## Question 9:

**Regression line of y on x:**

Calculate gradient $b_1$ in $y - \bar{y} = b_1(x - \bar{x})$:
$S_{xy} = 24879 - 472 \times 400/8 = 1279$ | 2 marks | M1 A1 |
$S_{xx} = 29950 - 472^2/8 = 2102$
$b_1 = S_{xy}/S_{xx} = 0.6085$ (3 s.f.)

Find regression line of $y$ on $x$:
$y = 400/8 + b_1(x - 472/8) = 50 + 0.6085(x-59) = 0.6085x + 14.1$ | 2 marks | M1 A1 |

Find $y$ when $x = 72$: $= 57.9$ or $58$
Allow use of regression line of $x$ on $y$ (since neither variable clearly independent):
$S_{yy} = 21226 - 400^2/8 = 1226$
$b_2 = S_{xy}/S_{yy} = 1.043$ | 2 marks | (M1 A1) |

$x = 472/8 + b_2(y - 400/8)$ | 2 marks | (M1 A1) |

$= 1.043y + 6.85$
$Y = 62.5$ or $62$ | 1 mark | A1 |

**Total: 5 marks**

**Product moment correlation coefficient:**

$r = 1279/\sqrt{(2102 \times 1226)}$
or $\sqrt{(0.6085 \times 1.043)} = 0.797$ | 2 marks | M1 A1* |

State both hypotheses (B0 for $r\ldots$):
$H_0: \rho = 0$, $H_1: \rho \neq 0$ | 1 mark | B1 |

State or use correct tabular two-tail $r$-value:
$r_{8,5\%} = 0.707$ | 1 mark | B1* |

State or imply valid method for conclusion:
Reject $H_0$ if $|r| >$ tab. value (AEF) | 1 mark | M1 |

Correct conclusion: There is non-zero correlation | 1 mark | DA1 | (AEF, dep A1*, B1*)

**Total: 6 marks**

---

## Question 10E:

Find MI of lamina about $Q$:
$I_{\text{lamina}} = \frac{1}{3}m\{(3a)^2 + (3a/2)^2\} + m(9a/2)^2$ | 2 marks | M1 A1 | $[= (15/4 + 81/4)ma^2 = 24ma^2]$

State or find MI of rod about $Q$:
$I_{\text{rod}} = (\frac{1}{3}+1)M(3a/2)^2\ [= 3Ma^2]$ | 1 mark | B1 |

Sum to find MI of object about $Q$:
$I_1 = 24ma^2 + 3Ma^2 = 3(8m+M)a^2$ | 1 mark | A1 | A.G.

Find MI of object about mid-point of $PQ$:
$I_2 = (15/4 + 3^2)ma^2 + \frac{1}{3}M(3a/2)^2$
$= (51/4)ma^2 + \frac{3}{4}Ma^2 = \frac{3}{4}(17m+M)a^2$ | 2 marks | M1 A1 | A.G.

Use equation of circular motion to find $\frac{d^2\theta}{dt^2}$ for axis $l_1$:
$[-]I_1\frac{d^2\theta}{dt^2} = mg \times (9a/2)\sin\theta + Mg \times (3a/2)\sin\theta$
$[= (9m/2 + 3M/2)ga\sin\theta]$ | 2 marks | M1 A1 |

Approximate $\sin\theta$ by $\theta$ and find $\omega_1^2$ in SHM equation:
$\omega_1^2 = (3m+M)g/2(8m+M)a$ | 1 mark | M1 |

Find period $T_1$ for axis $l_1$ from $2\pi/\omega_1$:
$T_1 = 2\pi\sqrt{\{2(8m+M)a/(3m+M)g\}}$ | 1 mark | A1 | (AEF)

Use equation of circular motion to find $\frac{d^2\theta}{dt^2}$ for axis $l_2$:
$[-]I_2\frac{d^2\theta}{dt^2} = mg \times 3a\sin\theta$ | 1 mark | M1 |

