| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed hypothesis test (lower tail, H₁: p < p₀) |
| Difficulty | Standard +0.3 This is a straightforward application of binomial hypothesis testing with standard procedures: identifying the distribution, calculating a probability using tables/formula, performing a one-tailed test at 5%, and finding a critical value at 1%. All steps are routine S2 content requiring no novel insight, though the multi-part structure and need for careful hypothesis statement elevate it slightly above the most basic exercises. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(X \sim B(10, p)\) | B1, B1 | Binomial(10, 0.75) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X=6) = 0.9219 - 0.7759 = 0.1460\) | M1 | \(P(X \le 6) - P(X \le 5)\) |
| A1 | 0.1460 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: p = 0.75\) (or \(p = 0.25\)) | B1 | Correct \(H_0\) |
| \(H_1: p < 0.75\) (or \(p > 0.25\)) | B1 | One tailed \(H_1\) |
| Under \(H_0\), \(X \sim B(20, 0.75)\) (or \(Y \sim B(20, 0.25)\)) | B1 | Implied |
| \(P(X \le 13) = 1 - 0.7858 = 0.2142\) (or \(P(Y \ge 7)\)) | M1, A1 | \(P(X \le 13)\) and \(1 -, 0.2142\) |
| Insufficient evidence to reject \(H_0\) as \(0.2412 > 0.05\). Doctor's belief is not supported by the sample | A1 | Context |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \le c) \le 0.01\) for \(p = 0.75\) (or \(P(Y \ge 20-c) \le 0.01\) for \(p = 0.25\)) | M1, A1 | |
| \(P(X \le 9) = 1 - 0.9961 = 0.0039\) (or \(P(Y \ge 11)\)) | 0.9961 or 0.9981 | |
| \(P(X \le 10) = 1 - 0.9861 = 0.0139\) (or \(P(Y \ge 10)\)) | B1 | |
| 13 outside critical region (or 7) | B1 | |
| C. R. is [0, 9], so greatest no. of patients is 9 |
**7(a)**
| $X \sim B(10, p)$ | B1, B1 | Binomial(10, 0.75) |
**7(b)**
| $P(X=6) = 0.9219 - 0.7759 = 0.1460$ | M1 | $P(X \le 6) - P(X \le 5)$ |
| | A1 | 0.1460 |
**7(c)**
| $H_0: p = 0.75$ (or $p = 0.25$) | B1 | Correct $H_0$ |
| $H_1: p < 0.75$ (or $p > 0.25$) | B1 | One tailed $H_1$ |
| Under $H_0$, $X \sim B(20, 0.75)$ (or $Y \sim B(20, 0.25)$) | B1 | Implied |
| $P(X \le 13) = 1 - 0.7858 = 0.2142$ (or $P(Y \ge 7)$) | M1, A1 | $P(X \le 13)$ and $1 -, 0.2142$ |
| Insufficient evidence to reject $H_0$ as $0.2412 > 0.05$. Doctor's belief is not supported by the sample | A1 | Context |
**7(d)**
| $P(X \le c) \le 0.01$ for $p = 0.75$ (or $P(Y \ge 20-c) \le 0.01$ for $p = 0.25$) | M1, A1 | |
| $P(X \le 9) = 1 - 0.9961 = 0.0039$ (or $P(Y \ge 11)$) | | 0.9961 or 0.9981 |
| $P(X \le 10) = 1 - 0.9861 = 0.0139$ (or $P(Y \ge 10)$) | | B1 |
| 13 outside critical region (or 7) | | B1 |
| C. R. is [0, 9], so greatest no. of patients is 9 | | |7. A drugs company claims that $75 \%$ of patients suffering from depression recover when treated with a new drug.
A random sample of 10 patients with depression is taken from a doctor's records.
\begin{enumerate}[label=(\alph*)]
\item Write down a suitable distribution to model the number of patients in this sample who recover when treated with the new drug.
Given that the claim is correct,
\item find the probability that the treatment will be successful for exactly 6 patients.
The doctor believes that the claim is incorrect and the percentage who will recover is lower. From her records she took a random sample of 20 patients who had been treated with the new drug. She found that 13 had recovered.
\item Stating your hypotheses clearly, test, at the $5 \%$ level of significance, the doctor's belief.
\item From a sample of size 20, find the greatest number of patients who need to recover for the test in part (c) to be significant at the $1 \%$ level.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2005 Q7 [14]}}