| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Two-tailed test critical region |
| Difficulty | Standard +0.3 This is a standard S2 hypothesis testing question requiring a two-tailed binomial test, finding critical regions from tables, and applying normal approximation to a larger sample. While it has multiple parts (9 marks total), each step follows routine procedures taught in S2 with no novel problem-solving required—slightly easier than average A-level difficulty. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: p = 0.2, \quad H_1: p \neq 0.2\) | B1B1 | |
| \(P(X \geq 9) = 1 - P(X \leq 8)\) | M1 | or attempt critical value/region |
| \(= 1 - 0.9900 = 0.01\), \(\quad\) CR \(X \geq 9\) | ||
| \(0.01 < 0.025\) or \(9 \geq 9\) or \(0.99 > 0.975\) or \(0.02 < 0.05\) or lies in interval with correct interval stated | A1 | |
| Evidence that the percentage of pupils that read Deano is not 20% | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim \text{Bin}(20, 0.2)\) | B1 | may be implied or seen in (i) or (ii) |
| So 0 or \([9, 20]\) make test significant | B1B1B1 | 0, 9; between "their 9" and 20 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: p = 0.2, \quad H_1: p \neq 0.2\) | B1 | |
| \(W \sim \text{Bin}(100, 0.2)\) | ||
| \(W \sim N(20, 16)\) | B1; B1 | normal; 20 and 16 |
| \(P(X \leq 18) = P\!\left(Z \leq \dfrac{18.5 - 20}{4}\right)\) or \(\dfrac{x + \frac{1}{2} - 20}{4} = \pm 1.96\) | M1M1A1 | \(\pm\)cc, standardise; or use z value, standardise |
| \(= P(Z \leq -0.375)\) | ||
| \(= 0.352 - 0.354\) | A1 | CR \(X < 12.16\) or \(11.66\) for \(\frac{1}{2}\) |
| \([0.352 > 0.025\) or \(18 > 12.16\) therefore insufficient evidence to reject \(H_0]\) | ||
| Combined numbers of Deano readers suggests 20% of pupils read Deano | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Conclusion that they are different | B1 | |
| Either large sample size gives better result; Or looks as though they are not all drawn from the same population | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: p = 0.2, \quad H_1: p > 0.2\) | B1B0 | |
| \(P(X \geq 9) = 1 - P(X \leq 8) = 0.01\), CR \(X \geq 8\) | M1, A0 | |
| \(0.01 < 0.05\) or \(9 \geq 8\), evidence percentage of pupils that read Deano is not 20% | A1 | |
| \(X \sim \text{Bin}(20, 0.2)\); So 0 or \([8,20]\) make test significant | B1; B1B0B1 | 0, 9; between "their 8" and 20 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: p = 0.2, \quad H_1: p < 0.2\) | B1\(\checkmark\) | |
| \(W \sim N(20, 16)\) | B1; B1 | normal; 20 and 16 |
| \(P(X \leq 18) = P\!\left(Z \leq \dfrac{18.5-20}{4}\right)\) or \(\dfrac{x-20}{4} = -1.6449\) | M1M1A1 | \(\pm\)cc, standardise; or standardise, use z value |
| \(= P(Z \leq -0.375) = 0.3520\) | A1 | CR \(X < 13.4\) or \(12.9\); awrt \(0.352\) |
| \([0.352 > 0.05\) or \(18 > 13.4\) therefore insufficient evidence to reject \(H_0]\) | ||
| Combined numbers of Deano readers suggests 20% of pupils read Deano | A1 |
## Question 7(a)(i) [Two-tailed test]:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.2, \quad H_1: p \neq 0.2$ | B1B1 | |
| $P(X \geq 9) = 1 - P(X \leq 8)$ | M1 | or attempt critical value/region |
| $= 1 - 0.9900 = 0.01$, $\quad$ CR $X \geq 9$ | | |
| $0.01 < 0.025$ or $9 \geq 9$ or $0.99 > 0.975$ or $0.02 < 0.05$ or lies in interval with correct interval stated | A1 | |
