| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Two-tailed hypothesis test |
| Difficulty | Moderate -0.3 This is a standard S2 hypothesis test question with routine application of binomial-to-normal approximation. Part (a) requires recall of binomial conditions, part (b) is a textbook one-tailed test with clear parameters (n=250, p=0.1, x=40), and part (c) asks for standard validity checks (np>5, n(1-p)>5). The calculations are straightforward with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Probability of success/failure is constant | B1 | |
| Trials are independent | B1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(p\) represent proportion of students who can distinguish between brands | ||
| \(H_0: p = 0.1\); \(H_1: p > 0.1\) | B1 | (both) |
| \(\alpha = 0.01\); CR: \(z > 2.3263\) | B1 | 2.3263 |
| \(np = 25\); \(npq = 22.5\) | B1 | both, can be implied |
| \(z = \frac{39.5 - 25}{\sqrt{22.5}} = 3.056\ldots\) | M1 | Standardisation with \(\pm 0.5\) and their \(\sqrt{npq}\) |
| A1 | AWRT 3.06 | |
| Reject \(H_0\): claim cannot be accepted | A1\(\checkmark\) | Based on clear evidence from \(z\)-test; (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. \(np\), \(nq\) both \(> 75\) — true so acceptable | B1 | |
| \(p\) close to \(0.5\) — not true, assumption not met; success/failure not clear cut necessarily; independence — one student influences another | B1 | (2 marks) |
| Alternative (b): \(z = 3.06 \Rightarrow p = 0.9989 > 0.99\) or \(p = 0.0011 < 0.01\) | B1 | equate to 2.3263 |
## Question 4:
**(a)**
Probability of success/failure is constant | B1 |
Trials are independent | B1 | (2 marks)
**(b)**
Let $p$ represent proportion of students who can distinguish between brands | |
$H_0: p = 0.1$; $H_1: p > 0.1$ | B1 | (both)
$\alpha = 0.01$; CR: $z > 2.3263$ | B1 | 2.3263
$np = 25$; $npq = 22.5$ | B1 | both, can be implied
$z = \frac{39.5 - 25}{\sqrt{22.5}} = 3.056\ldots$ | M1 | Standardisation with $\pm 0.5$ and their $\sqrt{npq}$
| A1 | AWRT 3.06
Reject $H_0$: claim cannot be accepted | A1$\checkmark$ | Based on clear evidence from $z$-test; (6 marks)
**(c)**
e.g. $np$, $nq$ both $> 75$ — true so acceptable | B1 |
$p$ close to $0.5$ — not true, assumption not met; success/failure not clear cut necessarily; independence — one student influences another | B1 | (2 marks)
**Alternative (b):** $z = 3.06 \Rightarrow p = 0.9989 > 0.99$ or $p = 0.0011 < 0.01$ | B1 | equate to 2.3263
---4. In an experiment, there are 250 trials and each trial results in a success or a failure.
\begin{enumerate}[label=(\alph*)]
\item Write down two other conditions needed to make this into a binomial experiment.
It is claimed that $10 \%$ of students can tell the difference between two brands of baked beans. In a random sample of 250 students, 40 of them were able to distinguish the difference between the two brands.
\item Using a normal approximation, test at the $1 \%$ level of significance whether or not the claim is justified. Use a one-tailed test.
\item Comment on the acceptability of the assumptions you needed to carry out the test.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2005 Q4 [10]}}