OCR S2 — Question 3 7 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Marks7
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Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeTwo-tailed test critical region
DifficultyModerate -0.3 This is a straightforward binomial hypothesis test with clear parameters (n=8, x=5, p=0.3) requiring standard procedure: state hypotheses, calculate P(X≥5) under H₀, compare to 5% significance level. The calculation is routine with small numbers, though students must recognize it's actually a two-tailed test despite the question saying 'not 30%'. Slightly easier than average due to simple arithmetic and standard method application.
Spec2.05b Hypothesis test for binomial proportion

3 The manufacturers of a brand of chocolates claim that, on average, \(30 \%\) of their chocolates have hard centres. In a random sample of 8 chocolates from this manufacturer, 5 had hard centres. Test, at the \(5 \%\) significance level, whether there is evidence that the population proportion of chocolates with hard centres is not \(30 \%\), stating your hypotheses clearly. Show the values of any relevant probabilities.

Answer: \(H_0: p = 0.3\)
\(H_1: p \neq 0.3\)
\(\text{B}(8, 0.3)\)
\(\text{P}(\leq 4) = 0.9420\); \(\text{P}(> 4) = 0.0580\)
\(\text{P}(\leq 5) = 0.9887\); \(\text{P}(> 5) = 0.0113\)
Compare 0.025 or critical value 6
Do not reject \(H_0\)
AnswerMarks Guidance
Insufficient evidence that manufacturer's claim is wrongMarks: B1 B1 M1 A1 M1 M1 A1 A1 Guidance: NH stated, must be this form (or \(\pi\)). AH stated, must be this form (or \(\pi\)) [\(\mu\): B1 both]. \(\text{B}(8, 0.3)\) stated or implied. Any one of these four probabilities seen. Either compare P(≤ 5) & 0.025 / P(≤ 4) & 0.975 OR critical region ≥ 6 with 5. \(H_0\) not rejected, can be implied, needs essentially correct method. Correct conclusion in context. [SR: Normal, Poisson: can get B2M1A0M0M1A1; P(≤ 5): first 4 marks. P(= 5): first 3 marks only.]
**Answer:** $H_0: p = 0.3$
$H_1: p \neq 0.3$
$\text{B}(8, 0.3)$
$\text{P}(\leq 4) = 0.9420$; $\text{P}(> 4) = 0.0580$
$\text{P}(\leq 5) = 0.9887$; $\text{P}(> 5) = 0.0113$
Compare 0.025 or critical value 6
Do not reject $H_0$
Insufficient evidence that manufacturer's claim is wrong | **Marks:** B1 B1 M1 A1 M1 M1 A1 A1 | **Guidance:** NH stated, must be this form (or $\pi$). AH stated, must be this form (or $\pi$) [$\mu$: B1 both]. $\text{B}(8, 0.3)$ stated or implied. Any one of these four probabilities seen. Either compare P(≤ 5) & 0.025 / P(≤ 4) & 0.975 OR critical region ≥ 6 with 5. $H_0$ not rejected, can be implied, needs essentially correct method. Correct conclusion in context. [SR: Normal, Poisson: can get B2M1A0M0M1A1; P(≤ 5): first 4 marks. P(= 5): first 3 marks only.]
3 The manufacturers of a brand of chocolates claim that, on average, $30 \%$ of their chocolates have hard centres. In a random sample of 8 chocolates from this manufacturer, 5 had hard centres. Test, at the $5 \%$ significance level, whether there is evidence that the population proportion of chocolates with hard centres is not $30 \%$, stating your hypotheses clearly. Show the values of any relevant probabilities.

\hfill \mbox{\textit{OCR S2  Q3 [7]}}