## Question 10:

### Part 10(i):
| $H_0: \mu_x - \mu_y = 1$, $H_1: \mu_x - \mu_y > 1$ | B1 | State both hypotheses (B0 for $\bar{x}$...) |
| $d_i$: $1.1\ 2.1\ 0.7\ 0.4\ 2.0\ 2.8\ 0.2\ 1.9\ 1.4$ | M1 | Consider differences $d_i$, e.g. $x_i - y_i$ |
| $\bar{d} = 12.6/9 = 1.4$ | B1 | Find sample mean |
| $s^2 = (23.72 - 12.6^2/9)/8\ [= 0.76$ or $0.8718^2]$ | M1 | Estimate population variance (allow biased: $[0.6756$ or $0.8219^2]$) |
| $t_{8,\ 0.9} = 1.397$ or $1.40$ | B1 | State or use correct tabular $t$-value |
| $t = (\bar{d} - 1)/(s/\sqrt{9}) = 1.38$ [Accept $H_0$:] No evidence for coach's belief | M1 A1 | Find $t$ (or compare $\bar{d} - 1 = 0.4$ with $t_{8,0.9}\ s/\sqrt{9} = 0.406$); Consistent conclusion |
| or time will not decrease by more than 1s | $B1\sqrt{}$ | |

**Total: 8 marks**

### Part 10(ii):
| $s_y^2 = (11395.79 - 319.5^2/9)/8 = 6.693$ or $2.587^2$ | B1 | Estimate population variance using $y$'s sample |
| $319.5/9 \pm t\sqrt{(s_y^2/9)}$ | M1 | Find confidence interval (M0 if $s^2$ not $s_y^2$) |
| $t_{8,\ 0.975} = 2.306$ or $2.31$ | A1 | State or use correct tabular value of $t$ |
| $35.5 \pm 1.99$ or $[33.5,\ 37.5]$ (allow $35.5 \pm 2.0$) | A1 | Evaluate C.I.; SC B1 (max 1/4) for $1.4 \pm 0.67$ or $[0.73,\ 2.07]$ |

**Total: 4 marks**

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## Question 11A:

### Part 11A(i):
| $I_{lamina} = \frac{1}{3}kM((\frac{1}{2}a)^2 + (\frac{1}{2}a)^2)\ [= (k/6)Ma^2]$ | B1 | Find or state MI of lamina about axis at its centre |
| $I'_{lamina} = I_{lamina} + kM((3a/2)^2 + (3a/2)^2)\ [= (14k/3)Ma^2]$ | M1 A1 | Find MI of lamina about axis at $A$ |
| $I_{AB} = \frac{1}{3}Ma^2 + Ma^2\ [= (4/3)Ma^2]$ | B1 | Find or state MI of $AB$ (or $AD$) about axis at $A$ |
| $I_{BC} = \frac{1}{3}Ma^2 + M(a^2 + (2a)^2)\ [= (16/3)Ma^2]$ | M1 A1 | Find or state MI of $BC$ (or $DC$) about axis at $A$ |
| OR: $I_{rod} = \frac{1}{3}Ma^2 + Ma^2\ [= (4/3)Ma^2]$ | (B1) | Find or state MI of any rod about axis at centre of frame |
| $I_{frame} = 4 \times I_{rod} + 4M(a^2 + a^2)\ [= (40/3)Ma^2]$ | (M1 A1) | Find or state MI of frame about axis at $A$ |
| $I = (14k/3 + 2 \times 4/3 + 2 \times 16/3)Ma^2$ or $(14k/3 + 40/3)Ma^2$ | M1 | Verify MI of system about axis at $A$ |
| $= \frac{2}{3}Ma^2(7k + 20)$ | A1 AG | A0 if inadequate explanation |

**Total: 8 marks**

### Part 11A(ii):
| $\frac{1}{2}I\omega^2 = kMg \times 3a + 2Mg \times 2a + Mg \times 4a$ or $kMg \times 3a + 4Mg \times 2a$ | M1 A2 | Find $\omega^2$ or angular speed $\omega$ when $D$ below $B$ by energy; Award A1 if error in only term |
| $\omega^2 = \{3(3k+8)/(7k+20)\}\ g/a$; $4 \times 3(3k+8) = 5(7k+20)$, $k=4$ | A1 | Find $k$ by equating $\omega^2$ to $\{\frac{1}{2}\sqrt{(5g/a)}\}^2$ |
| | M1 A1 | |

**Total: 6 marks**

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## Question 11B:

### Part 11B(i):
| $\bar{x} = 469/250 = 1.876$ and $\sigma^2 = (1195 - 469^2/250)/249 = 1.266$ | B1 AG / B1 AG | Verify mean and unbiased variance of data, showing method |
| $6 \times 0.313 = 1.878$ and $6 \times 0.313 \times 0.687 = 1.29[0]$ | M1 A1 | Find mean $np$ and variance $npq$ of binomial distribution |
| Means and variances are similar for $X$ and $B(6,\ 0.313)$ | A1 | Valid explanation (AEF; needs $1.878$, $1.290$ correct to $0.002$) |

**Total: 4 marks**

### Part 11B(ii):
| $250 \times\ ^6C_4 \times 0.313^4 \times 0.687^2\ [= 250 \times 0.06795 = 16.99]$; Allow $49.7 \times (^6C_4 \times 0.313)/(^6C_3 \times 0.687)$ | M1 A1 | Show how expected frequency for $x = 4$ is found (AEF for $^6C_4 = 15$) |

**Total: 2 marks**

### Part 11B(iii):
| $H_0$: [Binomial] distribution fits data | B1 | State (at least) null hypothesis |
| $O_i$: $22\ 83\ 72\ 53\ \underline{20}$; $E_i$: $26.3\ 71.9\ 81.8\ 49.7\ \underline{20.3}$ | *M1 A1 | Combine last 3 cells so that all exp. value $\geq 5$ |
| $X^2 = 0.703 + 1.714 + 1.174 + 0.219 + 0.004 = 3.81$ | M1 A1 | Find value of $\chi^2$ from $\Sigma(E_i - O_i)^2/E_i$ |
| No. $n$ of cells: $7\ 6\ \underline{5}$; $\chi^2_{n-2,\ 0.95}$: $11.07\ 9.488\ \underline{7.815}$ | $B1\sqrt{}$ | State or use consistent tabular value $\chi^2_{n-2,\ 0.95}$ (to 3 s.f.) |
| Accept $H_0$ if $X^2 <$ tabular value | M1 | State or imply valid method for conclusion |
| $3.81\ [\pm 0.01] < 7.815$ so distribution fits or scientist's belief is correct | A1 | Conclusion (AEF; requires both values approx. correct) |

**Total: 8 marks**