| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed hypothesis test (upper tail, H₁: p > p₀) |
| Difficulty | Moderate -0.3 This is a standard one-tailed binomial hypothesis test with clearly stated context and significance level. Students must formulate H₀: p=0.3 vs H₁: p>0.3, find P(X≥18) where X~B(40,0.3), and compare to 0.05. While it requires proper hypothesis setup and binomial probability calculation (likely using tables), it's a routine S2 textbook exercise with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: p = 0.3\) | B1 | Must use \(p\) or \(\pi\) |
| \(H_1: p > 0.3\) | B1 | Must use \(p\) and be one tail |
| Let X represent the number of tomatoes greater than 4 cm: \(X \sim B(40, 0.3)\) | B1 | May be implied by calculation |
| \(P(X \geq 18) = 1 - P(X \leq 17)\) | M1 | Attempt to find \(1 - P(X \leq 17)\) or get a correct probability. For CR method must attempt to find \(P(X \geq 18)\) or give correct critical region |
| \(= 0.0320\) | A1 | awrt 0.032 or correct CR: \(X \geq 18\) |
| \(0.0320 < 0.05\), so reject \(H_0\) / significant | M1 | Correct statement based on probability, \(H_1\) and 0.05, or correct contextualised statement |
| New fertiliser has increased the probability of a tomato being greater than 4 cm / Dhriti's claim is true | B1 dep | Must use words: increased, tomato, and reference to size or diameter. Dependent on previous M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(X\) represent the number of sunflower plants more than 1.5m high; \(X \sim Po(10)\), \(\mu = 10\) | B1 | Mean = 10, may be implied in (i) or (ii) |
| \(P(8 \leq X \leq 13) = P(X \leq 13) - P(X \leq 7)\) | M1 | Attempting to find \(P(X \leq 13) - P(X \leq 7)\) |
| \(= 0.8645 - 0.2202 = 0.6443\) | A1 | awrt 0.644 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim N(10, 7.5)\) | A1 | |
| \(\sigma^2 = 7.5\) | B1 | May be implied by correct standardised formula |
| \(P(7.5 \leq X \leq 13.5) = P\!\left(\dfrac{7.5-10}{\sqrt{7.5}} \leq X \leq \dfrac{13.5-10}{\sqrt{7.5}}\right)\) | M1 | Using 7.5 or 8.5 or 12.5 or 13.5 |
| \(= P(-0.913 \leq X \leq 1.278)\) | M1 | Standardising using correct values and mean and SD. A1 for each z-value: awrt \(-0.91\) and awrt \(1.28\) |
| \(= 0.8997 - (1 - 0.8186)\) | M1 | Finding correct area; prob \(> 0.5\) – prob \(< 0.5\) or prob \(< 0.5\) + prob \(< 0.5\) |
| \(= 0.7183\) | A1 A1 | awrt 0.718 or 0.719. Dependent on all three method marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Normal approximation (not Poisson) | M1 | |
| Since \(n\) is large and \(p\) close to half, or \(np = 10\), \(npq = 7.5\) so mean \(\neq\) variance, or \(np (= 10)\) and \(nq (= 30)\) both \(> 5\) | A1 | Do not allow \(np > 5\) and \(npq > 5\) |
| or exact binomial \(= 0.7148\) | B1 dep |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| A hypothesis test is a mathematical procedure to examine a value of a population parameter proposed by the null hypothesis compared with an alternative hypothesis | B1 | Method for deciding between 2 hypotheses |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The critical region is the range of values or a test statistic or region where the test is significant | B1g | Range of values — may be implied by other words. Not "region" on its own |
| that would lead to the rejection of \(H_0\) | B1h | Which leads you to reject \(H_0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let X represent the number of incoming calls: \(X \sim Po(9)\) | B1 | Using \(P_o(9)\) |
| \(P(X \geq 16) = 0.0220\) | M1 A1 | Attempting to find \(P(X \geq 16)\) or \(P(x \leq 3)\); A1: 0.0220 or \(P(X \geq 16)\) |
| \(P(x \leq 3) = 0.0212\) | A1 | 0.0212 or \(P(x \leq 3)\) |
| Critical region: \(x \leq 3\) or \(x \geq 16\) | B1 | Correct critical region. A completely correct CR gets all 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Significance level \(= 0.0220 + 0.0212 = 0.0432\) or \(4.32\%\) | B1 | awrt 0.0432 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \lambda = 0.45\); \(H_1: \lambda < 0.45\) (accept \(H_0: \lambda = 4.5\); \(H_1: \lambda < 4.5\)) | B1 | May use \(\lambda\) or \(\mu\). Needs both \(H_0\) and \(H_1\) |
| Using \(X \sim Po(4.5)\) | M1 | Using \(P_o(4.5)\) |
| \(P(X \leq 1) = 0.0611\); CR: \(X \leq 0\) | A1 | Correct probability or CR only |
| \(0.0611 > 0.05\); \(1 \geq 0\) or 1 not in the critical region | M1 | Correct statement based on probability, \(H_1\) and 0.05, or correct contextualised statement |
| There is evidence to accept \(H_0\) / not significant | B1 cao | Must see word calls in conclusion |
| There is no evidence that there are fewer calls during school holidays | Conclusion in context |
# Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.3$ | B1 | Must use $p$ or $\pi$ |
| $H_1: p > 0.3$ | B1 | Must use $p$ and be one tail |
