| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed hypothesis test (upper tail, H₁: p > p₀) |
| Difficulty | Standard +0.3 This is a straightforward application of binomial hypothesis testing with clear structure: calculate probabilities using B(15, 1/6), then perform two one-tailed tests at 10% significance level. The question guides students through each step methodically, requiring only standard techniques with no novel insight or complex reasoning beyond textbook procedures. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X \sim B\!\left(15, \frac{1}{6}\right)\) | ||
| \(P(X=0) = \left(\frac{5}{6}\right)^{15} = 0.065\) | M1 A1 cao | \(\left(\frac{5}{6}\right)^{15}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X=4) = \binom{15}{4} \times \left(\frac{1}{6}\right)^4 \times \left(\frac{5}{6}\right)^{11} = 0.142\) | M1 M1 A1 cao | \(\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right)^{11}\); multiply by \(\binom{15}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X>3) = 1 - P(X \leq 3) = 1 - 0.7685 = 0.232\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Let \(p =\) probability of a six on any throw | B1 | Definition of \(p\) |
| \(H_0: p = \frac{1}{6}\), \(H_1: p < \frac{1}{6}\) | B1 | Both hypotheses |
| \(X \sim B\!\left(15, \frac{1}{6}\right)\), \(P(X=0) = 0.065\) | M1 | 0.065 |
| \(0.065 < 0.1\) so reject \(H_0\) | M1 dep | Comparison |
| Sufficient evidence at 10% level that dice are biased against sixes | E1 dep |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: p = \frac{1}{6}\), \(H_1: p > \frac{1}{6}\) | B1 | Both hypotheses |
| \(P(X \geq 5) = 1 - P(X \leq 4) = 1 - 0.910 = 0.09\) | M1 M1 dep | 0.09; Comparison |
| \(0.09 < 0.1\) so reject \(H_0\) | E1 dep | |
| Sufficient evidence at 10% level that dice are biased in favour of sixes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Conclusions contradictory | E1 | Contradictory |
| Even if null hypothesis is true, it will be rejected 10% of the time purely by chance (or other sensible comments) | E1 | By chance |
## Question 7:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim B\!\left(15, \frac{1}{6}\right)$ | | |
| $P(X=0) = \left(\frac{5}{6}\right)^{15} = 0.065$ | M1 A1 cao | $\left(\frac{5}{6}\right)^{15}$ |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X=4) = \binom{15}{4} \times \left(\frac{1}{6}\right)^4 \times \left(\frac{5}{6}\right)^{11} = 0.142$ | M1 M1 A1 cao | $\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right)^{11}$; multiply by $\binom{15}{4}$ |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X>3) = 1 - P(X \leq 3) = 1 - 0.7685 = 0.232$ | M1 A1 | |
### Part (iv)(A)
| Answer | Mark | Guidance |
|--------|------|----------|
| Let $p =$ probability of a six on any throw | B1 | Definition of $p$ |
| $H_0: p = \frac{1}{6}$, $H_1: p < \frac{1}{6}$ | B1 | Both hypotheses |
| $X \sim B\!\left(15, \frac{1}{6}\right)$, $P(X=0) = 0.065$ | M1 | 0.065 |
| $0.065 < 0.1$ so reject $H_0$ | M1 dep | Comparison |
| Sufficient evidence at 10% level that dice are biased against sixes | E1 dep | |
### Part (iv)(B)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: p = \frac{1}{6}$, $H_1: p > \frac{1}{6}$ | B1 | Both hypotheses |
| $P(X \geq 5) = 1 - P(X \leq 4) = 1 - 0.910 = 0.09$ | M1 M1 dep | 0.09; Comparison |
| $0.09 < 0.1$ so reject $H_0$ | E1 dep | |
| Sufficient evidence at 10% level that dice are biased in favour of sixes | | |
### Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| Conclusions contradictory | E1 | Contradictory |
| Even if null hypothesis is true, it will be rejected 10% of the time purely by chance (or other sensible comments) | E1 | By chance |7 A game requires 15 identical ordinary dice to be thrown in each turn.\\
Assuming the dice to be fair, find the following probabilities for any given turn.
\begin{enumerate}[label=(\roman*)]
\item No sixes are thrown.
\item Exactly four sixes are thrown.
\item More than three sixes are thrown.
David and Esme are two players who are not convinced that the dice are fair. David believes that the dice are biased against sixes, while Esme believes the dice to be biased in favour of sixes.
In his next turn, David throws no sixes. In her next turn, Esme throws 5 sixes.
\item Writing down your hypotheses carefully in each case, decide whether\\
(A) David's turn provides sufficient evidence at the $10 \%$ level that the dice are biased against sixes,\\
(B) Esme's turn provides sufficient evidence at the $10 \%$ level that the dice are biased in favour of sixes.
\item Comment on your conclusions from part (iv).
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2005 Q7 [16]}}