OCR MEI S1 — Question 2 17 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeOne-tailed hypothesis test (upper tail, H₁: p > p₀)
DifficultyModerate -0.3 This is a straightforward S1 hypothesis testing question with standard binomial calculations. Parts (i) and (ii) are routine expectation and probability calculations, while part (iii) requires setting up and performing a one-tailed test at 5% level—all standard textbook procedures with no novel insight required. Slightly easier than average due to clear structure and familiar context.
Spec2.04b 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

2 Mark is playing solitaire on his computer. The probability that he wins a game is 0.2 , independently of all other games that he plays.
  1. Find the expected number of wins in 12 games.
  2. Find the probability that
    (A) he wins exactly 2 out of the next 12 games that he plays,
    (B) he wins at least 2 out of the next 12 games that he plays.
  3. Mark's friend Ali also plays solitaire. Ali claims that he is better at winning games than Mark. In a random sample of 20 games played by Ali, he wins 7 of them. Write down suitable hypotheses for a test at the \(5 \%\) level to investigate whether Ali is correct. Give a reason for your choice of alternative hypothesis. Carry out the test.

Question 2:
Part (i)
AnswerMarks Guidance
AnswerMarks Guidance
\(E(X) = np = 12 \times 0.2 = 2.4\)M1 For product. If wrong \(n\) used consistently throughout, allow M marks only
Do not allow subsequent roundingA1 CAO NB If they round to 2, even if they have obtained 2.4 first they get M1A0. For answer of '2.4 or 2 if rounded up' allow M1A0. Answer of 2 without working gets M0A0. If they attempt \(E(X)\) by summing products \(xp\) give no marks unless answer is fully correct
Total: 2 marks
Part (ii)(A)
AnswerMarks Guidance
AnswerMarks Guidance
\(X \sim B(12, 0.2)\)
\(P(\text{Wins exactly } 2) = \binom{12}{2} \times 0.2^2 \times 0.8^{10} = 0.2835\)M1 \(0.2^2 \times 0.8^{10}\). With \(p + q = 1\)
M1\(\binom{12}{2} \times p^2 q^{10}\). Also for \(66 \times 0.004295\)
A1 CAOAllow answers within the range 0.283 to 0.284 with or without working or 0.28 to 0.283 if working shown
OR from tables: \(0.5583 - 0.2749 = 0.2834\)OR: M2 for \(0.5583 - 0.2749\) A1 CAO
Total: 3 marks
Part (ii)(B)
AnswerMarks Guidance
AnswerMarks Guidance
\(P(\text{Wins at least } 2) = 1 - P(X \leq 1) = 1 - 0.2749 = 0.7251\)M1 \(P(X \leq 1)\). M1 0.2749 seen
M1\(1 - P(X \leq 1)\). M1 \(1 - 0.2749\) seen
A1 CAOAllow 0.725 to 0.73 but not 0.72
Point probability method:
- \(P(1) = 12 \times 0.2 \times 0.8^{11} = 0.2062\), \(P(0) = 0.8^{12} = 0.0687\)
- So \(P(X \leq 1) = 0.2749\) gets M1 then mark as per scheme
SC1 for \(1 - P(X \leq 2) = 1 - 0.5583 = 0.4417\)
For misread of tables value of 0.2749, allow 0 in (A) but MAX M1M1 in (B)
Total: 3 marks
Part (iii)
AnswerMarks Guidance
AnswerMarks Guidance
Let \(p\) = probability that Ali wins a gameB1 For definition of \(p\) in context. Minimum needed for B1 is \(p\) = probability that Ali wins. Allow \(p\) = P(Ali wins) for B1. Definition of \(p\) must include word probability (or chance or proportion or percentage but NOT possibility)
\(H_0: p = 0.2\)B1 Allow \(p=20\%\), allow \(\theta\) or \(\pi\) and \(\rho\) but not \(x\). However allow any single symbol if defined. Allow \(H_0 = p=0.2\), Allow \(H_0: p = \frac{2}{10}\). Do not allow \(H_0: P(X=x) = 0.2\), \(H_1: P(X=x) > 0.2\). Do not allow \(H_0\): \(=0.2\), \(=20\%\), \(P(0.2)\), \(p(0.2)\), \(p(x)=0.2\), \(x=0.2\) (unless \(x\) correctly defined as a probability). Do not allow \(H_1: p \geq 0.2\). Do not allow \(H_0\) and \(H_1\) reversed for B marks but can still get E1
\(H_1: p > 0.2\)B1 Allow NH and AH in place of \(H_0\) and \(H_1\)
\(H_1\) has this form as Ali claims that he is better at winning games than Mark isE1 For hypotheses given in words allow Maximum B0B1B1E1. Hypotheses in words must include probability (or chance or proportion or percentage) and the figure 0.2 oe
EITHER Probability method:
AnswerMarks Guidance
AnswerMarks Guidance
\(P(X \geq 7) = 1 - P(X \leq 6) = 1 - 0.9133 = 0.0867 > 5\%\)B1 For \(P(X \geq 7)\). Do not award first B1 for poor symbolic notation such as \(P(X=7) = 0.0867\)
B1For 0.0867 Or \(1 - 0.9133\) seen
M1For comparison with 5% dep on B1 for 0.0867
So not significant, so there is not enough evidence to reject the null hypothesis and we conclude that there is not enough evidence to suggest that Ali is better at winning games than MarkA1 For not significant or 'accept \(H_0\)' or 'cannot reject \(H_0\)' or 'reject \(H_1\)'. Must include 'not enough evidence' or something similar for E1
E1 dep on M1A1Zero for use of point prob \(P(X=7) = 0.0546\)
OR Critical region method:
AnswerMarks Guidance
AnswerMarks Guidance
\(X \sim B(20, 0.2)\)
\(P(X \geq 7) = 1 - P(X \leq 6) = 1 - 0.9133 = 0.0867 > 5\%\)B1 For 0.0867. Allow any form of statement of CR e.g. \(X \geq 8\), 8 to 20, 8 or above, \(X > 8\), \(\{8, \ldots\}\), annotated number line, etc but not \(P(X \geq 8)\)
\(P(X \geq 8) = 1 - P(X \leq 7) = 1 - 0.9679 = 0.0321 < 5\%\)B1 For 0.0321. \(\{8,9,10,11,12\}\) gets max B2M1A0 – tables stop at 8
So critical region is \(\{8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}\)M1 For at least one comparison with 5%. NB USE OF POINT PROBABILITIES gets B0B0M0A0
7 does not lie in the critical region, so not significantA1 CAO For critical region and not significant or 'accept \(H_0\)' or 'cannot reject \(H_0\)' or 'reject \(H_1\)' dep on M1 and at least one B1. Providing there is sight of 95%, allow B1 for 0.9133, B1 for 0.9679, M1 for comparison with 95% A1CAO for correct CR
So there is not enough evidence to reject the null hypothesis and we conclude that there is not enough evidence to suggest that Ali is better at winning games than MarkE1 dep on M1A1 See additional notes below the scheme for other possibilities. PLEASE CHECK THAT THERE IS NO EXTRA WORKING ON THE SECOND PAGE IN THE ANSWER BOOKLET
Total: 5 marks
TOTAL: 17 marks
# Question 2:

