| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | Specimen |
| Marks | 13 |
| 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 standard S2 hypothesis testing question with routine binomial calculations in parts (a)-(c), requiring only textbook methods. Part (d) requires finding a sample size by trial, which adds mild problem-solving but remains algorithmic. Slightly easier than average A-level due to straightforward application of learned procedures. |
| Spec | 2.04d Normal approximation to binomial5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02d Binomial: mean np and variance np(1-p)5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X=5) = {}^{20}C_5(0.3)^5(0.7)^{15}\) or \(0.4164 - 0.2375\) | M1 | \({}^{20}C_5(p)^5(1-p)^{15}\) or using \(P(X\leq 5)-P(X\leq 4)\) |
| \(= 0.17886...\) awrt 0.179 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mean \(= 6\) | B1 | |
| \(\text{sd} = \sqrt{20 \times 0.7 \times 0.3}\) | M1 | Use of \(20\times0.7\times0.3\) (with or without square root) |
| \(= 2.049...\) awrt 2.05 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: p=0.3\), \(H_1: p>0.3\) | B1 | Both hypotheses correct (\(p\) or \(\pi\)) |
| \(X \sim B(20, 0.3)\) | M1 | May be implied by 0.7723, 0.2277, 0.8867 or 0.1133 |
| \(P(X \geq 8) = 0.2277\) or \(P(X \geq 10)=0.0480\), so CR \(X \geq 10\) | A1 | Awrt 0.228 or CR \(X \geq 10\) |
| Insufficient evidence to reject \(H_0\) or Not Significant or 8 does not lie in the critical region | dM1 | Correct comment dependent on previous M1 |
| There is no evidence to support the Director (of Studies') belief/There is no evidence that the proportion of parents that do not support the new curriculum is greater than 30% | A1 cso | Cso requires correct contextual conclusion with underlined words and all previous marks in (c) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim B(2n, 0.25)\) | ||
| \(X \sim B(8, 0.25)\): \(P(X \geq 4) = 0.1138\); \(X \sim B(10, 0.25)\): \(P(X \geq 5) = 0.0781\) | M1 | For 0.1138 or 0.0781 or 0.8862 or 0.9219 seen |
| \(2n = 10\) | A1 | B(10, 0.25) selected (may be implied by \(n=10\) or \(2n=10\) or \(n=5\)) |
| \(n = 5\) | A1 | An answer of 5 with no incorrect working scores 3/3 |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X=5) = {}^{20}C_5(0.3)^5(0.7)^{15}$ or $0.4164 - 0.2375$ | M1 | ${}^{20}C_5(p)^5(1-p)^{15}$ or using $P(X\leq 5)-P(X\leq 4)$ |
| $= 0.17886...$ awrt 0.179 | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mean $= 6$ | B1 | |
| $\text{sd} = \sqrt{20 \times 0.7 \times 0.3}$ | M1 | Use of $20\times0.7\times0.3$ (with or without square root) |
| $= 2.049...$ awrt 2.05 | A1 | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p=0.3$, $H_1: p>0.3$ | B1 | Both hypotheses correct ($p$ or $\pi$) |
| $X \sim B(20, 0.3)$ | M1 | May be implied by 0.7723, 0.2277, 0.8867 or 0.1133 |
| $P(X \geq 8) = 0.2277$ or $P(X \geq 10)=0.0480$, so CR $X \geq 10$ | A1 | Awrt 0.228 or CR $X \geq 10$ |
| Insufficient evidence to reject $H_0$ or Not Significant or 8 does not lie in the critical region | dM1 | Correct comment dependent on previous M1 |
| There is no evidence to support the Director (of Studies') belief/There is no evidence that the proportion of parents that do not support the new curriculum is greater than 30% | A1 cso | Cso requires correct contextual conclusion with underlined words and all previous marks in (c) |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim B(2n, 0.25)$ | | |
| $X \sim B(8, 0.25)$: $P(X \geq 4) = 0.1138$; $X \sim B(10, 0.25)$: $P(X \geq 5) = 0.0781$ | M1 | For 0.1138 or 0.0781 or 0.8862 or 0.9219 seen |
| $2n = 10$ | A1 | B(10, 0.25) selected (may be implied by $n=10$ or $2n=10$ or $n=5$) |
| $n = 5$ | A1 | An answer of 5 with no incorrect working scores 3/3 |6. The Headteacher of a school claims that $30 \%$ of parents do not support a new curriculum. In a survey of 20 randomly selected parents, the number, $X$, who do not support the new curriculum is recorded.
Assuming that the Headteacher's claim is correct, find
\begin{enumerate}[label=(\alph*)]
\item the probability that $X = 5$
\item the mean and the standard deviation of $X$
The Director of Studies believes that the proportion of parents who do not support the new curriculum is greater than $30 \%$. Given that in the survey of 20 parents 8 do not support the new curriculum,
\item test, at the $5 \%$ level of significance, the Director of Studies' belief. State your hypotheses clearly.
The teachers believe that the sample in the original survey was biased and claim that only $25 \%$ of the parents are in support of the new curriculum. A second random sample, of size $2 n$, is taken and exactly half of this sample supports the new curriculum.
A test is carried out at a $10 \%$ level of significance of the teachers' belief using this sample of size $2 n$
Using the hypotheses $\mathrm { H } _ { 0 } : p = 0.25$ and $\mathrm { H } _ { 1 } : p > 0.25$
\item find the minimum value of $n$ for which the outcome of the test is that the teachers' belief is rejected.
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\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2018 Q6 [13]}}