| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| 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 with clear parameters (n=8, p=0.36, significance level 5%). Part (i) requires standard procedure: stating hypotheses, calculating P(X≥7), and comparing to 5%. Part (ii) tests understanding of Type II error definition and requires calculating β, which is slightly beyond routine but still mechanical. The small sample size makes calculations manageable, and the question structure is typical for S2 level with no novel problem-solving 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 proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(H_0 : p = 0.36\), \(H_1 : p > 0.36\) | B1 | Both hypotheses correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum P = 0.00429 < 0.05\) | M1, A1, M1 | Evaluating P(7) or P(8); Correct answer for both; Comparing their prob sum to 0.05 or |
| Accept driving instructor's claim | B1 | Correct conclusion two no contradictions |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Type I error}) = 0.0292\) or \(0.0293\) | B1, M1, B1, A1 | Correct answer; Evaluating P(6); Correct P(5) and showing this is not in the CR either by \(\sum P > 0.05\) or P(5) > 0.05; Correct answer. NB Marks for part (ii) may be awarded in part (i) but not vice versa. |
(i) $H_0 : p = 0.36$, $H_1 : p > 0.36$ | B1 | Both hypotheses correct
$P(7) = {}^8C_7 \times (0.36)^7 (0.64)^1 = 0.00401$
$P(8) = (0.36)^8 = 0.000282$
$\sum P = 0.00429 < 0.05$ | M1, A1, M1 | Evaluating P(7) or P(8); Correct answer for both; Comparing their prob sum to 0.05 or
Accept driving instructor's claim | B1 | Correct conclusion two no contradictions
**[5]**
(ii) Type I error: $P(6) = {}^8C_6 \times (0.36)^6 (0.64)^2 = 0.02496$
$P(5) = {}^8C_5 (0.36)^5(0.64)^3 = 0.08876, > 0.05$
$P(\text{Type I error}) = 0.0292$ or $0.0293$ | B1, M1, B1, A1 | Correct answer; Evaluating P(6); Correct P(5) and showing this is not in the CR either by $\sum P > 0.05$ or P(5) > 0.05; Correct answer. NB Marks for part (ii) may be awarded in part (i) but not vice versa.
**[4]**4 In a certain city it is necessary to pass a driving test in order to be allowed to drive a car. The probability of passing the driving test at the first attempt is 0.36 on average. A particular driving instructor claims that the probability of his pupils passing at the first attempt is higher than 0.36 . A random sample of 8 of his pupils showed that 7 passed at the first attempt.\\
(i) Carry out an appropriate hypothesis test to test the driving instructor's claim, using a significance level of $5 \%$.\\
(ii) In fact, most of this random sample happened to be careful and sensible drivers. State which type of error in the hypothesis test (Type I or Type II) could have been made in these circumstances and find the probability of this type of error when a sample of size 8 is used for the test.
\hfill \mbox{\textit{CAIE S2 2009 Q4 [9]}}