## Question 10:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0$: No difference in preferences [or independent] | B1 | State (at least) null hypothesis |
| $28.57 \quad 37.71 \quad 13.71$ | | Find expected values (to 1 d.p.) |
| $21.43 \quad 28.29 \quad 10.29$ | M1 A1 | (lose A1 if rounded to integers) |
| $\chi^2 = 0.411 + 0.078 + 0.214 + 0.549 + 0.104 + 0.286 = 1.64$ (or $1.61$ if 1 d.p. used) | M1 A1 | Calculate value of $\chi^2$ |
| $\chi^2_{2,\,0.95} = 5.99_{[1]}$ | B1 | State or use correct tabular $\chi^2$ value |
| Accept $H_0$ if $\chi^2 \leq$ tabular value | M1 | Valid method for reaching conclusion |
| No difference in preferences | A1 | Conclusion consistent with correct values (AEF) |
| **Part total: 8** | | |
| $\chi^2_{new} = n \times \chi^2$ | M1 | Calculate new value $\chi_{new}^2$ |
| $\chi^2_{2,\,0.95} = 5.99_{[1]}$ | B1 | State or use correct tabular $\chi^2$ value |
| $n > 5.99/1.64$, $n_{min} = 4$ | M1 A1 | Find $n_{min}$ |
| **Part total: 4** | **[Total: 12]** | |

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## Question 11a:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mga(1-\cos\theta)$ $[v^2 = u^2 + 2ga(1-\cos\theta)]$ | M1 A1 | Use conservation of energy ($v^2 = \ldots$) |
| $R = mg\cos\theta - mv^2/a$ | M1 A1 | Equate radial forces ($R$ may be taken as zero) |
| $0 = mg(3\cos\theta - 2) - mu^2/a$ | | Eliminate $v^2$ to find $\cos\theta$ when $R=0$ |
| $\cos\theta_1 = \frac{1}{3}(2 + 2/5) = 4/5$ | M1 A1 | **A.G.** (denoting this $\theta$ by $\theta_1$) |
| $v_1^2 = 2ga/5 + 2ga/5 = 4ga/5$ | B1 | Find $v^2$ at this point |
| $v_2^2 = v_1^2\sin^2\theta_1 + 2ga(1 + \cos\theta_1)$ $= (4/5 \times 3^2/5^2 + 2 \times 9/5)ga = 486ga/125$ or $3.89ga$ | M1, A1 | Find vertical comp. $v_2$ of $v$ at plane (using $u$ in place of $v_1$ can earn M1 A0) |
| $v_3 = (5/9)v_2$ $[v_3^2 = 6ga/5]$ | M1 | Find vertical comp. $v_3$ of rebound speed |
| $v_3^2/2g = (5/9)^2(486ga/125)/2g = 3a/5$ | M1, A1 | Find vertical height reached |
| **Part total: 6+6 = 12** | | |

## Question 11b:

**Hypotheses:**
$H_0: \mu_X = \mu_Y$, $H_1: \mu_X < \mu_Y$ | B1 |

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**Estimate population variance using X's sample:**

$s_X^2 = (85.8 - 58.2^2/60) / 59$

$[= 0.4974 \text{ or } 0.7053^2]$ | M1 A1 | Allow use of biased: $\sigma_{X,60}^2 = 0.4891$ or $0.6994^2$

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**Estimate population variance using Y's sample:**

$s_Y^2 = (188.6 - 97.6^2/80) / 79$

$[= 0.8801 \text{ or } 0.9381^2]$ | M1 A1 | Allow use of biased: $\sigma_{Y,80}^2 = 0.8691$ or $0.9323^2$

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**Estimate population variance for combined sample:**

$s^2 = s_X^2/60 + s_Y^2/80$

$= 0.01929 \text{ or } 0.1389^2$ | |

$(or \ 0.01901 \text{ or } 0.1379^2)$ | M1 A1 | Allow use of $\sigma_{X,60}^2$, $\sigma_{Y,80}^2$

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**Calculate value of z (to 2 d.p., either sign):**

$z = (0.97 - 1.22) / s$

$= -0.25/0.1389 = -1.80$

$(or -1.813)$ | M1, A1 |

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**Find $\Phi(z)$:**

$\Phi(z) = 0.9641 \text{ (or } 0.9651)$ | M1 |

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**Find corresponding significance level:**

$3.59 \text{ (or } 3.49)$ | M1 |

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**Find set of possible values of $\alpha$ (to 1 d.p.):**

$\alpha > 3.6$ | A1 | **12**

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**S.R.** Allow (implicit) assumption of equal variances as follows, but deduct A1 if not explicit:

**Find pooled estimate of common variance $s^2$:**

$(60\sigma_{X,60}^2 + 80\sigma_{Y,80}^2)/138$

$= 0.7165 \text{ or } 0.8465^2$ | |

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**Calculate value of z (to 2 d.p.):**

$z = (0.97 - 1.22)/s\sqrt{(1/60 + 1/80)}$

$= -1.73$ | |

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**Find set of possible values of $\alpha$ (to 1 d.p.):**

$\Phi(z) = 0.9582$, $\alpha > 4.2$ | | **[12]**