| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Two-tailed test critical region |
| Difficulty | Standard +0.3 This is a standard S2 hypothesis testing question requiring table lookups and basic probability calculations. Part (i) is routine critical region finding, (ii) tests understanding of significance level, (iii) is a straightforward power calculation, and (iv) requires careful enumeration of cases but uses only basic probability rules. All techniques are standard textbook exercises with no novel insight required. |
| Spec | 2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(R \sim B(25, 0.8)\) | B1 | B(25, 0.8) stated or implied, e.g. from N(20, 4) |
| \(P(R \leq 16) = 0.0468, P(R \leq 17) = 0.1091\) | M1 | One relevant probability seen [Normal: M0A0] |
| \(k = 16\) | A1 3 | Answer \(k = 16\) only [SR: unsupported 16, B1M0B1] |
| (ii) \(20p = 0.9936\) | M1 | \(20 \times\) their \(p\) or \(20 \times 0.05\) |
| A1 2 | Answer, a.r.t. 0.936, 1.S.M. | |
| (iii) \(P(R \leq 16 | p = 0.6) = 0.7265\) | M1 |
| A1 2 | Answer 0.7265 or 0.727 | |
| (iv) \(\alpha: \quad p' = 0.5 \times 0.0468 + 0.5 \times 0.7265 = 0.38665\) | M1 | "Tree diagram" probability, any sensible \(p\) |
| \(2 \times p' \times (1 - p') = 0.474\) | M1 | Value in range [0.38, 0.39] |
| A1 4 | Correct formula, including 2, any \(p'\) (with 0.47, 0.48) | |
| Answer (0.47, 0.48) | ||
| or \(\beta: \quad 0.8 A \quad 0.8 R \quad .5^2 \times .9532 \times .0468 = .0112\) | M1 | \(p_1q_2 + p_2q_1\), etc (0.5 not needed) |
| \(0.8 R \quad 0.8 A \quad .5^2 \times .0468 \times .9532 = .0112\) | A1 | 4 cases, √ on their ps and qs, 0.5 not needed |
| \(0.6 A \quad 0.8 R \quad .5^2 \times .2735 \times .0468 = .0032\) | M1 | e.g. 2p₁q₂ + p₂q₁) |
| \(0.6 R \quad 0.8 A \quad .5^2 \times 2265 \times .9532 = .1731\) | Completely correct list of cases and probabilities, | |
| \(0.8 A \quad 0.6 R \quad .5^2 \times .0468 \times .2735 = .0032\) | including 0.5 | |
| \(0.6 R \quad 0.6 A \quad .5^2 \times .2735 \times .7265 = .0497\) | A1 | Answer in range [0.47, 0.48] |
| \(0.6 A \quad 0.6 R \quad .5^2 \times .7265 \times .2735 = .0497\) |
**(i)** $R \sim B(25, 0.8)$ | B1 | B(25, 0.8) stated or implied, e.g. from N(20, 4)
$P(R \leq 16) = 0.0468, P(R \leq 17) = 0.1091$ | M1 | One relevant probability seen [Normal: M0A0]
$k = 16$ | A1 3 | Answer $k = 16$ only [SR: unsupported 16, B1M0B1]
**(ii)** $20p = 0.9936$ | M1 | $20 \times$ their $p$ or $20 \times 0.05$
| A1 2 | Answer, a.r.t. 0.936, 1.S.M.
**(iii)** $P(R \leq 16 | p = 0.6) = 0.7265$ | M1 | Find $P(R \leq k | p = 0.6)$
| A1 2 | Answer 0.7265 or 0.727
**(iv)** $\alpha: \quad p' = 0.5 \times 0.0468 + 0.5 \times 0.7265 = 0.38665$ | M1 | "Tree diagram" probability, any sensible $p$
$2 \times p' \times (1 - p') = 0.474$ | M1 | Value in range [0.38, 0.39]
| A1 4 | Correct formula, including 2, any $p'$ (with 0.47, 0.48)
| | Answer (0.47, 0.48)
or $\beta: \quad 0.8 A \quad 0.8 R \quad .5^2 \times .9532 \times .0468 = .0112$ | M1 | $p_1q_2 + p_2q_1$, etc (0.5 not needed)
$0.8 R \quad 0.8 A \quad .5^2 \times .0468 \times .9532 = .0112$ | A1 | 4 cases, √ on their ps and qs, 0.5 not needed
$0.6 A \quad 0.8 R \quad .5^2 \times .2735 \times .0468 = .0032$ | M1 | e.g. 2p₁q₂ + p₂q₁)
$0.6 R \quad 0.8 A \quad .5^2 \times 2265 \times .9532 = .1731$ | | Completely correct list of cases and probabilities,
$0.8 A \quad 0.6 R \quad .5^2 \times .0468 \times .2735 = .0032$ | | including 0.5
$0.6 R \quad 0.6 A \quad .5^2 \times .2735 \times .7265 = .0497$ | A1 | Answer in range [0.47, 0.48]
$0.6 A \quad 0.6 R \quad .5^2 \times .7265 \times .2735 = .0497$ |6 A factory makes chocolates of different types. The proportion of milk chocolates made on any day is denoted by $p$. It is desired to test the null hypothesis $\mathrm { H } _ { 0 } : p = 0.8$ against the alternative hypothesis $\mathrm { H } _ { 1 } : p < 0.8$. The test consists of choosing a random sample of 25 chocolates. $\mathrm { H } _ { 0 }$ is rejected if the number of milk chocolates is $k$ or fewer. The test is carried out at a significance level as close to $5 \%$ as possible.\\
(i) Use tables to find the value of $k$, giving the values of any relevant probabilities.\\
(ii) The test is carried out 20 times, and each time the value of $p$ is 0.8 . Each of the tests is independent of all the others. State the expected number of times that the test will result in rejection of the null hypothesis.\\
(iii) The test is carried out once. If in fact the value of $p$ is 0.6 , find the probability of rejecting $\mathrm { H } _ { 0 }$.\\
(iv) The test is carried out twice. Each time the value of $p$ is equally likely to be 0.8 or 0.6 . Find the probability that exactly one of the two tests results in rejection of the null hypothesis.
\hfill \mbox{\textit{OCR S2 2005 Q6 [11]}}