| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Expected value and most likely value |
| Difficulty | Moderate -0.3 This is a straightforward S2 binomial hypothesis testing question with standard bookwork components. Part (a) is trivial algebra (E(X)=np), parts (b)(i-ii) are direct binomial probability calculations, and part (c) is a textbook one-tailed hypothesis test with clearly stated hypotheses. All steps are routine applications of standard methods with no problem-solving insight required, making it slightly easier than average. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.05n = 3\) | M1 | Using \(0.05n\) |
| \(n = 60\) | A1 | cao; for 60 with no incorrect working award M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R \sim B(20, 0.05)\) | B1 | Using or writing \(B(20, 0.05)\) in (i) or (ii) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(R=4) = {}^{20}C_4(0.05)^4(0.95)^{16}\) OR \(P(R=4) = P(R \leq 4) - P(R \leq 3) = 0.9974 - 0.9841\) | M1 | Writing or using \(P(R \leq 4) - P(R \leq 3)\) or using \({}^{20}C_4(p)^4(1-p)^{16}\) |
| \(= 0.0133\) | A1 | awrt 0.0133 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(R \geq 4) = 1 - P(R \leq 3) = 1 - 0.9841 = 0.0159\) | M1 | Writing or using \(1 - P(R \leq 3)\) |
| \(= 0.0159\) | A1 | awrt 0.0159 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: p = 0.05\), \(H_1: p > 0.05\) | B1 | Both hypotheses correct and labelled \(H_0\) and \(H_1\); must use \(p\) or \(\pi\); do not allow \(p(x)\) |
| \(P(R \geq 4) = 1 - P(R \leq 3)\) | M1 | Writing or using \(B(50, 0.05)\) AND writing or using \(1 - P(R \leq 3)\) or \(P(R \leq 3) = 0.7604\) on its own or one of: \(P(R \geq 7) = 0.0118\); \(P(R \leq 6) = 0.9882\); \(P(R \geq 8) = 0.0032\); \(P(R \leq 7) = 0.9968\) |
| \(= 0.2396\), CR \(R \geq 8\) | A1 | awrt 0.240 or 0.24 or \(R \geq 8\) oe or 0.7604 |
| Insufficient evidence to reject \(H_0\), Not Significant. Accept \(H_0\). 4 does not lie in the Critical Region. | M1d | Dependent on previous M; correct statement; follow through their probability/CR; ignore their comparison in all cases; mentally compare probability: for prob \(< 0.5\) compare to 0.01 (1-tail) or 0.005 (2-tail); for prob \(> 0.5\) compare to 0.99 (1-tail) or 0.995 (2-tail) |
| No evidence to support Patrick's claim. Or: no evidence that people in *Reddman* have a probability greater than 5% of having red hair | A1cso | Fully correct solution and correct contextual statement containing the word Patrick if writing about the claim, or red hair if full context |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.05n = 3$ | M1 | Using $0.05n$ |
| $n = 60$ | A1 | cao; for 60 with no incorrect working award M1A1 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R \sim B(20, 0.05)$ | B1 | Using or writing $B(20, 0.05)$ in (i) or (ii) |
### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(R=4) = {}^{20}C_4(0.05)^4(0.95)^{16}$ **OR** $P(R=4) = P(R \leq 4) - P(R \leq 3) = 0.9974 - 0.9841$ | M1 | Writing or using $P(R \leq 4) - P(R \leq 3)$ or using ${}^{20}C_4(p)^4(1-p)^{16}$ |
| $= 0.0133$ | A1 | awrt 0.0133 |
### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(R \geq 4) = 1 - P(R \leq 3) = 1 - 0.9841 = 0.0159$ | M1 | Writing or using $1 - P(R \leq 3)$ |
| $= 0.0159$ | A1 | awrt 0.0159 |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.05$, $H_1: p > 0.05$ | B1 | Both hypotheses correct and labelled $H_0$ and $H_1$; must use $p$ or $\pi$; do not allow $p(x)$ |
| $P(R \geq 4) = 1 - P(R \leq 3)$ | M1 | Writing or using $B(50, 0.05)$ AND writing or using $1 - P(R \leq 3)$ **or** $P(R \leq 3) = 0.7604$ on its own **or** one of: $P(R \geq 7) = 0.0118$; $P(R \leq 6) = 0.9882$; $P(R \geq 8) = 0.0032$; $P(R \leq 7) = 0.9968$ |
| $= 0.2396$, CR $R \geq 8$ | A1 | awrt 0.240 or 0.24 or $R \geq 8$ oe or 0.7604 |
| Insufficient evidence to reject $H_0$, Not Significant. Accept $H_0$. 4 does not lie in the Critical Region. | M1d | Dependent on previous M; correct statement; follow through their probability/CR; ignore their comparison in all cases; mentally compare probability: for prob $< 0.5$ compare to 0.01 (1-tail) or 0.005 (2-tail); for prob $> 0.5$ compare to 0.99 (1-tail) or 0.995 (2-tail) |
| No evidence to support **Patrick's** claim. Or: no evidence that people in *Reddman* have a probability greater than 5% of having **red hair** | A1cso | Fully correct solution and correct contextual statement containing the word **Patrick** if writing about the claim, or **red hair** if full context |
---\begin{enumerate}
\item In a region of the UK, $5 \%$ of people have red hair. In a random sample of size $n$, taken from this region, the expected number of people with red hair is 3\\
(a) Calculate the value of $n$.
\end{enumerate}
A random sample of 20 people is taken from this region. Find the probability that\\
(b) (i) exactly 4 of these people have red hair,\\
(ii) at least 4 of these people have red hair.
Patrick claims that Reddman people have a probability greater than $5 \%$ of having red hair. In a random sample of 50 Reddman people, 4 of them have red hair.\\
(c) Stating your hypotheses clearly, test Patrick's claim. Use a $1 \%$ level of significance.\\
\href{http://PhysicsAndMathsTutor.com}{PhysicsAndMathsTutor.com}
\hfill \mbox{\textit{Edexcel S2 2016 Q2 [12]}}