| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Calculate Type II error probability |
| Difficulty | Standard +0.3 This is a standard hypothesis test question covering routine procedures: one-tailed binomial test with critical value/p-value calculation, defining Type I error conceptually, and computing Type II error probability. All parts follow textbook methods with small sample size (n=9) making calculations straightforward. Slightly easier than average due to simple numbers and direct application of standard techniques. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c Significance levels: one-tail and two-tail5.02b Expectation and variance: discrete random variables5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: P(10) = 0.1\), \(H_1: P(10) > 0.1\) | B1 | Both. Allow '\(p\)' for \(P(10)\) |
| \(B(9, 0.1)\), \(P(X \geqslant 3) = 1 - (0.9^9 + 9 \times 0.9^8 \times 0.1 + {}^9C_2 \times 0.9^7 \times 0.1^2)\) | M1 | Allow one extra term in bracket |
| \(= 0.05297\ldots\) or \(0.053(0)\) | A1 | |
| comp \(0.01\) | M1 | Valid comparison. (comparison with 0.99 can recover previous M1 A1 for 0.9470) |
| No evidence (at 1% level) to reject \(H_0\). Claim not justified | A1ft | No contradictions |
| Answer | Marks |
|---|---|
| \(H_0\) not rejected oe | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \geqslant 4) = ``0.05297" - {}^9C_3 \times 0.9^6 \times 0.1^3\) | M1 | or \(1-(0.9^9 + 9 \times 0.9^8 \times 0.1 + {}^9C_2 \times 0.9^7 \times 0.1^2 + {}^9C_3 \times 0.9^6 \times 0.1^3)\) |
| \(= 0.00833\) | A1 | Note: 0.05297 and 0.00833 both needed in (i) or (iii) to justify CV |
| Hence crit value is 4 | B1 | Allow without working. Or in (i) may be implied by attempt at \(P(X < 4)\) below |
| \(B(9, 0.5)\), \(P(X < 4)\) | M1 | Stated or implied |
| \(= 0.5^9 + 9 \times 0.5^8 \times 0.5 + {}^9C_2 \times 0.5^7 \times 0.5^2 + {}^9C_3 \times 0.5^6 \times 0.5^3\) | M1 | Attempt \(P(X < 4)\) with \(p = 0.5\) |
| \(P(\text{Type II}) = 0.254\) (3 sf) | A1 |
## Question 7(i):
| $H_0: P(10) = 0.1$, $H_1: P(10) > 0.1$ | B1 | Both. Allow '$p$' for $P(10)$ |
| $B(9, 0.1)$, $P(X \geqslant 3) = 1 - (0.9^9 + 9 \times 0.9^8 \times 0.1 + {}^9C_2 \times 0.9^7 \times 0.1^2)$ | M1 | Allow one extra term in bracket |
| $= 0.05297\ldots$ or $0.053(0)$ | A1 | |
| comp $0.01$ | M1 | Valid comparison. (comparison with 0.99 can recover previous M1 A1 for 0.9470) |
| No evidence (at 1% level) to reject $H_0$. Claim not justified | A1ft | No contradictions |
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## Question 7(ii):
| $H_0$ not rejected oe | B1 | |
---
## Question 7(iii):
| $P(X \geqslant 4) = ``0.05297" - {}^9C_3 \times 0.9^6 \times 0.1^3$ | M1 | or $1-(0.9^9 + 9 \times 0.9^8 \times 0.1 + {}^9C_2 \times 0.9^7 \times 0.1^2 + {}^9C_3 \times 0.9^6 \times 0.1^3)$ |
| $= 0.00833$ | A1 | Note: 0.05297 and 0.00833 both needed in (i) or (iii) to justify CV |
| Hence crit value is 4 | B1 | Allow without working. Or in (i) may be implied by attempt at $P(X < 4)$ below |
| $B(9, 0.5)$, $P(X < 4)$ | M1 | Stated or implied |
| $= 0.5^9 + 9 \times 0.5^8 \times 0.5 + {}^9C_2 \times 0.5^7 \times 0.5^2 + {}^9C_3 \times 0.5^6 \times 0.5^3$ | M1 | Attempt $P(X < 4)$ with $p = 0.5$ |
| $P(\text{Type II}) = 0.254$ (3 sf) | A1 | |7 A ten-sided spinner has edges numbered $1,2,3,4,5,6,7,8,9,10$. Sanjeev claims that the spinner is biased so that it lands on the 10 more often than it would if it were unbiased. In an experiment, the spinner landed on the 10 in 3 out of 9 spins.\\
(i) Test at the $1 \%$ significance level whether Sanjeev's claim is justified.\\
(ii) Explain why a Type I error cannot have been made.\\
In fact the spinner is biased so that the probability that it will land on the 10 on any spin is 0.5 .\\
(iii) Another test at the $1 \%$ significance level, also based on 9 spins, is carried out. Calculate the probability of a Type II error.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE S2 2018 Q7 [12]}}