| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Find sample size for test |
| Difficulty | Standard +0.3 This is a straightforward binomial hypothesis test with standard procedures. Part (i) requires routine application of the binomial test at a given significance level. Part (ii) adds mild problem-solving by requiring students to work backwards to find the critical sample size, but this is a predictable extension that follows directly from understanding the test mechanics. The question is slightly above average difficulty due to the reverse-engineering in part (ii), but remains well within standard S2 territory. |
| Spec | 2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(H_0: p = 0.35\) | B1 | Each hypothesis correct, B1+B1, allow p: .35 if .35 used |
| \(H_1: p < 0.35\) | B1 | |
| \(B(14, 0.35)\) | M1 | Correct distribution stated or implied, can be implied by \(N(4.9, \ldots)\), but not \(\text{Po}(4.9)\) |
| \(\alpha: P(≤ 5) = 0.0839 > 0.025\) | A1 | 0.0839 seen, or \(P(≤ 1) = 0.0205\) if clearly using CR |
| \(\beta: CR ≤ 1\), probability 0.0205 | B1 | Compare binomial tail with 0.025, or = 2 binomial CR |
| Do not reject \(H_0\). Insufficient evidence that proportion that can receive Channel C is less than 35%. | M1 | Do not reject \(H_0\), √ on their probability, not from N or Po or \(P(< 2)\); Contextualised conclusion √ |
| (ii) \(B(8, 0.35): P(0) = 0.0319\) | M1 | Attempt to find \(P(0)\) from \(B(n, 0.35)\) |
| A1 | One correct probability \([\text{Po}(2) = .0236, n = 18: M1A1]\) | |
| A1 | 4 | Both probabilities correct |
| or | \(0.65^n > 0.025; n \text{ in } 0.65 > \ln 0.025\) | |
| A1A1 | In range [8.5, 8.6]; answer 8 or ≤ 8 only | |
| \(8.56; \text{ largest value of } n = \mathbf{8}\) |
(i) $H_0: p = 0.35$ | B1 | Each hypothesis correct, B1+B1, allow p: .35 if .35 used
$H_1: p < 0.35$ | B1 |
$B(14, 0.35)$ | M1 | Correct distribution stated or implied, can be implied by $N(4.9, \ldots)$, but not $\text{Po}(4.9)$
$\alpha: P(≤ 5) = 0.0839 > 0.025$ | A1 | 0.0839 seen, or $P(≤ 1) = 0.0205$ if clearly using CR
$\beta: CR ≤ 1$, probability 0.0205 | B1 | Compare binomial tail with 0.025, or = 2 binomial CR
Do not reject $H_0$. Insufficient evidence that proportion that can receive Channel C is less than 35%. | M1 | Do not reject $H_0$, √ on their probability, not from N or Po or $P(< 2)$; Contextualised conclusion √
(ii) $B(8, 0.35): P(0) = 0.0319$ | M1 | Attempt to find $P(0)$ from $B(n, 0.35)$
| | | A1 | One correct probability $[\text{Po}(2) = .0236, n = 18: M1A1]$
| | | A1 | 4 | Both probabilities correct
or | | $0.65^n > 0.025; n \text{ in } 0.65 > \ln 0.025$ | MIMI | Answer 8 or $≤ 8$ only, needs minimum M1A1
| | | A1A1 | In range [8.5, 8.6]; answer 8 or ≤ 8 only
| | $8.56; \text{ largest value of } n = \mathbf{8}$ | |7 A television company believes that the proportion of households that can receive Channel C is 0.35 .\\
(i) In a random sample of 14 households it is found that 2 can receive Channel C. Test, at the $2.5 \%$ significance level, whether there is evidence that the proportion of households that can receive Channel C is less than 0.35.\\
(ii) On another occasion the test is carried out again, with the same hypotheses and significance level as in part (i), but using a new sample, of size $n$. It is found that no members of the sample can receive Channel C. Find the largest value of $n$ for which the null hypothesis is not rejected. Show all relevant working.
\hfill \mbox{\textit{OCR S2 2007 Q7 [11]}}