| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed hypothesis test (lower tail, H₁: p < p₀) |
| Difficulty | Standard +0.3 This is a straightforward one-tailed binomial hypothesis test at a standard significance level with clear parameters (n=20, p=0.25), requiring only calculation of P(X≤1) and comparison to 0.025. Part (b) tests understanding of binomial assumptions (independence), which is a standard conceptual point. The question is slightly easier than average as it's a textbook application with no unusual complications. |
| 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 |
|---|---|---|
| \(H_0: p = 0.25\) where \(p = P(\text{a packet contains gift})\) | B1 (1.1) | or \(p =\) proportion of packets containing gift |
| \(H_1: p < 0.25\) | B1 (2.5) | One error e.g. undefined \(p\): B1B0 |
| \(B(20, 0.25)\) & \(X = 1\) | M1 (3.3) | soi |
| \(P(X \leq 1) = 0.0243\) | A1 (3.4) | Condone \(P(X=1) = 0.0243\) but not \(P(X=1) = 0.0211\) or other incorrect |
| comp \(0.025\) | A1 (1.1) | dep \(0.0243\) and \(0.025\) |
| Reject \(H_0\) | M1 (1.1) | Allow e.g. "\(H_0\) is incorrect". Dep \(0.0243\) or \(P(X \leq 1)\) stated or \(0.0211\). Can be implied by correct conclusion as for A1 below |
| Sufficient evidence that proportion containing gift is less than \(0.25\) | A1 (2.2b) | In context, not definite, e.g. not "Proportion is less" |
| Answer | Marks | Guidance |
|---|---|---|
| EITHER: whether a packet contains a free gift is not independent of whether other nearby packets contain the free gift. OR: The probability that a packet contains a gift is not the same for each packet, or The proportion of packets with gifts in each box is not constant. OR: Free gifts not distributed randomly | B1 (3.5b) | Allow: The probability of packet containing a gift is not independent. Explanation in context of why either the independence condition or the constant probability condition is not met. NOT: The number of gifts in each box is not constant |
# Question 10(a):
$H_0: p = 0.25$ where $p = P(\text{a packet contains gift})$ | B1 (1.1) | or $p =$ proportion of packets containing gift
$H_1: p < 0.25$ | B1 (2.5) | One error e.g. undefined $p$: B1B0
$B(20, 0.25)$ & $X = 1$ | M1 (3.3) | soi
$P(X \leq 1) = 0.0243$ | A1 (3.4) | Condone $P(X=1) = 0.0243$ but not $P(X=1) = 0.0211$ or other incorrect
comp $0.025$ | A1 (1.1) | dep $0.0243$ and $0.025$
Reject $H_0$ | M1 (1.1) | Allow e.g. "$H_0$ is incorrect". Dep $0.0243$ or $P(X \leq 1)$ stated or $0.0211$. Can be implied by correct conclusion as for A1 below
Sufficient evidence that proportion containing gift is less than $0.25$ | A1 (2.2b) | In context, not definite, e.g. not "Proportion is less"
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# Question 10(b):
EITHER: whether a packet contains a free gift is not independent of whether other nearby packets contain the free gift. OR: The probability that a packet contains a gift is not the same for each packet, or The proportion of packets with gifts in each box is not constant. OR: Free gifts not distributed randomly | B1 (3.5b) | Allow: The probability of packet containing a gift is not independent. Explanation in context of why either the independence condition or the constant probability condition is not met. NOT: The number of gifts in each box is not constant
---10 Some packets of a certain kind of biscuit contain a free gift. The manufacturer claims that the proportion of packets containing a free gift is 1 in 4 . Marisa suspects that this claim is not true, and that the true proportion is less than 1 in 4 . She chooses 20 packets at random and finds that exactly 1 contains the free gift.
\begin{enumerate}[label=(\alph*)]
\item Use a binomial model to test the manufacturer's claim, at the $2.5 \%$ significance level.
The packets are packed in boxes, with each box containing 40 packets. Marisa chooses three boxes at random and finds that one box contains 19 packets with the free gift and the other two boxes contain no packets with the free gift.
\item Give a reason why this suggests that the binomial model used in part (a) may not be appropriate.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q10 [8]}}