| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2004 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Critique inappropriate sampling methods |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question covering standard S2 content. Part (i) requires basic understanding of sampling bias (non-random selection), part (ii) is a routine binomial test with critical region calculation, part (iii) asks for standard Type II error definition, and part (iv) involves a simple probability calculation. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Not random, could be more light etc. | B1 1 | Any sensible reason |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| One-tailed test | B1 | For correct answer |
| \(H_0: p = 0.35\) | B1 | For \(H_0\) and \(H_1\) |
| \(H_1: p > 0.35\) | ||
| \(P(8) = 0.35^8 = 0.000225\) | M1* | For attempt at any Binomial expression \(P(0) - P(8)\) |
| \(P(7) = 0.35^7 \times 0.65^1 \times {}_8C_7 = 0.0033456\) | M1 | For summing probabilities starting at \(P(8)\) and working backwards until \(> 0.05\) (or equiv.) |
| \(P(6) = 0.35^6 \times 0.65^2 \times {}_8C_6 = 0.02174\) | ||
| \(P(5) = 0.35^5 \times 0.65^3 \times {}_8C_5 = 0.08077\) | ||
| Critical region is \(6, 7, 8\) survive | A1 | For correct answer |
| \(4\) is not in CR (OR \(\Pr(\geq 4) = 0.294\) and comparison \(0.5\)/or equiv.) | M1*dep | For deciding whether 4 is in their CR or not OR finding relevant prob and showing comparison |
| \(\Rightarrow\) no significant improvement in survival rate | A1ft 7 | For correct conclusion (ft from their critical region) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Saying no improvement when there is | B1 1 | Or equivalent, relating to the question |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Need \(P(0, 1, 2, 3, 4, 5)\) or \(1 - P(6, 7, 8)\) | M1 | For identifying type II error |
| \(P(8) = 0.4^8\ (= 0.0006554)\) | ||
| \(P(7) = 0.4^7 \times 0.6 \times {}_8C_7\ (= 0.007864)\) | ||
| \(P(6) = 0.4^6 \times 0.6^2 \times {}_8C_6\ (= 0.04128)\) | ||
| \(1 - (0.4^8 + 0.4^7 \times 0.6 \times {}_8C_7 + 0.4^6 \times 0.6^2 \times {}_8C_6)\) | ||
| \(= 0.950\) | A1 2 | For correct answer |
# Question 7:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Not random, could be more light etc. | B1 **1** | Any sensible reason |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| One-tailed test | B1 | For correct answer |
| $H_0: p = 0.35$ | B1 | For $H_0$ and $H_1$ |
| $H_1: p > 0.35$ | | |
| $P(8) = 0.35^8 = 0.000225$ | M1* | For attempt at any Binomial expression $P(0) - P(8)$ |
| $P(7) = 0.35^7 \times 0.65^1 \times {}_8C_7 = 0.0033456$ | M1 | For summing probabilities starting at $P(8)$ and working backwards until $> 0.05$ (or equiv.) |
| $P(6) = 0.35^6 \times 0.65^2 \times {}_8C_6 = 0.02174$ | | |
| $P(5) = 0.35^5 \times 0.65^3 \times {}_8C_5 = 0.08077$ | | |
| Critical region is $6, 7, 8$ survive | A1 | For correct answer |
| $4$ is not in CR (**OR** $\Pr(\geq 4) = 0.294$ and comparison $0.5$/or equiv.) | M1*dep | For deciding whether 4 is in their CR or not OR finding relevant prob and showing comparison |
| $\Rightarrow$ no significant improvement in survival rate | A1ft **7** | For correct conclusion (ft from their critical region) |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Saying no improvement when there is | B1 **1** | Or equivalent, relating to the question |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Need $P(0, 1, 2, 3, 4, 5)$ or $1 - P(6, 7, 8)$ | M1 | For identifying type II error |
| $P(8) = 0.4^8\ (= 0.0006554)$ | | |
| $P(7) = 0.4^7 \times 0.6 \times {}_8C_7\ (= 0.007864)$ | | |
| $P(6) = 0.4^6 \times 0.6^2 \times {}_8C_6\ (= 0.04128)$ | | |
| $1 - (0.4^8 + 0.4^7 \times 0.6 \times {}_8C_7 + 0.4^6 \times 0.6^2 \times {}_8C_6)$ | | |
| $= 0.950$ | A1 **2** | For correct answer |7 In a research laboratory where plants are studied, the probability of a certain type of plant surviving was 0.35 . The laboratory manager changed the growing conditions and wished to test whether the probability of a plant surviving had increased.\\
(i) The plants were grown in rows, and when the manager requested a random sample of 8 plants to be taken, the technician took all 8 plants from the front row. Explain what was wrong with the technician's sample.\\
(ii) A suitable sample of 8 plants was taken and 4 of these 8 plants survived. State whether the manager's test is one-tailed or two-tailed and also state the null and alternative hypotheses. Using a $5 \%$ significance level, find the critical region and carry out the test.\\
(iii) State the meaning of a Type II error in the context of the test in part (ii).\\
(iv) Find the probability of a Type II error for the test in part (ii) if the probability of a plant surviving is now 0.4.
\hfill \mbox{\textit{CAIE S2 2004 Q7 [11]}}