# Question 2:

## Part (i)(A)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $X \sim B(15, 0.85)$ | | |
| $P(\text{exactly 12 germinate}) = \binom{15}{12} \times 0.85^{12} \times 0.15^3$ | M1 | For $0.85^{12} \times 0.15^3$ |
| | M1 | For $\binom{15}{12} \times p^{12} \times q^3$ |
| $= 0.2184$ | A1 | CAO |
| **OR** from tables: $0.3958 - 0.1773 = 0.2185$ | M2, A1 | For $0.3958 - 0.1773$; CAO |

## Part (i)(B)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X < 12) = P(X \leq 11) = 0.1773$ | M1 | For $P(X \leq 11)$ or $P(\leq 11)$ (with no extras). CAO as final answer. May see alternative method: $0.3958 - 0.2185 = 0.1773$; $0.3958 - $ their wrong answer to part (i) scores M1A0 |
| | A1 | |

## Part (ii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $p$ = probability of a seed germinating (for the population) | B1 | For definition of $p$. See below for additional notes |
| $H_0: p = 0.85$ | B1 | For $H_0$ |
| $H_1: p < 0.85$ | B1 | For $H_1$. Dep on $< 0.85$ used in $H_1$. Do not allow just 'Germination rate will be lower' or similar. For use of 0.15 as P(not germinating), contact team leader. E0 for simply stating $H_1$ in words |
| $H_1$ has this form because the test is to investigate whether the proportion of seeds which germinate is lower | E1 | |

## Part (iii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $X \sim B(20, 0.85)$; $P(X \leq 13) = 0.0219$ | M1* | For probability (provided not as part of finding $P(X = 13)$). Ignore notation. No further marks if point probs used — $P(X = 13) = 0.0160$. DO NOT FT wrong $H_1$, but see extra notes |
| $0.0219 > 1\%$ | M1* dep | For comparison. Allow 'accept $H_0$' or 'reject $H_1$'. Must include 'sufficient evidence' or something similar such as 'to suggest that' ie an element of doubt either in the A or E mark. |
| So not enough evidence to reject $H_0$. Not significant. | A1* | For not significant oe |
| Conclude that there is not enough evidence to indicate that the proportion of seeds which have germinated has decreased | E1* dep | For conclusion in context. Must mention decrease, not just change |

**Alternative (Critical Region) Method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Lower tail: $P(X \leq 13) = 0.0219 > 1\%$; $P(X \leq 12) = 0.0059 < 1\%$ | M1 | For either probability. No marks if CR not justified. Condone $\{0,1,2,\ldots,12\}$, $X \leq 12$, oe but not $P(X \leq 12)$ etc. Could get M1A0A1E1 if poor notation for CR. Do not allow just '13 not in CR' — must say 'not significant' or accept $H_0$ or similar |
| So critical region is $\{0,1,2,3,4,5,6,7,8,9,10,11,12\}$ | A1 | cao dep on at least one correct comparison with 1% |
| 13 not in CR so not significant | A1* | |
| There is insufficient evidence to indicate that the proportion of seeds which have germinated has decreased | E1* dep | |

## Part (iv)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $33 < 35$ | M1 | For comparison. Allow '33 lies in the CR'. Must include 'sufficient evidence' or similar such as 'to suggest that'. Do not FT wrong $H_1$: in part (iv) ignore any interchanged $H_0$ and $H_1$ seen in part (ii) |
| So there is sufficient evidence to reject $H_0$ | A1* | If use a calculator to find $P(X \leq 33) = 0.000661$ and compare with 1% then B2 for $P(X \leq 33) = 0.000661 < 0.01$ so reject $H_0$ then final E1 as per scheme |
| Conclude that there is enough evidence to indicate that the proportion of seeds which have germinated has decreased | E1* dep | For conclusion in context. Must mention decrease, not just change |

## Part (v)

| Answer | Marks | Guidance |
|--------|-------|----------|
| For $n = 3$, $P(X \leq 0) = 0.0034 < 0.01$ | M1 | For $P(X \leq 0) = 0.0034$. Allow 0.003 |
| For $n = 2$, $P(X \leq 0) = 0.0225 > 0.01$ | M1 | For $P(X \leq 0) = 0.0225$. Condone just '$n = 3$' for final A mark dep on both M marks |
| So the least value of $n$ for which the critical region is not empty and thus $H_0$ could be rejected is 3 | A1 | CAO. If wrong $H_1$ allow max M2A0 if correct probabilities seen |
| **Alternative using logs:** $0.15^n < 0.01$ | M1 | |
| $n > \log 0.01 / \log 0.15$ | M1 | |
| $n > 2.427$, least $n = 3$ | A1 | |