| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 17 |
| 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 straightforward S2 hypothesis testing question with standard binomial calculations. Part (a) is basic distribution identification, (b) requires simple binomial probability calculation, (c) is a routine one-tailed test following standard procedure, and (d) involves finding a new probability with p=0.6. All steps are textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks |
|---|---|
| Binomial, \(n = 4\), \(p = \frac{1}{2}\) | B2 |
| Fixed no. of coins flipped, 2 outcomes, \(p\) fixed | B2 |
| \(H \sim B(4, \frac{1}{2})\) | M1 |
| \(P(\text{more heads}) = P(H \geq 3)\) | M1 A1 |
| \(= 4(\frac{1}{2})^3(\frac{1}{2}) + (\frac{1}{2})^4\) | M1 A1 |
| \(= \frac{4}{16} + \frac{1}{16} = \frac{5}{16}\) | A1 |
| Let \(X =\) no. of times get more heads \(\therefore X \sim B(5, \frac{5}{16})\) | M1 |
| \(H_0 : p = \frac{5}{16}\) \(H_1 : p > \frac{5}{16}\) | B1 |
| \(P(X \geq 4) = 5(\frac{5}{16})^4(\frac{11}{16}) + (\frac{5}{16})^5\) | M1 |
| \(= 0.0358\) (3sf) | A1 |
| Less than 5% \(\therefore\) significant, evidence of higher prob. | A1 |
| \(P(\text{head}) : P(\text{tail}) = 1.5 : 1 = 3 : 2 \therefore P(\text{head}) = \frac{3}{5}\) | M1 |
| \(\therefore H \sim B(4, \frac{3}{5})\) | M1 A1 |
| \(P(H \geq 3) = 4(\frac{3}{5})^3(\frac{2}{5}) + (\frac{3}{5})^4\) | M1 A1 |
| \(= \frac{297}{625}\) or 0.4752 (4sf) | A1 |
Binomial, $n = 4$, $p = \frac{1}{2}$ | B2 |
Fixed no. of coins flipped, 2 outcomes, $p$ fixed | B2 |
$H \sim B(4, \frac{1}{2})$ | M1 |
$P(\text{more heads}) = P(H \geq 3)$ | M1 A1 |
$= 4(\frac{1}{2})^3(\frac{1}{2}) + (\frac{1}{2})^4$ | M1 A1 |
$= \frac{4}{16} + \frac{1}{16} = \frac{5}{16}$ | A1 |
Let $X =$ no. of times get more heads $\therefore X \sim B(5, \frac{5}{16})$ | M1 |
$H_0 : p = \frac{5}{16}$ $H_1 : p > \frac{5}{16}$ | B1 |
$P(X \geq 4) = 5(\frac{5}{16})^4(\frac{11}{16}) + (\frac{5}{16})^5$ | M1 |
$= 0.0358$ (3sf) | A1 |
Less than 5% $\therefore$ significant, evidence of higher prob. | A1 |
$P(\text{head}) : P(\text{tail}) = 1.5 : 1 = 3 : 2 \therefore P(\text{head}) = \frac{3}{5}$ | M1 |
$\therefore H \sim B(4, \frac{3}{5})$ | M1 A1 |
$P(H \geq 3) = 4(\frac{3}{5})^3(\frac{2}{5}) + (\frac{3}{5})^4$ | M1 A1 |
$= \frac{297}{625}$ or 0.4752 (4sf) | A1 |
---5. Four coins are flipped together and the random variable $H$ represents the number of heads obtained. Assuming that the coins are fair,
\begin{enumerate}[label=(\alph*)]
\item suggest with reasons a suitable distribution for modelling $H$ and give the value of any parameters needed,
\item show that the probability of obtaining more heads than tails is $\frac { 5 } { 16 }$.
The four coins are flipped 5 times and more heads are obtained than tails 4 times.
\item Stating your hypotheses clearly, test at the $5 \%$ level of significance whether or not there is evidence of the probability of getting more heads than tails being more than $\frac { 5 } { 16 }$.
Given that the four coins are all biased such that the chance of each one showing a head is 50\% more than the chance of it showing a tail,
\item find the probability of obtaining more heads than tails when the four coins are flipped together.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q5 [17]}}