| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Multiple binomial probability calculations |
| Difficulty | Standard +0.3 This is a straightforward S1 hypothesis testing question with standard binomial calculations. Part (i) involves routine binomial probability computations and expectation. Parts (ii)-(iv) follow a textbook template for one-tailed hypothesis tests: stating hypotheses, finding a critical region using tables, and drawing a conclusion. All steps are procedural with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| (i) (A) \(P(\text{2 faulty tiles}) = \binom{18}{2} \times 0.1^2 \times 0.9^{16} = 0.2835\) | M1 \(0.1^2 \times 0.9^{16}\), M1 \(\binom{18}{2} \times p^2 q^{16}\), A1 CAO | |
| OR from tables \(0.7338 - 0.4503 = 0.2835\) | OR: M2 for \(0.7338 - 0.4503\) A1 CAO | 3 marks |
| (B) \(P(\text{More than 2 faulty tiles}) = 1 - 0.7338 = 0.2662\) | M1 \(P(X \leq 2)\), M1 dep for \(1 - P(X \leq 2)\), A1 CAO | 3 marks |
| (C) \(E(X) = np = 18 \times 0.1 = 1.8\) | M1 for product \(18 \times 0.1\), A1 CAO | 2 marks |
| (ii) (A) Let \(p =\) probability that a randomly selected tile is faulty | B1 for definition of \(p\) in context | |
| \(H_0: p = 0.1\) | B1 for \(H_0\) | |
| \(H_1: p > 0.1\) | B1 for \(H_1\) | 3 marks |
| (B) \(H_1\) has this form as the manufacturer believes that the number of faulty tiles may increase. | E1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \geq 5) = 1 - P(X \leq 4) = 1 - 0.9718 = 0.0282 < 5\%\) | B1 for 0.0982, B1 for 0.0282 | |
| So critical region is \(\{5,6,7,8,9,10,11,12,13,14,15,16,17,18\}\) | M1 for at least one comparison with 5%, A1 CAO for critical region dep on M1 and at least one B1 | 4 marks |
| (iv) 4 does not lie in the critical region, (so there is insufficient evidence to reject the null hypothesis and we conclude that there is not enough evidence to suggest that the number of faulty tiles has increased. | M1 for comparison, A1 for conclusion in context | 2 marks |
**(i) (A)** $P(\text{2 faulty tiles}) = \binom{18}{2} \times 0.1^2 \times 0.9^{16} = 0.2835$ | M1 $0.1^2 \times 0.9^{16}$, M1 $\binom{18}{2} \times p^2 q^{16}$, A1 CAO |
OR from tables $0.7338 - 0.4503 = 0.2835$ | OR: M2 for $0.7338 - 0.4503$ A1 CAO | 3 marks
**(B)** $P(\text{More than 2 faulty tiles}) = 1 - 0.7338 = 0.2662$ | M1 $P(X \leq 2)$, M1 dep for $1 - P(X \leq 2)$, A1 CAO | 3 marks
**(C)** $E(X) = np = 18 \times 0.1 = 1.8$ | M1 for product $18 \times 0.1$, A1 CAO | 2 marks
**(ii) (A)** Let $p =$ probability that a randomly selected tile is faulty | B1 for definition of $p$ in context |
$H_0: p = 0.1$ | B1 for $H_0$ |
$H_1: p > 0.1$ | B1 for $H_1$ | 3 marks
**(B)** $H_1$ has this form as the manufacturer believes that the number of faulty tiles may increase. | E1 | 1 mark
**(iii)** $P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.9018 = 0.0982 > 5\%$
$P(X \geq 5) = 1 - P(X \leq 4) = 1 - 0.9718 = 0.0282 < 5\%$ | B1 for 0.0982, B1 for 0.0282 |
So critical region is $\{5,6,7,8,9,10,11,12,13,14,15,16,17,18\}$ | M1 for at least one comparison with 5%, A1 CAO for critical region dep on M1 and at least one B1 | 4 marks
**(iv)** 4 does not lie in the critical region, (so there is insufficient evidence to reject the null hypothesis and we conclude that there is not enough evidence to suggest that the number of faulty tiles has increased. | M1 for comparison, A1 for conclusion in context | 2 marks
**TOTAL: 18 marks**
---6 A manufacturer produces tiles. On average 10\% of the tiles produced are faulty. Faulty tiles occur randomly and independently.
A random sample of 18 tiles is selected.
\begin{enumerate}[label=(\roman*)]
\item (A) Find the probability that there are exactly 2 faulty tiles in the sample.\\
(B) Find the probability that there are more than 2 faulty tiles in the sample.\\
(C) Find the expected number of faulty tiles in the sample.
A cheaper way of producing the tiles is introduced. The manufacturer believes that this may increase the proportion of faulty tiles. In order to check this, a random sample of 18 tiles produced using the cheaper process is selected and a hypothesis test is carried out.
\item (A) Write down suitable null and alternative hypotheses for the test.\\
(B) Explain why the alternative hypothesis has the form that it does.
\item Find the critical region for the test at the $5 \%$ level, showing all of your calculations.
\item In fact there are 4 faulty tiles in the sample. Complete the test, stating your conclusion clearly.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2010 Q6 [18]}}