| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed hypothesis test (lower tail, H₁: p < p₀) |
| Difficulty | Standard +0.3 This is a straightforward application of the sign test with clear structure: part (i) requires simple binomial probability calculations to show impossibility of rejection (routine verification), while part (ii) is a standard one-tailed hypothesis test with given data. Both parts follow textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 5.07b Sign test: and Wilcoxon signed-rank |
| Answer | Marks | Guidance |
|---|---|---|
| 0.0625 > 0.05 so \(H_0\) cannot be rejected | M1, A1 | or 0.9375< 0.95 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0:m = 30, H_1:m < 30\) | B1, B1 | Allow 'data above or below median', Both hypotheses, median or \(m\) May be implied |
| Answer | Marks | Guidance |
|---|---|---|
| \(X \sim B(72, \frac{1}{2})\) | M1, M1 | No, or wrong CC (27.5 or 44.5), -1.886 or -2.003 M0, Any other CC M0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.0385\) or \(0.0386\) | M1, M1, A1 | \(=(-)-1.767\) or \(CV=(-)-1.645\) |
| Answer | Marks | Guidance |
|---|---|---|
| There is sufficient evidence to accept that the median time for Elena's swims is less than 30s | A2, M1, A1ft | Using calculator procedure or -1.767<-1.645 not over-assertive |
| [9] |
**(i)** For $n = 4$ P(X = 0) or P(X = 4) $= 2^{-4}= 0.0625$
0.0625 > 0.05 so $H_0$ cannot be rejected | M1, A1 | or 0.9375< 0.95
| [2] |
**(ii)** Sample of times considered random
$H_0:m = 30, H_1:m < 30$ | B1, B1 | Allow 'data above or below median', Both hypotheses, median or $m$ May be implied
Use sign test
$X \sim B(72, \frac{1}{2})$ | M1, M1 | No, or wrong CC (27.5 or 44.5), -1.886 or -2.003 M0, Any other CC M0
$P(X \leq 28) = $ (from N(36,18))
$\Phi(28.5or 43.5 – 36)/18^{1.2}$
$= 0.0385$ or $0.0386$ | M1, M1, A1 | $=(-)-1.767$ or $CV=(-)-1.645$ | 0.0297 or 0.0227 A1
Or from B(72, ½ ) = 0.0382
Compare with 0.05 and reject $H_0$
There is sufficient evidence to accept that the median time for Elena's swims is less than 30s | A2, M1, A1ft | Using calculator procedure or -1.767<-1.645 not over-assertive | 0.03818457, No, or wrong, CC M1A1ft
| [9] |5 A one-tail sign test of a population median is to be carried out at the $5 \%$ significance level using a sample of size $n$.\\
(i) Show by calculation that the test can never result in rejection of the null hypothesis when $n = 4$.
The coach of a college swimming team expects Elena, the best 50 m freestyle swimmer, to have a median time less than 30 seconds. Elena found from records of her previous 72 swims that 44 were less than 30 seconds and 28 were greater than 30 seconds.\\
(ii) Stating a necessary assumption, test at the $5 \%$ significance level whether Elena's median time for the 50 m freestyle is less than 30 seconds.
\hfill \mbox{\textit{OCR S4 2012 Q5 [11]}}