Approximate $\sin\theta$ by $\theta$ and find $\omega_2^2$ in SHM equation:
$\omega_2^2 = 4mg/(17m+M)a$ | 1 mark | M1 |

Find period $T_2$ for axis $l_2$ from $2\pi/\omega_2$:
$T_2 = 2\pi\sqrt{\{(17m+M)a/4mg\}}$ | 1 mark | A1 | (AEF)

Verify that $T_1 = T_2$ when $m = M$ (AEF):
$T_1 = 2\pi\sqrt{(18a/4g)} = T_2$ | 1 mark | B1 | [Taking $m=M$ throughout 2nd part can earn: M1 A1 M1 A0 M1 M1 A0 B1 (max 6/8)]

**Total: 8 marks**

---

## Question 10O:

State hypotheses (B0 for $\bar{x}\ldots$):
$H_0: \mu_X = \mu_Y$, $H_1: \mu_X \neq \mu_Y$ | 1 mark | B1 |

State assumption: Distributions have equal variances | 1 mark | B1 | (AEF)

Find sample mean and estimate population variances:
$\bar{x} = 4.2$, $\bar{y} = 4.8$
$s_X^2 = (180 - 42^2/10)/9 = 0.4$ or $0.6325^2$
$s_Y^2 = (281.5 - 57.6^2/12)/11 = 0.4564$ or $251/550$ or $0.6755^2$ | 1 mark | M1 | (allow biased: $0.36$ or $0.6^2$); (allow biased: $0.4183$ or $0.6468^2$)

Estimate (pooled) common variance:
$s^2 = (9s_X^2 + 11s_Y^2)/20$
or $(180 - 42^2/10 + 281.5 - 57.6^2/12)/20 = 0.431$ or $0.6565^2$ | 2 marks | M1 A1 | (AEF) (note $s_X^2$ and $s_Y^2$ not needed explicitly)

Calculate value of $t$ (to 3 s.f.):
$[-]t = (\bar{y}-\bar{x})/s\sqrt{(1/10+1/12)} = 2.13$ | 2 marks | M1 A1 |

State or use correct tabular $t$ value:
$t_{20,\,0.975} = 2.086$ [allow 2.09] | 1 mark | B1* | (or can compare $\bar{y}-\bar{x} = 0.6$ with $0.586$)

Correct conclusion:
$t > 2.09$ so mean masses not same | 1 mark | DB1$\checkmark$ | (AEF, $\checkmark$ on $t$, dep *B1)

**S.R.** Implicitly taking $s_X^2$, $s_Y^2$ as population variances:
$z = (\bar{y}-\bar{x})/\sqrt{(s_X^2/10 + s_Y^2/12)} = 0.6/\sqrt{0.078} = 2.15$ | 1 mark | (B1) | (may also earn first B1 B1 M1)

**Total: 9 marks**

Comparison with $z_{0.975}$ and conclusion:
$2.15 > 1.96$ so mean masses not same | 1 mark | (B1) | (can earn at most 5/9)

State hypotheses (B0 for $\bar{x}\ldots$):
$H_0: \mu_X = 3.8$, $H_1: \mu_X > 3.8$
or $H_0: \mu_X = \mu_Z$, $H_1: \mu_X > \mu_Z$ | 1 mark | B1 |

Calculate value of $t$ using $s_X$ from above:
$t = (4.2-3.8)/(s_X/\sqrt{10}) = 2.0$ | 2 marks | M1 A1 |

State or use correct tabular $t$ value:
$t_{9,\,0.95} = 1.833$ [allow 1.83] | 1 mark | B1* | (or can compare $0.4$ with $0.367$)

Correct conclusion:
$t > 1.833$, so claim is justified
or mean mass of Royals $>$ mean mass of Crowns | 1 mark | DB1$\checkmark$ | (A.E.F., $\checkmark$ on $t$, dep *B1)

**Total: 5 marks**