Edexcel S2 2013 January — Question 6 10 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2013
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeOne-tailed hypothesis test (lower tail, H₁: p < p₀)
DifficultyStandard +0.3 This is a standard S2 hypothesis testing question with routine binomial calculations. Parts (a) and (b) test definitions, part (c) is a straightforward one-tailed test with given values, and part (d) requires iterating to find n where P(X=0)<0.01, which is methodical rather than conceptually challenging. Slightly easier than average due to the definitional components and standard procedure.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail

6. (a) Explain what you understand by a hypothesis.
(b) Explain what you understand by a critical region. Mrs George claims that 45\% of voters would vote for her.
In an opinion poll of 20 randomly selected voters it was found that 5 would vote for her.
(c) Test at the \(5 \%\) level of significance whether or not the opinion poll provides evidence to support Mrs George's claim. In a second opinion poll of \(n\) randomly selected people it was found that no one would vote for Mrs George.
(d) Using a \(1 \%\) level of significance, find the smallest value of \(n\) for which the hypothesis \(\mathrm { H } _ { 0 } : p = 0.45\) will be rejected in favour of \(\mathrm { H } _ { 1 } : p < 0.45\)

Question 6:
(a)
AnswerMarks Guidance
AnswerMark Guidance
A statement concerning a population parameterB1 Must include words "population parameter"
(b)
AnswerMarks Guidance
AnswerMark Guidance
A critical region is the range/set of values/answers or a test statistic or region/area or values where the test is significantB1
that would lead to the rejection of H0 / acceptance of H1B1
(c)
AnswerMarks Guidance
AnswerMark Guidance
\(H_0: p=0.45\), \(H_1: p < 0.45\) (or \(p \neq 0.45\))
\(X \sim B(20, 0.45)\)M1 Using normal approximation to binomial is M0
\(P(X \leq 5) = 0.0553\), CR \(X \leq 4\)A1 Allow 0.9447 if not using CR
Accept \(H_0\). Not significant. 5 does not lie in the Critical RegionM1d Dependent on previous M
There is no evidence that the proportion who voted for Mrs George is not 45% or there is evidence to support Mrs George's claimA1cso Conclusion must contain words Mrs George. No incorrect working seen
(d)
AnswerMarks Guidance
AnswerMark Guidance
\(B(8, 0.45)\): \(P(0) = 0.0084\)M1 Attempt to find \(P(0)\) from \(B(n,0.45)\) or \((0.55)^n < 0.01\)
\(B(7, 0.45)\): \(P(0) = 0.0152\)A1 \(P(0)=0.0084\) and \(P(0)=0.0152\) or getting 7.7
Smallest value of \(n\) is 8B1 cso. \(n=8\) should not come from incorrect working. Answer of 8 alone gains M1A1B1
Alternative: \((0.55)^n < 0.01\)M1
\(n\log 0.55 < \log 0.01\)
\(n > 7.7\ldots\)A1
Smallest value of \(n\) is 8B1cso
## Question 6:

**(a)**
| Answer | Mark | Guidance |
|--------|------|----------|
| A statement concerning a **population parameter** | B1 | Must include words "population parameter" |

**(b)**
| Answer | Mark | Guidance |
|--------|------|----------|
| A critical region is the range/set of values/answers or a test statistic or region/area or values where the test is significant | B1 | |
| that would lead to the rejection of H0 / acceptance of H1 | B1 | |

**(c)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: p=0.45$, $H_1: p < 0.45$ (or $p \neq 0.45$) | | |
| $X \sim B(20, 0.45)$ | M1 | Using normal approximation to binomial is M0 |
| $P(X \leq 5) = 0.0553$, CR $X \leq 4$ | A1 | Allow 0.9447 if not using CR |
| Accept $H_0$. Not significant. 5 does not lie in the Critical Region | M1d | Dependent on previous M |
| There is no evidence that the proportion who voted for Mrs George is not 45% or there is evidence to support Mrs George's claim | A1cso | Conclusion must contain words **Mrs George**. No incorrect working seen |

**(d)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $B(8, 0.45)$: $P(0) = 0.0084$ | M1 | Attempt to find $P(0)$ from $B(n,0.45)$ or $(0.55)^n < 0.01$ |
| $B(7, 0.45)$: $P(0) = 0.0152$ | A1 | $P(0)=0.0084$ and $P(0)=0.0152$ or getting 7.7 |
| Smallest value of $n$ is **8** | B1 | cso. $n=8$ should not come from incorrect working. Answer of 8 alone gains M1A1B1 |
| **Alternative:** $(0.55)^n < 0.01$ | M1 | |
| $n\log 0.55 < \log 0.01$ | | |
| $n > 7.7\ldots$ | A1 | |
| Smallest value of $n$ is **8** | B1cso | |

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6. (a) Explain what you understand by a hypothesis.\\
(b) Explain what you understand by a critical region.

Mrs George claims that 45\% of voters would vote for her.\\
In an opinion poll of 20 randomly selected voters it was found that 5 would vote for her.\\
(c) Test at the $5 \%$ level of significance whether or not the opinion poll provides evidence to support Mrs George's claim.

In a second opinion poll of $n$ randomly selected people it was found that no one would vote for Mrs George.\\
(d) Using a $1 \%$ level of significance, find the smallest value of $n$ for which the hypothesis $\mathrm { H } _ { 0 } : p = 0.45$ will be rejected in favour of $\mathrm { H } _ { 1 } : p < 0.45$\\

\hfill \mbox{\textit{Edexcel S2 2013 Q6 [10]}}