| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Calculate Type II error probability |
| Difficulty | Challenging +1.2 This question requires understanding of hypothesis testing terminology (Type I/II errors), finding critical regions using binomial tables, and calculating probabilities under alternative hypotheses. While conceptually more demanding than routine calculations, it follows standard S2 procedures with straightforward binomial probability lookups and calculations, making it moderately above average difficulty. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(0) = 0.5^{10} = 0.000976 = P(10)\) \(P(1) = ^{10}C_1 \times 0.5^1 \times 0.5^9 = 0.00976 = P(9)\) \(\sum 4 \text{ probs} = 0.02147 < 10\%\) \(P(2) = ^{10}C_2 \times 0.5^2 \times 0.5^8 = 0.0439 = P(8)\) \(\sum 6 \text{ probs} = 0.1093 > 10\%\) Values of \(h\) are \(2, 3, 4, 5, 6, 7, 8\) | M1, M1, M1, A1 [4] | Evaluating two Binomial end probs of 0 and 1, (or 10 and 9). Summing either/both ends and comparing to 10% (doesn't have to be two-tailed here). Summing a third/last end probability and comparing to 10% (needn't be two-tailed here). Correct answer only. A1 dep on all 3 Ms. No errors seen. |
| (ii) \(P(\text{Type II error}) = P(2, 3, 4, 5, 6, 7, 8) (= 1 - P(0, 1, 9, 10)) = 1 - \left(\frac{(0.3)^{10} + ^{10}C_1(0.7)(0.3)^9}{+ ^{10}C_9(0.7)^9(0.3) + (0.7)^{10}}\right) = 1 - 0.149 = 0.851\) | B1 ft, M1, A1 [3] | Identifying correct probs B1 their ans (i). Binomial expression for their P(Type II error) with powers summing to 10, \(^{10}C\) sometimes and 0.3 and 0.7 seen. Correct final answer. |
(i) $P(0) = 0.5^{10} = 0.000976 = P(10)$ $P(1) = ^{10}C_1 \times 0.5^1 \times 0.5^9 = 0.00976 = P(9)$ $\sum 4 \text{ probs} = 0.02147 < 10\%$ $P(2) = ^{10}C_2 \times 0.5^2 \times 0.5^8 = 0.0439 = P(8)$ $\sum 6 \text{ probs} = 0.1093 > 10\%$ Values of $h$ are $2, 3, 4, 5, 6, 7, 8$ | M1, M1, M1, A1 [4] | Evaluating two Binomial end probs of 0 and 1, (or 10 and 9). Summing either/both ends and comparing to 10% (doesn't have to be two-tailed here). Summing a third/last end probability and comparing to 10% (needn't be two-tailed here). Correct answer only. A1 dep on all 3 Ms. No errors seen.
(ii) $P(\text{Type II error}) = P(2, 3, 4, 5, 6, 7, 8) (= 1 - P(0, 1, 9, 10)) = 1 - \left(\frac{(0.3)^{10} + ^{10}C_1(0.7)(0.3)^9}{+ ^{10}C_9(0.7)^9(0.3) + (0.7)^{10}}\right) = 1 - 0.149 = 0.851$ | B1 ft, M1, A1 [3] | Identifying correct probs B1 their ans (i). Binomial expression for their P(Type II error) with powers summing to 10, $^{10}C$ sometimes and 0.3 and 0.7 seen. Correct final answer.4 It is not known whether a certain coin is fair or biased. In order to perform a hypothesis test, Raj tosses the coin 10 times and counts the number of heads obtained. The probability of obtaining a head on any throw is denoted by $p$.\\
(i) The null hypothesis is $p = 0.5$. Find the acceptance region for the test, given that the probability of a Type I error is to be at most 0.1 .\\
(ii) Calculate the probability of a Type II error in this test if the actual value of $p$ is 0.7 .
\hfill \mbox{\textit{CAIE S2 2009 Q4 [7]}}