## Question 10(i):

| $H_0: \mu_P = \mu_Q,\ H_1: \mu_P > \mu_Q$ | B1 | State hypotheses (B0 for $\bar{x}\ ...$) |
|---|---|---|
| $s_P^2 = (138\,200 - 2620^2/50)/49 = 18.61$ or $912/49$ and $s_Q^2 = (157\,000 - 3060^2/60)/59 = 15.93$ or $940/59$ | M1 A1 | Estimate both population variances (to 3 s.f.) (allow biased: $18.24$ and $15.67$) |
| $s^2 = s_P^2/50 + s_Q^2/60 = 0.6378$ or $0.7986^2$ | M1 A1 | Estimate combined variance (to 3 s.f.; may be implicit) |
| $z_{0.9} = 1.28[2]$ | *B1 | State or use correct tabular $z$ value |
| $z = (\bar{y} - \bar{x})/s = (52.4 - 51)/s = 1.75$, $z >$ tabular value so [accept $H_1$ and] | M1 A1 | Calculate value of $z$ (or $-z$) (or can compare $\bar{y} - \bar{x} = 1.02$ with $1.4$) |
| college $P$ students take more time | $\text{B1}\sqrt{}$ (AEF) | Correctly stated conclusion (FT on $z$, dep *B1) |
| [SC: $s^2 = (912+940)/108 = 17.15$ or $4.141^2$; $z = 1.4/s\sqrt{1/50+1/60} = 1.77$] | | SC: Using pooled estimate of common variance can earn B1 M1 A1 (may be implied) M0 *B1 M0 $\text{B1}\sqrt{}$ (max 5/9) |

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## Question 10(ii):

| $\bar{x} - \bar{y} \pm zs$ (or $\bar{y} - \bar{x} \pm zs$) | M1 | Find confidence interval for difference |
|---|---|---|
| $z_{0.95} = 1.64[5]$ | A1 | Use appropriate tabular value |
| $1.4 \pm 1.31$ or $[0.09,\ 2.71]$, or $-1.4 \pm 1.31$ or $[-2.71,\ -0.09]$ [SC: e.g. $\bar{x} - \bar{y} \pm zs\sqrt{1/50+1/60} = 1.4 \pm 1.30$] | A1 | Evaluate confidence interval (either form). SC: Allow M1 A1 if common variance used (max 2/3) |

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## Question 11A(i):

| $\tfrac{1}{2}mv^2 = \tfrac{1}{2}mu^2 + mga\cos\theta$ and $T = mv^2/a + mg\cos\theta$, $T = mu^2/a + 3mg\cos\theta$ | M1, M1 A1 | Use conservation of energy (A0 if no $m$) and find tension $T$ at $P_1$ by using $F = ma$ (A1 for both correct). Combine to verify $\cos\theta$ |
|---|---|---|
| $(11/5)mg = 2mg/5 + 3mg\cos\theta$, $\cos\theta = 3/5$ | M1 A1 (AG) | By substituting $T = (11/5)mg$ and $u = \sqrt{2ag/5}$ |
| $v^2 = u^2 + 2ag\cos\theta = (2/5 + 6/5)ag = 8ag/5$; $v = \sqrt{8ag/5}$ or $\sqrt{1.6ag}$ or $4\sqrt{a}$ | B1 (AEF) | Find $v$ at $P_1$ (can assume $\cos\theta = 3/5$) |

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## Question 11A(ii):

| $t = (a\sin\theta)/(v\cos\theta) = 4a/3v \quad [= \tfrac{1}{3}\sqrt{a}]$ | M1 A1 | Find time $t$ to $P_2$ by considering horizontal motion |
|---|---|---|
| $h = (v\sin\theta)t + \tfrac{1}{2}gt^2 = 16a/15 + 5a/9 = 73a/45$ or $1.62a$ | M1 A1 | Find height fallen at $P_2$ by considering vertical motion |
| $OP_2 = h + a\cos\theta = 20a/9$ or $2.22a$ | M1 A1 | Find $OP_2$ |

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## Question 11B(i):

| $\sum x = 48,\ \sum y = 29+2p,\ \sum xy = 242+17p,\ \sum x^2 = 450,\ [\sum y^2 = 219+2p^2]$; $S_{xy} = 242+17p - 48(29+2p)/6 = 10+p$; $S_{xx} = 450 - 48^2/6 = 66$; $[S_{yy} = 219+2p^2-(29+2p)^2/6 = (8p^2-116p+473)/6]$ | M1 A1 | Find required values |
|---|---|---|
| $0.25 = S_{xy}/S_{xx} = (10+p)/66,\quad p = 6.5$ | M1 A1 | Find $p$ from gradient in equation of regression line |
| $(29+2p)/6 = 0.25 \times 48/6 + k,\quad k = 7 - 2 = 5$ | M1 A1 | Find $k$ from means and regression line |

## Question 11B(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum y = 42$, $\sum y^2 = 303.5$, $S_{yy} = 303.5 - \frac{42^2}{6} = 9.5$ | M1 | Find correlation coefficient $r$ |
| $r = \frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}} = \frac{16.5}{\sqrt{66 \times 9.5}} = 0.659$ | A1 | |
| | **2** | |

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## Question 11B(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \rho = 0$, $H_1: \rho > 0$ | B1 | State both hypotheses ($B0$ for $r \ldots$) |
| $EITHER:$ $r_{6,\, 5\%} = 0.729$ | *B1 | State or use correct tabular one-tail $r$-value |
| Accept $H_0$ if $\|r\| <$ tab. $r$-value $\quad$ (AEF) | M1 | State or imply valid method for conclusion |
| $OR:$ $t_r = \frac{r\sqrt{(n-2)}}{(1-r^2)} = 1.75$, $t_{4,\,0.95} = 2.132$ | (*B1) | (Rarely seen) |
| Accept $H_0$ if $\|t_r\| <$ tab. $t$-value $\quad$ (AEF) | (M1) | |
| No positive correlation $\quad$ (AEF) | A1 | Correct conclusion (dep \*A1, \*B1) |
| | **4** | |