## Part (i)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = np = 12 \times 0.2 = 2.4$ | M1 | For product. If wrong $n$ used consistently throughout, allow M marks only |
| Do not allow subsequent rounding | A1 CAO | NB If they round to 2, even if they have obtained 2.4 first they get M1A0. For answer of '2.4 or 2 if rounded up' allow M1A0. Answer of 2 without working gets M0A0. If they attempt $E(X)$ by summing products $xp$ give no marks unless answer is fully correct |

**Total: 2 marks**

## Part (ii)(A)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $X \sim B(12, 0.2)$ | | |
| $P(\text{Wins exactly } 2) = \binom{12}{2} \times 0.2^2 \times 0.8^{10} = 0.2835$ | M1 | $0.2^2 \times 0.8^{10}$. With $p + q = 1$ |
| | M1 | $\binom{12}{2} \times p^2 q^{10}$. Also for $66 \times 0.004295$ |
| | A1 CAO | Allow answers within the range 0.283 to 0.284 with or without working or 0.28 to 0.283 if working shown |
| OR from tables: $0.5583 - 0.2749 = 0.2834$ | OR: M2 for $0.5583 - 0.2749$ | A1 CAO |

**Total: 3 marks**

## Part (ii)(B)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Wins at least } 2) = 1 - P(X \leq 1) = 1 - 0.2749 = 0.7251$ | M1 | $P(X \leq 1)$. M1 0.2749 seen |
| | M1 | $1 - P(X \leq 1)$. M1 $1 - 0.2749$ seen |
| | A1 CAO | Allow 0.725 to 0.73 but not 0.72 |

**Point probability method:**
- $P(1) = 12 \times 0.2 \times 0.8^{11} = 0.2062$, $P(0) = 0.8^{12} = 0.0687$
- So $P(X \leq 1) = 0.2749$ gets M1 then mark as per scheme

**SC1** for $1 - P(X \leq 2) = 1 - 0.5583 = 0.4417$

For misread of tables value of 0.2749, allow 0 in (A) but MAX M1M1 in (B)

