| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| 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 application of binomial probability and hypothesis testing with standard calculations. Part (i) involves routine binomial probability computations and expectation. Parts (ii)-(iii) require setting up a one-tailed test and finding a critical region, which are standard S1 procedures with no novel problem-solving required. Slightly above average difficulty due to the hypothesis testing component, but all steps are textbook-standard. |
| 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 |
|---|---|---|
| Or from tables: \(0.6482 - 0.3980 = 0.2502\) | M1 \(0.2^3 \times 0.8^{12}\), M1 \(\binom{15}{3}\times p^3 q^{12}\), A1 CAO / OR: M2 for \(0.6482-0.3980\), A1 CAO | 3 marks |
| (B) \(P(X \geq 3) = 1 - 0.3980 = 0.6020\) | M1 \(P(X\leq2)\), M1 \(1-P(X\leq2)\), A1 CAO | 2 marks *(note: marked as 3 then 2)* |
| (C) \(E(X) = np = 15 \times 0.2 = 3.0\) | M1 for product, A1 CAO |
| Answer | Marks | Guidance |
|---|---|---|
| (B) \(H_1\) has this form as the proportion who eat at least 5 a day is expected to increase | B1 definition of \(p\) in context, B1 \(H_0\), B1 \(H_1\), E1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| 7 lies in the critical region, so we reject null hypothesis and conclude that there is evidence to suggest that the proportion who eat at least five a day has increased. | B1 for 0.1642, B1 for 0.0611, M1 for at least one comparison with 10%, A1 CAO for critical region dep on M1 and at least one B1 / M1 dep comparison, A1 dep decision, A1 conclusion in context | Total: 18 marks |
# Question 8:
## Part (i):
$X \sim B(15,\ 0.2)$
**(A)** $P(X=3) = \binom{15}{3} \times 0.2^3 \times 0.8^{12} = 0.2501$
Or from tables: $0.6482 - 0.3980 = 0.2502$ | M1 $0.2^3 \times 0.8^{12}$, M1 $\binom{15}{3}\times p^3 q^{12}$, A1 CAO / OR: M2 for $0.6482-0.3980$, A1 CAO | **3 marks**
**(B)** $P(X \geq 3) = 1 - 0.3980 = 0.6020$ | M1 $P(X\leq2)$, M1 $1-P(X\leq2)$, A1 CAO | **2 marks** *(note: marked as 3 then 2)*
**(C)** $E(X) = np = 15 \times 0.2 = 3.0$ | M1 for product, A1 CAO |
## Part (ii):
(A) Let $p$ = probability of a randomly selected child eating at least 5 a day
$H_0: p = 0.2$
$H_1: p > 0.2$
(B) $H_1$ has this form as the proportion who eat at least 5 a day is expected to increase | B1 definition of $p$ in context, B1 $H_0$, B1 $H_1$, E1 | **4 marks**
## Part (iii):
Let $X \sim B(15,\ 0.2)$
$P(X \geq 5) = 1 - P(X \leq 4) = 1 - 0.8358 = 0.1642 > 10\%$
$P(X \geq 6) = 1 - P(X \leq 5) = 1 - 0.9389 = 0.0611 < 10\%$
So critical region is $\{6,7,8,9,10,11,12,13,14,15\}$
7 lies in the critical region, so we reject null hypothesis and conclude that there is evidence to suggest that the proportion who eat at least five a day has increased. | B1 for 0.1642, B1 for 0.0611, M1 for at least one comparison with 10%, A1 CAO for critical region dep on M1 and at least one B1 / M1 dep comparison, A1 dep decision, A1 conclusion **in context** | **Total: 18 marks**8 The Department of Health 'eat five a day' advice recommends that people should eat at least five portions of fruit and vegetables per day. In a particular school, $20 \%$ of pupils eat at least five a day.
\begin{enumerate}[label=(\roman*)]
\item 15 children are selected at random.\\
(A) Find the probability that exactly 3 of them eat at least five a day.\\
(B) Find the probability that at least 3 of them eat at least five a day.\\
(C) Find the expected number who eat at least five a day.
A programme is introduced to encourage children to eat more portions of fruit and vegetables per day. At the end of this programme, the diets of a random sample of 15 children are analysed. A hypothesis test is carried out to examine whether the proportion of children in the school who eat at least five a day has increased.
\item (A) Write down suitable null and alternative hypotheses for the test.\\
(B) Give a reason for your choice of the alternative hypothesis.
\item Find the critical region for the test at the $10 \%$ significance level, showing all of your calculations. Hence complete the test, given that 7 of the 15 children eat at least five a day.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2009 Q8 [18]}}