| Evidence that the percentage of pupils that read Deano is not 20% | A1 | |
## Question 7(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim \text{Bin}(20, 0.2)$ | B1 | may be implied or seen in (i) or (ii) |
| So 0 or $[9, 20]$ make test significant | B1B1B1 | 0, 9; between "their 9" and 20 |
**Total part (a): (9)**
## Question 7(b) [Two-tailed, Normal Approximation]:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.2, \quad H_1: p \neq 0.2$ | B1 | |
| $W \sim \text{Bin}(100, 0.2)$ | | |
| $W \sim N(20, 16)$ | B1; B1 | normal; 20 and 16 |
| $P(X \leq 18) = P\!\left(Z \leq \dfrac{18.5 - 20}{4}\right)$ or $\dfrac{x + \frac{1}{2} - 20}{4} = \pm 1.96$ | M1M1A1 | $\pm$cc, standardise; or use z value, standardise |
| $= P(Z \leq -0.375)$ | | |
| $= 0.352 - 0.354$ | A1 | CR $X < 12.16$ or $11.66$ for $\frac{1}{2}$ |
| $[0.352 > 0.025$ or $18 > 12.16$ therefore insufficient evidence to reject $H_0]$ | | |
| Combined numbers of Deano readers suggests 20% of pupils read Deano | A1 | |
**Total part (b): (8)**
## Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Conclusion that they are different | B1 | |
| Either large sample size gives better result; Or looks as though they are not all drawn from the same population | B1 | |
**Total part (c): (2)**
---
## Question 7(a)(i) [One-tailed version]:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.2, \quad H_1: p > 0.2$ | B1B0 | |
| $P(X \geq 9) = 1 - P(X \leq 8) = 0.01$, CR $X \geq 8$ | M1, A0 | |
| $0.01 < 0.05$ or $9 \geq 8$, evidence percentage of pupils that read Deano is not 20% | A1 | |
| $X \sim \text{Bin}(20, 0.2)$; So 0 or $[8,20]$ make test significant | B1; B1B0B1 | 0, 9; between "their 8" and 20 |
## Question 7(b) [One-tailed version]:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.2, \quad H_1: p < 0.2$ | B1$\checkmark$ | |
| $W \sim N(20, 16)$ | B1; B1 | normal; 20 and 16 |
| $P(X \leq 18) = P\!\left(Z \leq \dfrac{18.5-20}{4}\right)$ or $\dfrac{x-20}{4} = -1.6449$ | M1M1A1 | $\pm$cc, standardise; or standardise, use z value |
| $= P(Z \leq -0.375) = 0.3520$ | A1 | CR $X < 13.4$ or $12.9$; awrt $0.352$ |
| $[0.352 > 0.05$ or $18 > 13.4$ therefore insufficient evidence to reject $H_0]$ | | |
| Combined numbers of Deano readers suggests 20% of pupils read Deano | A1 | |
**Total: 19 marks**7. A teacher thinks that $20 \%$ of the pupils in a school read the Deano comic regularly.
He chooses 20 pupils at random and finds 9 of them read the Deano.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Test, at the $5 \%$ level of significance, whether or not there is evidence that the percentage of pupils that read the Deano is different from 20\%. State your hypotheses clearly.
\item State all the possible numbers of pupils that read the Deano from a sample of size 20 that will make the test in part (a)(i) significant at the $5 \%$ level.\\
(9)
The teacher takes another 4 random samples of size 20 and they contain 1, 3, 1 and 4 pupils that read the Deano.
\end{enumerate}\item By combining all 5 samples and using a suitable approximation test, at the $5 \%$ level of significance, whether or not this provides evidence that the percentage of pupils in the school that read the Deano is different from 20\%.
\item Comment on your results for the tests in part (a) and part (b).
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2006 Q7 [19]}}