OCR MEI AS Paper 2 2024 June — Question 12 6 marks

Exam BoardOCR MEI
ModuleAS Paper 2 (AS Paper 2)
Year2024
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeState hypotheses with additional parts
DifficultyModerate -0.8 This is a straightforward hypothesis test question requiring standard procedures: stating H₀ and H₁, calculating a binomial probability using given parameters (n=92, p=0.08, X≤3), and interpreting the result against a 5% significance level. All steps are routine textbook exercises with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure and need for contextual interpretation.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail

12 Data collected in the twentieth century showed that the probability of a randomly selected person having blue eyes was 0.08 . A medical researcher believes that the probability in 2024 is less than this so they decide to carry out a hypothesis test at the \(5 \%\) significance level.
  1. Write down suitable hypotheses for the test, defining the parameter used.
  2. Assuming that the probability that a person selected at random has blue eyes is still 0.08 , calculate the probability that 3 or fewer people in a random sample of 92 people have blue eyes. The researcher collects a random sample of 92 people and finds that 3 of them have blue eyes.
  3. Use your answer to part (b) to carry out the test, giving your conclusion in context.

Question 12:
Part (a):
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0: p = 0.08\) and \(H_1: p < 0.08\)B1 1.1 - Requires both. Can be stated in words. NOTE: \(H_0 = 0.08\) etc is B0
\(p\) is the probability that a person selected at random has blue eyesB1 2.5 - Accept 'proportion' instead of probability but not number/amount etc. Must have highlighted words to score B1
Total: [2]
Part (b):
AnswerMarks Guidance
AnswerMarks Guidance
\(0.057(43\ldots)\) BCB1 1.1 - By calculator - awrt 0.057
Total: [1]
Part (c):
AnswerMarks Guidance
AnswerMarks Guidance
\(0.057 > 0.05\). A CR approach is possible here too where the CR is \(X \leq 2\) then '3 not in CR' etc.M1 3.4 - Their 0.057 correctly compared with 0.05. NOTE: comparing \(p(X=3) = 0.03849\) with 0.05 scores 0 in part (c)
Do not reject \(H_0\) or accept \(H_0\) or reject \(H_1\)M1 1.1 - FT consistent conclusion with their probability
insufficient evidence to suggest that the probability that a person selected at random has blue eyes is less than 0.08 OR insufficient evidence to suggest that the probability that a person selected at random has blue eyes has decreased OR insufficient evidence to support the medical researchers' beliefA1 2.2b - Highlighted words needed. Must be a full contextual conclusion. Accept proportion instead of probability but NOT number/amount. No assertive statements e.g. 'shows that' or 'proves that' etc score A0. Accept 'not enough evidence' for 'insufficient evidence'. The researcher believed the true probability was less than 0.08 in the question so accept this equivalent statement
Total: [3]
## Question 12:

### Part (a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: p = 0.08$ **and** $H_1: p < 0.08$ | B1 | 1.1 - Requires both. Can be stated in words. NOTE: $H_0 = 0.08$ etc is B0 |
| $p$ is the **probability** that a person selected at random has **blue eyes** | B1 | 2.5 - Accept '**proportion**' instead of probability but not number/amount etc. Must have highlighted words to score B1 |

**Total: [2]**

### Part (b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.057(43\ldots)$ **BC** | B1 | 1.1 - By calculator - awrt 0.057 |

**Total: [1]**

### Part (c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.057 > 0.05$. A CR approach is possible here too where the CR is $X \leq 2$ then '3 not in CR' etc. | M1 | 3.4 - Their 0.057 correctly compared with 0.05. NOTE: comparing $p(X=3) = 0.03849$ with 0.05 scores 0 in part (c) |
| Do not reject $H_0$ or accept $H_0$ or reject $H_1$ | M1 | 1.1 - FT consistent conclusion with their probability |
| **insufficient evidence** to **suggest** that the **probability** that a person selected at random has blue eyes is **less than** 0.08 **OR** insufficient evidence to **suggest** that the **probability** that a person selected at random has blue eyes **has decreased** **OR** insufficient evidence to **support the medical researchers' belief** | A1 | 2.2b - Highlighted words needed. Must be a full contextual conclusion. Accept proportion instead of probability but NOT number/amount. No assertive statements e.g. 'shows that' or 'proves that' etc score A0. Accept 'not enough evidence' for 'insufficient evidence'. The researcher believed the true probability was less than 0.08 in the question so accept this equivalent statement |

**Total: [3]**
12 Data collected in the twentieth century showed that the probability of a randomly selected person having blue eyes was 0.08 . A medical researcher believes that the probability in 2024 is less than this so they decide to carry out a hypothesis test at the $5 \%$ significance level.
\begin{enumerate}[label=(\alph*)]
\item Write down suitable hypotheses for the test, defining the parameter used.
\item Assuming that the probability that a person selected at random has blue eyes is still 0.08 , calculate the probability that 3 or fewer people in a random sample of 92 people have blue eyes.

The researcher collects a random sample of 92 people and finds that 3 of them have blue eyes.
\item Use your answer to part (b) to carry out the test, giving your conclusion in context.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI AS Paper 2 2024 Q12 [6]}}