| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Calculate Type I error probability |
| Difficulty | Standard +0.3 This is a straightforward one-tailed binomial hypothesis test requiring standard procedure (state hypotheses, find critical region, compare result, conclude) plus basic understanding of Type I/II errors. The calculations are routine with small numbers (n=18, p=0.1), and part (iii) directly tests definition recall. Slightly above average difficulty due to the multi-part nature and error interpretation, but no novel insight required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion |
| Answer | Marks | Guidance |
|---|---|---|
| Part (i) | ||
| \(H_0\): population proportion = 0.1 oe | B1 | Allow "\(p = 0.1\)" and "\(p > 0.1\)" |
| \(H_1\): population proportion > 0.1 oe | ||
| \(P(X \geq 4) = 1 - P(X \leq 3) = 1 - \left(0.9^{18} + 18 \times 0.9^{17} \times 0.1 + \binom{18}{C_2} \times 0.9^{18} \times 0.1^2 + \binom{18}{C_3} \times 0.9^{15} \times 0.1^3\right)\) | M1 | Allow 1 − (one term omitted or extra or wrong) |
| \(= 0.0982\) (3 sf) | A1 | (note CR method 0.0982 and CR ⩾ 5 for A1) Dep M1M1 no contradictions "Accept \(H_0\)" provided \(H_0\) defined |
| Comp 0.08 | M1 | |
| No evidence that more reach 1m | A1\(\checkmark\) | [5] |
| Part (ii) | ||
| Not rejected \(H_0\) | B1\(\checkmark\) | Ft their (i) |
| Type II | B1dep | [2] If (i) "reject \(H_0\)" then ft gives Type I error |
| \(\checkmark\) | ||
| Part (iii) | ||
| \(P(X \geq 5) (= 0.0282)\) | M1 | Attempt \(P(X \geq 5)\) e.g. '0.0982' − \(^{18}C_4 \times 0.9^{14} \times 0.1^4\) oe. Valid comp of their ⩾ 5 (if CR method used, could be awarded in (i)) |
| \(0.0282 < 0.08\) | B1\(\checkmark\) | |
| \(P(\text{Type I error}) = 0.0282\) (3 sf) | A1 | [3] |
| Total | [10] |
| **Part (i)** |
|---|
| $H_0$: population proportion = 0.1 oe | B1 | Allow "$p = 0.1$" and "$p > 0.1$" |
| $H_1$: population proportion > 0.1 oe | | |
| $P(X \geq 4) = 1 - P(X \leq 3) = 1 - \left(0.9^{18} + 18 \times 0.9^{17} \times 0.1 + \binom{18}{C_2} \times 0.9^{18} \times 0.1^2 + \binom{18}{C_3} \times 0.9^{15} \times 0.1^3\right)$ | M1 | Allow 1 − (one term omitted or extra or wrong) |
| $= 0.0982$ (3 sf) | A1 | (note CR method 0.0982 and CR ⩾ 5 for A1) Dep M1M1 no contradictions "Accept $H_0$" provided $H_0$ defined |
| Comp 0.08 | M1 | |
| No evidence that more reach 1m | A1$\checkmark$ | [5] |
| **Part (ii)** |
|---|
| Not rejected $H_0$ | B1$\checkmark$ | Ft their (i) |
| Type II | B1dep | [2] If (i) "reject $H_0$" then ft gives Type I error |
| | $\checkmark$ | |
| **Part (iii)** |
|---|
| $P(X \geq 5) (= 0.0282)$ | M1 | Attempt $P(X \geq 5)$ e.g. '0.0982' − $^{18}C_4 \times 0.9^{14} \times 0.1^4$ oe. Valid comp of their ⩾ 5 (if CR method used, could be awarded in (i)) |
| $0.0282 < 0.08$ | B1$\checkmark$ | |
| $P(\text{Type I error}) = 0.0282$ (3 sf) | A1 | [3] |
| **Total** | [10] |5 It is known that when seeds of a certain type are planted, on average $10 \%$ of the resulting plants reach a height of 1 metre. A gardener wishes to investigate whether a new fertiliser will increase this proportion. He plants a random sample of 18 seeds of this type, using the fertiliser, and notes how many of the resulting plants reach a height of 1 metre.\\
(i) In fact 4 of the 18 plants reach a height of 1 metre. Carry out a hypothesis test at the $8 \%$ significance level.\\
(ii) Explain which of the errors, Type I or Type II, might have been made in part (i).
Later, the gardener plants another random sample of 18 seeds of this type, using the fertiliser, and again carries out a hypothesis test at the $8 \%$ significance level.\\
(iii) Find the probability of a Type I error.
\hfill \mbox{\textit{CAIE S2 2014 Q5 [10]}}