**Total: 3 marks**

## Part (iii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $p$ = probability that Ali wins a game | B1 | For definition of $p$ in context. Minimum needed for B1 is $p$ = probability that Ali wins. Allow $p$ = P(Ali wins) for B1. Definition of $p$ must include word probability (or chance or proportion or percentage but NOT possibility) |
| $H_0: p = 0.2$ | B1 | Allow $p=20\%$, allow $\theta$ or $\pi$ and $\rho$ but not $x$. However allow any single symbol if defined. Allow $H_0 = p=0.2$, Allow $H_0: p = \frac{2}{10}$. Do not allow $H_0: P(X=x) = 0.2$, $H_1: P(X=x) > 0.2$. Do not allow $H_0$: $=0.2$, $=20\%$, $P(0.2)$, $p(0.2)$, $p(x)=0.2$, $x=0.2$ (unless $x$ correctly defined as a probability). Do not allow $H_1: p \geq 0.2$. Do not allow $H_0$ and $H_1$ reversed for B marks but can still get E1 |
| $H_1: p > 0.2$ | B1 | Allow NH and AH in place of $H_0$ and $H_1$ |
| $H_1$ has this form as Ali claims that he is better at winning games than Mark is | E1 | For hypotheses given in words allow Maximum B0B1B1E1. Hypotheses in words must include probability (or chance or proportion or percentage) and the figure 0.2 oe |

**EITHER Probability method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X \geq 7) = 1 - P(X \leq 6) = 1 - 0.9133 = 0.0867 > 5\%$ | B1 | For $P(X \geq 7)$. Do not award first B1 for poor symbolic notation such as $P(X=7) = 0.0867$ |
| | B1 | For 0.0867 Or $1 - 0.9133$ seen |
| | M1 | For comparison with 5% dep on B1 for 0.0867 |
| So not significant, so there is not enough evidence to reject the null hypothesis and we conclude that there is not enough evidence to suggest that Ali is better at winning games than Mark | A1 | For not significant or 'accept $H_0$' or 'cannot reject $H_0$' or 'reject $H_1$'. Must include 'not enough evidence' or something similar for E1 |
| | E1 dep on M1A1 | Zero for use of point prob $P(X=7) = 0.0546$ |

**OR Critical region method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $X \sim B(20, 0.2)$ | | |
| $P(X \geq 7) = 1 - P(X \leq 6) = 1 - 0.9133 = 0.0867 > 5\%$ | B1 | For 0.0867. Allow any form of statement of CR e.g. $X \geq 8$, 8 to 20, 8 or above, $X > 8$, $\{8, \ldots\}$, annotated number line, etc but not $P(X \geq 8)$ |
| $P(X \geq 8) = 1 - P(X \leq 7) = 1 - 0.9679 = 0.0321 < 5\%$ | B1 | For 0.0321. $\{8,9,10,11,12\}$ gets max B2M1A0 – tables stop at 8 |
| So critical region is $\{8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}$ | M1 | For at least one comparison with 5%. NB USE OF POINT PROBABILITIES gets B0B0M0A0 |
| 7 does not lie in the critical region, so not significant | A1 CAO | For critical region and not significant or 'accept $H_0$' or 'cannot reject $H_0$' or 'reject $H_1$' dep on M1 and at least one B1. Providing there is sight of 95%, allow B1 for 0.9133, B1 for 0.9679, M1 for comparison with 95% A1CAO for correct CR |
| So there is not enough evidence to reject the null hypothesis and we conclude that there is not enough evidence to suggest that Ali is better at winning games than Mark | E1 dep on M1A1 | See additional notes below the scheme for other possibilities. **PLEASE CHECK THAT THERE IS NO EXTRA WORKING ON THE SECOND PAGE IN THE ANSWER BOOKLET** |

**Total: 5 marks**

**TOTAL: 17 marks**
2 Mark is playing solitaire on his computer. The probability that he wins a game is 0.2 , independently of all other games that he plays.
\begin{enumerate}[label=(\roman*)]
\item Find the expected number of wins in 12 games.
\item Find the probability that\\
(A) he wins exactly 2 out of the next 12 games that he plays,\\
(B) he wins at least 2 out of the next 12 games that he plays.
\item Mark's friend Ali also plays solitaire. Ali claims that he is better at winning games than Mark. In a random sample of 20 games played by Ali, he wins 7 of them. Write down suitable hypotheses for a test at the $5 \%$ level to investigate whether Ali is correct. Give a reason for your choice of alternative hypothesis. Carry out the test.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI S1  Q2 [17]}}