| Let X represent the number of tomatoes greater than 4 cm: $X \sim B(40, 0.3)$ | B1 | May be implied by calculation |
| $P(X \geq 18) = 1 - P(X \leq 17)$ | M1 | Attempt to find $1 - P(X \leq 17)$ or get a correct probability. For CR method must attempt to find $P(X \geq 18)$ or give correct critical region |
| $= 0.0320$ | A1 | awrt 0.032 or correct CR: $X \geq 18$ |
| $0.0320 < 0.05$, so reject $H_0$ / significant | M1 | Correct statement based on probability, $H_1$ and 0.05, or correct contextualised statement |
| New fertiliser has increased the probability of a tomato being greater than 4 cm / Dhriti's claim is true | B1 dep | Must use words: increased, tomato, and reference to size or diameter. Dependent on previous M1 |
**Two tail test would get: B1 B0 B1 M1 A1 M1 B0**
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# Question 6a(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $X$ represent the number of sunflower plants more than 1.5m high; $X \sim Po(10)$, $\mu = 10$ | B1 | Mean = 10, may be implied in (i) or (ii) |
| $P(8 \leq X \leq 13) = P(X \leq 13) - P(X \leq 7)$ | M1 | Attempting to find $P(X \leq 13) - P(X \leq 7)$ |
| $= 0.8645 - 0.2202 = 0.6443$ | A1 | awrt 0.644 |
# Question 6a(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim N(10, 7.5)$ | A1 | |
| $\sigma^2 = 7.5$ | B1 | May be implied by correct standardised formula |
| $P(7.5 \leq X \leq 13.5) = P\!\left(\dfrac{7.5-10}{\sqrt{7.5}} \leq X \leq \dfrac{13.5-10}{\sqrt{7.5}}\right)$ | M1 | Using 7.5 or 8.5 or 12.5 or 13.5 |
| $= P(-0.913 \leq X \leq 1.278)$ | M1 | Standardising using correct values and mean and SD. A1 for each z-value: awrt $-0.91$ and awrt $1.28$ |
| $= 0.8997 - (1 - 0.8186)$ | M1 | Finding correct area; prob $> 0.5$ – prob $< 0.5$ or prob $< 0.5$ + prob $< 0.5$ |
| $= 0.7183$ | A1 A1 | awrt 0.718 or 0.719. Dependent on all three method marks |
# Question 6b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Normal approximation (not Poisson) | M1 | |
| Since $n$ is large and $p$ close to half, **or** $np = 10$, $npq = 7.5$ so mean $\neq$ variance, **or** $np (= 10)$ and $nq (= 30)$ both $> 5$ | A1 | Do **not** allow $np > 5$ and $npq > 5$ |
| or exact binomial $= 0.7148$ | B1 dep | |
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# Question 7a(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| A hypothesis test is a mathematical procedure to examine a value of a population parameter proposed by the null hypothesis compared with an alternative hypothesis | B1 | Method for deciding between 2 hypotheses |
# Question 7a(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The critical region is the range of values or a test statistic or region where the test is significant | B1g | Range of values — may be implied by other words. Not "region" on its own |
| that would lead to the rejection of $H_0$ | B1h | Which leads you to reject $H_0$ |
# Question 7b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let X represent the number of incoming calls: $X \sim Po(9)$ | B1 | Using $P_o(9)$ |
| $P(X \geq 16) = 0.0220$ | M1 A1 | Attempting to find $P(X \geq 16)$ or $P(x \leq 3)$; A1: 0.0220 or $P(X \geq 16)$ |
| $P(x \leq 3) = 0.0212$ | A1 | 0.0212 or $P(x \leq 3)$ |
| Critical region: $x \leq 3$ or $x \geq 16$ | B1 | Correct critical region. A completely correct CR gets all 5 marks |
# Question 7c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Significance level $= 0.0220 + 0.0212 = 0.0432$ or $4.32\%$ | B1 | awrt 0.0432 |
# Question 7d:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \lambda = 0.45$; $H_1: \lambda < 0.45$ (accept $H_0: \lambda = 4.5$; $H_1: \lambda < 4.5$) | B1 | May use $\lambda$ or $\mu$. Needs both $H_0$ and $H_1$ |
| Using $X \sim Po(4.5)$ | M1 | Using $P_o(4.5)$ |
| $P(X \leq 1) = 0.0611$; CR: $X \leq 0$ | A1 | Correct probability or CR only |
| $0.0611 > 0.05$; $1 \geq 0$ or 1 not in the critical region | M1 | Correct statement based on probability, $H_1$ and 0.05, or correct contextualised statement |
| There is evidence to accept $H_0$ / not significant | B1 cao | Must see word **calls** in conclusion |
| There is no evidence that there are fewer calls during school holidays | | Conclusion in context |
\begin{enumerate}
\item Dhriti grows tomatoes. Over a period of time, she has found that there is a probability 0.3 of a ripe tomato having a diameter greater than 4 cm . She decides to try a new fertiliser. In a random sample of 40 ripe tomatoes, 18 have a diameter greater than 4 cm . Dhriti claims that the new fertiliser has increased the probability of a ripe tomato being greater than 4 cm in diameter.
\end{enumerate}
Test Dhriti's claim at the 5\% level of significance. State your hypotheses clearly.\\
\hfill \mbox{\textit{Edexcel S2 2008 Q5 [7]}}