/5)/(290-30^2/5)\}}$ | (A1) | |
| $= 40/\sqrt{(16\times110)}$ *or* $2.5\sqrt{(16/110)} = 0.953$ | A1 | |
| **Total: 3 marks** | | |

## Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(a)** State equation of actual regression line: $y = 0.25x - 1.5$ | B1 | |
| **(b)** State new value of $r$ or say unchanged: Same value as found in (iii) | B1 | |
| **Total: 2 marks** | | |

**Question total: [14]**