| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed hypothesis test (upper tail, H₁: p > p₀) |
| Difficulty | Moderate -0.3 This is a straightforward one-tailed binomial hypothesis test with clearly stated hypotheses (H₀: p=0.05, H₁: p>0.05), small sample size making calculations manageable, and standard 5% significance level. Students need to calculate P(X≥4) under H₀ and compare to 0.05, which is routine S1 content requiring only basic binomial probability calculations and standard test procedure. |
| Spec | 2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| If 95% seen anywhere | ||
| \(P(X \leq 3)\) | B1 | |
| 0.9891 | B1 | |
| Comparison with 95% | M1* | dep on B1 |
| Significant oe | A1* | |
| Conclusion | E1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Smallest CR that 4 could fall into is \(\{4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\}\) | B1 | |
| Size 0.0109 | B1 | |
| This is \(< 5\%\) | M1*, A1*, E1* | as per scheme; marks only awarded if 4 used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X \geq 3) = 1 - P(X \leq 2) > 5\%\) | ||
| \(P(X \geq 4) = 1 - P(X \leq 3) < 5\%\) | ||
| Either \(k=4\) or \(k-1=3\) so \(k=4\) | SC1 | |
| CR is \(\{4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\}\) | SC1 | |
| Conclusion | SC1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1 - 0.9891 < 5\%\) or \(0.0109 < 5\%\) | B0B1M1 | |
| \(P(X \leq k-1) = P(X \leq 3)\) so \(k-1=3\), \(k=4\) (or just \(k=8\)) | ||
| CR is \(\{4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\}\) and conclusion | A1E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Hypotheses | max B1B1B1 | |
| If compare with \(5\%\) | ignore lower tail; mark upper tail as per scheme; withhold A1E1 | |
| If compare with \(2.5\%\) | B0B0M0A0E0 | no marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Squiggly line between 3 and 4 (or on 4 exclusively) | B1 | |
| Arrow pointing to right | B1dep | |
| 0.0109 seen on diagram | M1 | from squiggly line or from 4 |
| Correct conclusion | A1E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: p=0.95\), \(H_1: p < 0.95\) where \(p\) = prob frame is faulty | B1B1B1 | |
| \(P(X \leq 14) = 0.0109 < 5\%\), so significant | B1B1M1A1E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| If \(H_0: p=0.5\), \(H_1: p>0.5\) seen but revert to 0.05 | allow marks for correct subsequent working | |
| If 0.5 used consistently | max B1 | for definition of \(p\); possibly B1 for notation \(P(X \geq 4)\) |
# Question 5 (Additional Notes):
## Comparison with 95% Method
| Answer/Working | Mark | Guidance |
|---|---|---|
| If 95% seen anywhere | | |
| $P(X \leq 3)$ | B1 | |
| 0.9891 | B1 | |
| Comparison with 95% | M1* | dep on B1 |
| Significant oe | A1* | |
| Conclusion | E1* | |
## Smallest Critical Region Method
| Answer/Working | Mark | Guidance |
|---|---|---|
| Smallest CR that 4 could fall into is $\{4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\}$ | B1 | |
| Size 0.0109 | B1 | |
| This is $< 5\%$ | M1*, A1*, E1* | as per scheme; marks only awarded if 4 used |
## Use of $k$ Method (No Probabilities Quoted)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X \geq 3) = 1 - P(X \leq 2) > 5\%$ | | |
| $P(X \geq 4) = 1 - P(X \leq 3) < 5\%$ | | |
| Either $k=4$ or $k-1=3$ so $k=4$ | SC1 | |
| CR is $\{4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\}$ | SC1 | |
| Conclusion | SC1 | |
## Use of $k$ Method (One Probability Quoted)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 - 0.9891 < 5\%$ or $0.0109 < 5\%$ | B0B1M1 | |
| $P(X \leq k-1) = P(X \leq 3)$ so $k-1=3$, $k=4$ (or just $k=8$) | | |
| CR is $\{4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\}$ and conclusion | A1E1 | |
## Two-Tailed Test (Correct $H_1: p > 0.05$)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Hypotheses | max B1B1B1 | |
| If compare with $5\%$ | | ignore lower tail; mark upper tail as per scheme; withhold A1E1 |
| If compare with $2.5\%$ | B0B0M0A0E0 | no marks |
## Line Diagram Method
| Answer/Working | Mark | Guidance |
|---|---|---|
| Squiggly line between 3 and 4 (or on 4 exclusively) | B1 | |
| Arrow pointing to right | B1dep | |
| 0.0109 seen on diagram | M1 | from squiggly line or from 4 |
| Correct conclusion | A1E1 | |
## Using $P(\text{Not Faulty})$ Method
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p=0.95$, $H_1: p < 0.95$ where $p$ = prob frame is faulty | B1B1B1 | |
| $P(X \leq 14) = 0.0109 < 5\%$, so significant | B1B1M1A1E1 | |
## NB
| Answer/Working | Mark | Guidance |
|---|---|---|
| If $H_0: p=0.5$, $H_1: p>0.5$ seen but revert to 0.05 | | allow marks for correct subsequent working |
| If 0.5 used consistently | max B1 | for definition of $p$; possibly B1 for notation $P(X \geq 4)$ |5 A manufacturer produces titanium bicycle frames. The bicycle frames are tested before use and on average $5 \%$ of them are found to be faulty. A cheaper manufacturing process is introduced and the manufacturer wishes to check whether the proportion of faulty bicycle frames has increased. A random sample of 18 bicycle frames is selected and it is found that 4 of them are faulty. Carry out a hypothesis test at the $5 \%$ significance level to investigate whether the proportion of faulty bicycle frames has increased.
\hfill \mbox{\textit{OCR MEI S1 2012 Q5 [8]}}