| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Calculate Type I error probability |
| Difficulty | Standard +0.3 Part (a) is a direct definition question asking for the significance level (Type I error = 5%), requiring only conceptual understanding. Part (b) requires explaining why a sample proportion alone doesn't constitute statistical evidence without considering variability, which is standard hypothesis testing interpretation. Both parts are straightforward applications of basic hypothesis testing concepts with no calculation complexity or novel insight required. |
| Spec | 2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X \sim B(600, 0.02)\) | M1 | soi, eg \(H_0: p = 0.02\) and \(B(600, p)\); Allow \(n = 600\), \(p = 0.02\); May be implied by \(0.0991\) or \(0.0202\) or \(0.9798\) or \(0.9009\) or correct values |
| Attempt \(P(X \geq n)\) for \(17 \leq n \leq 20\) | M1 | |
| \((P(X \geq 18) =)\ 0.0610\) or \(0.061\) (2 sf) | A1 | or \((P(X \leq 17) =)\ 0.939\) or \(0.94\) (2 sf) |
| \((P(X \geq 19) =)\ 0.0359\) or \(0.036\) (2 sf) | A1 | or \((P(X \leq 18) =)\ 0.964\) or \(0.96\) (2 sf); These two probabilities seen imply M1M1A1A1; Condone errors such as \(P(X > 18) = 0.0610\) |
| \(P(\text{concludes claim incorrect}) = 0.0359\) (3 sf) | A1 | Ignore hypotheses and/or "Reject \(H_0\)" or similar |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X \sim N(600 \times 0.02,\ 600 \times 0.02 \times 0.98)\) or \(X \sim N(12, 11.76)\) | M1 | soi. Can be scored either for \(N(12, 11.76)\) or \(B(600, 0.02)\) |
| Attempt \(P(X \geq n)\) for \(17 \leq n \leq 20\) | M1 | |
| \(P(X \geq 17) = 0.0724\) or \(0.072\) (2 sf) | A1 | \(P(x > a) = 0.05 \Rightarrow a = 17.64\) only gets M1 if a probability is calculated |
| \(P(X \geq 18) = 0.0401\) or \(0.040\) (2 sf) | A1 | |
| \(P(\text{concludes claim incorrect}) = 0.0401\) | A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X \sim N(600 \times 0.02,\ 600 \times 0.02 \times 0.98)\) or \(X \sim N(12, 11.76)\) | M1 | soi. Can be scored either for \(N(12, 11.76)\) or \(B(600, 0.02)\) |
| Attempt \(P(X \geq n)\) for \(17 \leq n \leq 20\) | M1 | |
| \(P(X \geq 18) = P(X \geq 17.5) = 0.054\) (2 sf) | A1 | |
| \(P(X \geq 19) = P(X \geq 18.5) = 0.0290\) (2 sf) | A1 | |
| \(P(\text{concludes claim incorrect}) = 0.0290\) | A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Incorrect because: You have to consider \(P(X \geq 18)\), or 18 is in the acceptance region (for 5% test), or critical region is \(\geq 19\), or CV is 19 | B1 | or 18 is under the significance level; Allow "You have to do a proper hypothesis test"; No other answers acceptable |
# Question 12:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim B(600, 0.02)$ | M1 | soi, eg $H_0: p = 0.02$ and $B(600, p)$; Allow $n = 600$, $p = 0.02$; May be implied by $0.0991$ or $0.0202$ or $0.9798$ or $0.9009$ or correct values |
| Attempt $P(X \geq n)$ for $17 \leq n \leq 20$ | M1 | |
| $(P(X \geq 18) =)\ 0.0610$ or $0.061$ (2 sf) | A1 | or $(P(X \leq 17) =)\ 0.939$ or $0.94$ (2 sf) |
| $(P(X \geq 19) =)\ 0.0359$ or $0.036$ (2 sf) | A1 | or $(P(X \leq 18) =)\ 0.964$ or $0.96$ (2 sf); These two probabilities seen imply M1M1A1A1; Condone errors such as $P(X > 18) = 0.0610$ |
| $P(\text{concludes claim incorrect}) = 0.0359$ (3 sf) | A1 | Ignore hypotheses and/or "Reject $H_0$" or similar |
**Unsupported answers:**
- $0.0359$: M1M1A1A0A0
- Critical region is $X \geq 19$: M1M1A1A0A0
## Part (a) ctd — Alternative method (normal with no cc):
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim N(600 \times 0.02,\ 600 \times 0.02 \times 0.98)$ or $X \sim N(12, 11.76)$ | M1 | soi. Can be scored either for $N(12, 11.76)$ or $B(600, 0.02)$ |
| Attempt $P(X \geq n)$ for $17 \leq n \leq 20$ | M1 | |
| $P(X \geq 17) = 0.0724$ or $0.072$ (2 sf) | A1 | $P(x > a) = 0.05 \Rightarrow a = 17.64$ only gets M1 if a probability is calculated |
| $P(X \geq 18) = 0.0401$ or $0.040$ (2 sf) | A1 | |
| $P(\text{concludes claim incorrect}) = 0.0401$ | A0 | |
## Part (a) ctd — Alternative method (normal with cc):
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim N(600 \times 0.02,\ 600 \times 0.02 \times 0.98)$ or $X \sim N(12, 11.76)$ | M1 | soi. Can be scored either for $N(12, 11.76)$ or $B(600, 0.02)$ |
| Attempt $P(X \geq n)$ for $17 \leq n \leq 20$ | M1 | |
| $P(X \geq 18) = P(X \geq 17.5) = 0.054$ (2 sf) | A1 | |
| $P(X \geq 19) = P(X \geq 18.5) = 0.0290$ (2 sf) | A1 | |
| $P(\text{concludes claim incorrect}) = 0.0290$ | A0 | |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Incorrect because: You have to consider $P(X \geq 18)$, or 18 is in the acceptance region (for 5% test), or critical region is $\geq 19$, or CV is 19 | B1 | or 18 is under the significance level; Allow "You have to do a proper hypothesis test"; No other answers acceptable |
---12 A firm claims that no more than $2 \%$ of their packets of sugar are underweight. A market researcher believes that the actual proportion is greater than $2 \%$. In order to test the firm's claim, the researcher weighs a random sample of 600 packets and carries out a hypothesis test, at the $5 \%$ significance level, using the null hypothesis $p = 0.02$.
\begin{enumerate}[label=(\alph*)]
\item Given that the researcher's null hypothesis is correct, determine the probability that the researcher will conclude that the firm's claim is incorrect.
\item The researcher finds that 18 out of the 600 packets are underweight. A colleague says\\
" 18 out of 600 is $3 \%$, so there is evidence that the actual proportion of underweight bags is greater than $2 \%$."
Criticise this statement.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2022 Q12 [6]}}