| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Single observation hypothesis test |
| Difficulty | Moderate -0.3 This is a standard S2 hypothesis testing question requiring routine application of binomial tables to find critical regions and significance levels. While it has multiple parts, each step follows a textbook procedure with no novel problem-solving required—students simply look up cumulative probabilities and compare values, making it slightly easier than average. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \leqslant 7) = 0.8883\) or \(P(X \leqslant 8) = 0.9644\) or \(P(X \geqslant 8) = 0.1117\) or \(P(X \geqslant 9) = 0.0356\) | M1 | For one of these 4 probabilities, may be implied by correct critical region |
| Critical Region is \(X \geqslant 9\) | A1 | For \(X \geqslant 9\) (allow \(X > 8\)), e.g. {9,12}, {9,10,11,12} etc. Must be \(X \geqslant 9\) for A1, do not award for just seeing \(P(X \geqslant 9)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((1 - 0.9644 =)\ 0.0356\) | B1cao | For 0.0356 or better. NB Calculator gives 0.03557486... |
| Answer | Marks | Guidance |
|---|---|---|
| Reject \(H_0\)/Significant or value of \(p\) is \(> 0.45\) | B1ft | ft their critical region. Must say "reject" and "\(H_0\)". No contradictory statements. Just saying "9 is not in the critical region" is not enough. Allow restart: \(P(X \geqslant 9) = 0.0356 < 0.05\) so significant. |
| Answer | Marks | Guidance |
|---|---|---|
| Conclusion would not change as \(H_0\) would still be rejected | B1 | If they score B0 in (c) then score B0B0 in (d) |
| Answer | Marks |
|---|---|
| Conclusion would change as \(H_0\) would not be rejected | B1 |
## Question 3:
### Part (a):
$P(X \leqslant 7) = 0.8883$ or $P(X \leqslant 8) = 0.9644$ or $P(X \geqslant 8) = 0.1117$ or $P(X \geqslant 9) = 0.0356$ | M1 | For one of these 4 probabilities, may be implied by correct critical region
Critical Region is $X \geqslant 9$ | A1 | For $X \geqslant 9$ (allow $X > 8$), e.g. {9,12}, {9,10,11,12} etc. Must be $X \geqslant 9$ for A1, do not award for just seeing $P(X \geqslant 9)$
### Part (b):
$(1 - 0.9644 =)\ 0.0356$ | B1cao | For 0.0356 or better. NB Calculator gives 0.03557486...
### Part (c):
Reject $H_0$/Significant or value of $p$ is $> 0.45$ | B1ft | ft their critical region. Must say "reject" **and** "$H_0$". No contradictory statements. Just saying "9 is not in the critical region" is not enough. Allow restart: $P(X \geqslant 9) = 0.0356 < 0.05$ so significant.
### Part (d)(i):
Conclusion would not change as $H_0$ would still be rejected | B1 | If they score B0 in (c) then score B0B0 in (d)
### Part (d)(ii):
Conclusion would change as $H_0$ would not be rejected | B1 |
**CR Notes:** (i) New CR is $X \geqslant 9$ but can treat any incorrect mention of CR as ISW. (ii) New CR is $X \geqslant 10$ but can treat any incorrect mention of CR as ISW.
---3. A single observation $x$ is to be taken from $X \sim \mathrm {~B} ( 12 , p )$
This observation is used to test $\mathrm { H } _ { 0 } : p = 0.45$ against $\mathrm { H } _ { 1 } : p > 0.45$
\begin{enumerate}[label=(\alph*)]
\item Using a $5 \%$ level of significance, find the critical region for this test.
\item State the actual significance level of this test.
The value of the observation is found to be 9
\item State the conclusion that can be made based on this observation.
\item State whether or not this conclusion would change if the same test was carried out at the
\begin{enumerate}[label=(\roman*)]
\item 10\% level of significance,
\item $1 \%$ level of significance.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2016 Q3 [6]}}