| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | November |
| Marks | 5 |
| 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 hypothesis testing question requiring standard binomial test setup, a single probability calculation to verify significance, and finding a sample size by trial (or solving an inequality). All techniques are routine for S2 level with no novel insight required, making it slightly easier than average. |
| Spec | 2.04c Calculate binomial probabilities2.05b Hypothesis test for binomial proportion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: P(6) = \frac{1}{6}\) \(\quad\) \(H_1: P(6) < \frac{1}{6}\) | B1 [1] | Allow \(H_0: p = \frac{1}{6}\) \(\quad\) \(H_1: p < \frac{1}{6}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(\frac{5}{6}\right)^{15}\) | M1 | |
| \(= 0.065 > 0.05\) | A1 [2] | Correct result and comparison needed for A1. SR if 2 tail test followed allow A1 for \(0.065 > 0.025\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(\frac{5}{6}\right)^{16} = 0.054\) and \(\left(\frac{5}{6}\right)^{17} = 0.045\) | M1 | both |
| Smallest \(n\) is 17 | A1 [2] | No errors seen |
| OR | ||
| \(\left(\frac{5}{6}\right)^n < 0.05\) and attempt to solve | M1 | |
| \(n\ln\left(\frac{5}{6}\right) < \ln 0.05\) | ||
| smallest \(n\) is 17 | A1 |
## Question 2:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: P(6) = \frac{1}{6}$ $\quad$ $H_1: P(6) < \frac{1}{6}$ | B1 [1] | Allow $H_0: p = \frac{1}{6}$ $\quad$ $H_1: p < \frac{1}{6}$ |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(\frac{5}{6}\right)^{15}$ | M1 | |
| $= 0.065 > 0.05$ | A1 [2] | Correct result and comparison needed for A1. SR if 2 tail test followed allow A1 for $0.065 > 0.025$ |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(\frac{5}{6}\right)^{16} = 0.054$ and $\left(\frac{5}{6}\right)^{17} = 0.045$ | M1 | both |
| Smallest $n$ is 17 | A1 [2] | No errors seen |
| **OR** | | |
| $\left(\frac{5}{6}\right)^n < 0.05$ and attempt to solve | M1 | |
| $n\ln\left(\frac{5}{6}\right) < \ln 0.05$ | | |
| smallest $n$ is 17 | A1 | |
---2 A die has six faces numbered $1,2,3,4,5,6$. Manjit suspects that the die is biased so that it shows a six on fewer throws than it would if it were fair. In order to test her suspicion, she throws the die a certain number of times and counts the number of sixes.\\
\begin{enumerate}[label=(\roman*)]
\item State suitable null and alternative hypotheses for Manjit's test.
\item There are no sixes in the first 15 throws. Show that this result is not significant at the $5 \%$ level.
\item Find the smallest value of $n$ such that, if there are no sixes in the first $n$ throws, this result is significant at the 5\% level.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2016 Q2 [5]}}