| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Expected value and most likely value |
| Difficulty | Challenging +1.2 This is a multi-part hypothesis testing question requiring normal approximation to binomial and critical region calculations. Part (a) is routine mean/SD calculation, part (b) is standard hypothesis test with continuity correction, but part (c) requires iterative trial of values to find the smallest n satisfying the critical region constraint at 1% significance, which elevates it above average difficulty. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) mean = \(np = 3.5\) | B1 | |
| (a)(ii) standard deviation = \(\sqrt{700 \times \frac{1}{200}(1 - \frac{1}{200})} = 1.86614\ldots\) awrt 1.866 | M1, A1 | M1 for a correct expression for standard deviation including root. NB Assuming Poisson will get \(\sqrt{3.5} = 1.87082\ldots\) but scores M0A0 [Ans only 2/2] |
| (b)(i) \(H_0: p = \frac{1}{200}\); \(H_1: p > \frac{1}{200}\) | B1, B1 | B1, B1 correct with \(p\) or \(\pi\) (may be seen in (ii)) [Use of \(\lambda\) or \(\mu = 2.5\) for (i) is B0] |
| (b)(ii) \(X \sim B(500, \frac{1}{200}) \to \text{Po}(2.5)\) | B1, M1, A1 | B1 for writing or using \(\text{Po}(2.5)\). 1st M1 for \(1 - P(X \leq 4)\) and use with bin or Poisson or for CR \(P(X \geq 6) = 0.042\) [< 0.05]. 1st A1 for awrt 0.109 or for CR \(X \geq 6\) |
| \(P(X > 5) = 1 - P(X \leq 4) = 1 - 0.8912 = 0.1088\) | ||
| [0.1088 > 0.05] therefore do not reject H₀, not significant, 5 does not lie in CR. The doctor's claim is not supported. or Past records are not out of date/reliable or Number/Proportion/Probability of people with allergy is not higher. | M1, A1, cso | 2nd M1 for a non-contradictory statement which follows from their probability/CR. 2nd A1 cso correct contextual statement and fully correct solution with all other marks scored in (b)(ii) |
| (c) \(Y \sim B(n, 0.30)\), \(P(Y = 0) = (0.70)^n < 0.005\), \(n \geq 14.85\), \(n = 15\) | M1, A1 cao, M1, A1 | M1 for correct expression for \(P(Y = 0)\) and comparison with 0.005 [allow inequality or equation]. Use of tables alone is M0. 1st A1 for \(n = 15\) cao [Answer only is M0A0 unless we see the 0.7ⁿ compared to 0.005]. 2nd M1 for using B("15", 0.30) to try and find an upper tail, need sight of one of given probs. 2nd A1 for \(w = 10\) [but \(w \geq 10\) is A0]. Allow \(Y \geq 10\) or \(Y > 9\) |
| \(P(Y \geq w) < 0.005\) | ||
| \(P(Y \geq 8) = 0.9848\) or \(P(Y \geq 9) = 0.0152\) [> 0.005] | ||
| \(P(Y \geq 9) = 0.9963\) or \(P(Y \geq 10) = 0.0037\) [< 0.005] | ||
| w = 10 |
**(a)(i)** mean = $np = 3.5$ | B1 |
**(a)(ii)** standard deviation = $\sqrt{700 \times \frac{1}{200}(1 - \frac{1}{200})} = 1.86614\ldots$ awrt **1.866** | M1, A1 | M1 for a correct expression for standard deviation including root. NB Assuming Poisson will get $\sqrt{3.5} = 1.87082\ldots$ but scores M0A0 [Ans only 2/2]
**(b)(i)** $H_0: p = \frac{1}{200}$; $H_1: p > \frac{1}{200}$ | B1, B1 | B1, B1 correct with $p$ or $\pi$ (may be seen in (ii)) [Use of $\lambda$ or $\mu = 2.5$ for (i) is B0]
**(b)(ii)** $X \sim B(500, \frac{1}{200}) \to \text{Po}(2.5)$ | B1, M1, A1 | B1 for writing or using $\text{Po}(2.5)$. 1st M1 for $1 - P(X \leq 4)$ and use with bin or Poisson or for CR $P(X \geq 6) = 0.042$ [< 0.05]. 1st A1 for awrt 0.109 or for CR $X \geq 6$
| $P(X > 5) = 1 - P(X \leq 4) = 1 - 0.8912 = 0.1088$ | |
| [0.1088 > 0.05] therefore do not reject H₀, not significant, 5 does not lie in CR. The doctor's claim is **not supported**. or Past records are not out of date/reliable or Number/Proportion/Probability of people with allergy is not higher. | M1, A1, cso | 2nd M1 for a non-contradictory statement which follows from their probability/CR. 2nd A1 cso correct contextual statement and fully correct solution with all other marks scored in (b)(ii)
**(c)** $Y \sim B(n, 0.30)$, $P(Y = 0) = (0.70)^n < 0.005$, $n \geq 14.85$, $n = 15$ | M1, A1 cao, M1, A1 | M1 for correct expression for $P(Y = 0)$ and comparison with 0.005 [allow inequality or equation]. Use of tables alone is M0. 1st A1 for $n = 15$ cao [Answer only is M0A0 unless we see the 0.7ⁿ compared to 0.005]. 2nd M1 for using B("15", 0.30) to try and find an upper tail, need sight of one of given probs. 2nd A1 for $w = 10$ [but $w \geq 10$ is A0]. Allow $Y \geq 10$ or $Y > 9$
| $P(Y \geq w) < 0.005$ | |
| $P(Y \geq 8) = 0.9848$ or $P(Y \geq 9) = 0.0152$ [> 0.005] | |
| $P(Y \geq 9) = 0.9963$ or $P(Y \geq 10) = 0.0037$ [< 0.005] | |
| **w = 10** | |
**Total: 13**
---4. In a large population, past records show that 1 in 200 adults has a particular allergy.
In a random sample of 700 adults selected from the population, estimate
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item the mean number of adults with the allergy,
\item the standard deviation of the number of adults with the allergy.
Give your answer to 3 decimal places.
A doctor claims that the past records are out of date and the proportion of adults with the allergy is higher than the records indicate.
A random sample of 500 adults is taken from the population and 5 are found to have the allergy.
A test of the doctor's claim is to be carried out at the 5\% level of significance.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item State the hypotheses for this test.
\item Using a suitable approximation, carry out the test.\\
(6)
It is also claimed that $30 \%$ of those with the allergy take medication for it daily.
To test this claim, a random sample of $n$ people with the allergy is taken. The random variable $Y$ represents the number of people in the sample who take medication for the allergy daily.
A two-tailed test, at the $1 \%$ level of significance, is carried out to see if the proportion differs from 30\%
The critical region for the test is $Y = 0$ or $Y \geqslant w$
\end{enumerate}\item Find the smallest possible value of $n$ and the corresponding value of $w$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2017 Q4 [13]}}