| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Calculate Type I error probability |
| Difficulty | Standard +0.3 This is a straightforward application of Type I and Type II error definitions in hypothesis testing with a binomial distribution. Students need to identify the rejection region, then calculate probabilities under H₀ and H₁ using standard binomial calculations (n=6, small enough for direct computation). The conceptual understanding required is basic, and the calculations are routine for S2 level. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05a Hypothesis testing language: null, alternative, p-value, significance |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Type I}) = 1 - P(\geq 4\) assuming \(p = 0.7)\) | M1 | or \(P(\leq 3\) assuming \(p = 0.7)\) May be implied '\(^nC_3 \times 0.7^3 + {^nC_2 \times 0.7^2 \times 0.3} + {^nC_1 \times 0.7 \times 0.3^3} + 0.3^n\)' |
| \(1-(^4C_4 \times 0.7^4 \times 0.3 + {^nC_3 \times 0.7^3} + {^nC_2 \times 0.7^2 \times 0.3} + 0.7^n)\) | M1 | Allow one end error |
| \((= 1 - 0.744)\) | ||
| \(= 0.256\) (3 s.f.) | A1 [3] | \(= 0.256\) (3 s.f.) SR if zero scored allow B1 for use of B(6, 0.7) in any two or more terms |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Type II}) = P(\geq 4\) assuming \(p = 0.35)\) | M1 | May be implied |
| \(= {^6C_4 \times 0.35^4 \times 0.65^2 +}\) | M1 | Allow one end error |
| \({^nC_5 \times 0.35^5 \times 0.65 + 0.35^n}\) | ||
| \(= 0.117\) | A1 [3] | SR if zero scored allow B1 for use of B(6, 0.35) in any two or more terms |
| Answer | Marks | Guidance |
|---|---|---|
| Type 1 | B1 | |
| They will reject Luigi's belief, although it might be true. | B1 [2] | In context |
**(i)**
$P(\text{Type I}) = 1 - P(\geq 4$ assuming $p = 0.7)$ | M1 | or $P(\leq 3$ assuming $p = 0.7)$ May be implied '$^nC_3 \times 0.7^3 + {^nC_2 \times 0.7^2 \times 0.3} + {^nC_1 \times 0.7 \times 0.3^3} + 0.3^n$'
$1-(^4C_4 \times 0.7^4 \times 0.3 + {^nC_3 \times 0.7^3} + {^nC_2 \times 0.7^2 \times 0.3} + 0.7^n)$ | M1 | Allow one end error
$(= 1 - 0.744)$ | |
$= 0.256$ (3 s.f.) | A1 [3] | $= 0.256$ (3 s.f.) SR if zero scored allow B1 for use of B(6, 0.7) in any two or more terms
**(ii)**
$P(\text{Type II}) = P(\geq 4$ assuming $p = 0.35)$ | M1 | May be implied
$= {^6C_4 \times 0.35^4 \times 0.65^2 +}$ | M1 | Allow one end error
${^nC_5 \times 0.35^5 \times 0.65 + 0.35^n}$ | |
$= 0.117$ | A1 [3] | SR if zero scored allow B1 for use of B(6, 0.35) in any two or more terms
**(iii)**
Type 1 | B1 |
They will reject Luigi's belief, although it might be true. | B1 [2] | In context6 At the last election, 70\% of people in Apoli supported the president. Luigi believes that the same proportion support the president now. Maria believes that the proportion who support the president now is $35 \%$. In order to test who is right, they agree on a hypothesis test, taking Luigi's belief as the null hypothesis. They will ask 6 people from Apoli, chosen at random, and if more than 3 support the president they will accept Luigi's belief.\\
(i) Calculate the probability of a Type I error.\\
(ii) If Maria's belief is true, calculate the probability of a Type II error.\\
(iii) In fact 2 of the 6 people say that they support the president. State which error, Type I or Type II, might be made. Explain your answer.
\hfill \mbox{\textit{CAIE S2 2013 Q6 [8]}}