## Question 10(i):

$\bar{x} = 0.91$, $\bar{y} = 1.205$; $s_x^2 = 19.56/49\ [= 489/1225\ \text{or}\ 0.3992]$ and $s_y^2 = 30.25/59\ [= 121/236\ \text{or}\ 0.5127]$ | B1, M1 | Find both sample means; Estimate both population variances (allow biased: $0.3912$ and $0.5042$)

EITHER: $s^2 = s_x^2/50 + s_y^2/60$ | (M1) | Estimate or imply combined variance

$= 0.01653$ or $0.1286^2$ (to 3 s.f. throughout) | A1 |

$[\pm](\bar{y} - \bar{x}) \pm zs$ | M1 | Find confidence interval for difference $Y - X$ or $X - Y$

$z_{0.95} = 1.645$; $[\pm]\ 0.295 \pm 0.211$ (allow $0.212$) | A1 | Use appropriate tabular value; Evaluate confidence interval (either form)

or $[\pm]\ [0.084,\ 0.506]$ (allow $[\pm][0.083,\ 0.507]$) | A1) |

OR: Assume equal [population] variances; $s^2 = (49s_x^2 + 59s_y^2)/108$ or $(19.56 + 30.25)/108$ | (B1) | State assumption; Find or imply pooled estimate of common variance

$= 4981/10800$ or $0.461$ or $0.679^2$ | B1 |

$[\pm](\bar{y} - \bar{x}) \pm zs\sqrt{1/50 + 1/60}$ | M1 | Find confidence interval for difference $Y - X$ or $X - Y$

$z_{0.95} = 1.645$ | A1 | Use appropriate tabular $z$-value (or $t$-value from calculator)

$[\pm]\ 0.295 \pm 0.214$ or $[\pm]\ [0.081,\ 0.509]$ | A1) | Evaluate confidence interval (either form)

**Total: 7**

## Question 10(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = (1.205 - 0.91)/s = 0.295/s = 2.29[4]$ [or $2.26[9]$] (to 3 s.f.) | M1 A1 | Find value of $z$ (either sign) |
| $\Phi(z) = 0.989[1]$ [or $0.988[4]$] | A1 | Find $\Phi(z)$ |
| $100 \times (1 - 0.989) \times 2 = 2.2$ [or $2.3$] (to 1 d.p.) | M1 A1 | Find limiting value for $\alpha$, based on two-tail test (**M0** for basing on one-tail test) |
| $\alpha < (or \leqslant) 2.2$ [or $2.3$] | A1 | Find set of possible values of $\alpha$ (Treat $\alpha$ instead of $\alpha\%$ as misread) |
| | **6** | |

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## Question 11A(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mga(\cos\alpha + \cos\beta)$ | M1 A1 | Find $v^2$ at $A'$ from conservation of energy (**A0** if no $m$) |
| $mv^2/a = mg\cos\beta$; $u^2 = ag\cos\beta + 2ag(1/16 + \cos\beta)$ | B1 | Use $F = ma$ radially at $A'$ with $R_{A'} = 0$; Use $\cos\alpha = 1/16$ and eliminate $v^2$ to verify $u^2$ |
| $= \frac{1}{8}ag(1 + 24\cos\beta)$ AG | M1 A1 | |
| | **5** | |

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## Question 11A(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}mw^2 = \frac{1}{2}mu^2 - mga(1 + \cos 2\beta)$ | B1 | Find $w^2$ at $B'$ from conservation of energy (**B0** if no $m$) |
| $mw^2/a = mg\cos 2\beta$; $u^2 = ag\cos 2\beta + 2ag(1 + \cos 2\beta)$ | B1 | Use $F = ma$ radially at $B'$ with $R_{B'} = 0$; Eliminate $w^2$ to find $u^2$ |
| $= ag(2 + 3\cos 2\beta)$ | M1 | |
| $1 + 24\cos\beta = 8(2 + 3\cos 2\beta) = 16 + 48\cos^2\beta - 24$; $16\cos^2\beta - 8\cos\beta - 3 = 0$ | M1 | Combine equations for $u^2$; Formulate and solve quadratic to find $\cos\beta$ |
| $\cos\beta = \frac{3}{4}$ [rejecting $-\frac{1}{4}$] | M1 A1 | |
| | **6** | |

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## Question 11A(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $u^2 = \frac{19}{8}ag$ or $2.375ag$; $R = mu^2/a + mg\cos\alpha = (19/8 + 1/16)mg$ | B1 | Find $u^2$ using value of $\cos\beta$; Use $F = ma$ radially at $A$ to find reaction $R$ at $A$ |
| $= \frac{39}{16}mg$ or $2.44mg$ | M1 A1 | |
| | **3** | |

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## Question 11B(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{xy} = 218.72 - 45.3 \times 40.5/9 = 14.87$ or $1.652$; $S_{xx} = 245.59 - 45.3^2/9 = 17.58$ or $1.953$; $S_{yy} = 195.11 - 40.5^2/9 = 12.86$ or $1.429$ | | Find required values |
| EITHER: $b_1 = S_{xy}/S_{xx} = 14.87/17.58 = 0.84585 = 0.846$; $(y - 40.5/9) = b_1(x - 45.3/9)$; $(y - 4.5) = 0.846(x - 5.033)$ | (M1 A1) | Find gradient in $y - \bar{y} = b_1(x - \bar{x})$ to 3 s.f.; Find equation of regression line to 3 s.f. |
| $y = 0.846x + 0.243$ | M1 A1 | |
| $x = 4.68$ | A1 | Find $x$ when $y = 4.2$ |
| OR: $b_2 = S_{xy}/S_{yy} = 14.87/12.86 = 1.1563 = 1.16$; $(x - 45.3/9) = b_2(y - 40.5/9)$; $(x - 5.033) = 1.16(y - 4.5)$ | (M1 A1) | Find gradient in $x - \bar{x} = b_2(y - \bar{y})$ to 3 s.f.; Find equation of regression line to 3 s.f. |
| $x = 1.16y - 0.170[04]$ | M1 A1 | |
| $x = 4.69$ | A1 | Find $x$ when $y = 4.2$ (**A0** for $x = 4.70$) |
| | **5** | |

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## Question 11B(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu_x - \mu_y = 0.3$, $H_1: \mu_x - \mu_y > 0.3$ (AEF) | B1 | State hypotheses (**B0** for $\bar{x}\ldots$) |
| $\bar{d} = 4.8/9$ or $8/15$ or $0.533$ (where $d = x - y$) | B1 | Find sample mean |
| $s^2 = (3.26 - 4.8^2/9)/8 = 7/80$ or $0.0875$ or $0.296^2$ | M1 A1 | Estimate population variance (allow biased: $7/90$ or $0.0778$ or $0.279^2$) |
| $t = (\bar{d} - 0.3)/(s/\sqrt{9}) = 2.37$ | M1 A1 | Find value of $t$ |
| $t_{8,\, 0.975} = 2.306$ or $2.31$ | B1 | State or use correct tabular $t$-value (or can compare $\bar{d}$ with $0.3 + t_{8,\,0.975}\, s/\sqrt{9} = 0.527$) |
| Reject $H_0$ if $t >$ tabular value; $2.37\ [\pm 0.1] > 2.31$ so accept belief (AEF) | M1 | Valid method for reaching conclusion; Correct conclusion from correct values |
| [of increase of more than $0.3$] (AEF) | A1 | SC: Wrong (hypothesis) test can earn only **B1** for hypotheses |
| | **9** | |