| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed test critical region |
| Difficulty | Moderate -0.3 This is a standard hypothesis testing question requiring application of binomial distribution tables to find a critical region at 5% significance level. While it involves multiple parts (finding critical region, Type I error probability, and drawing a conclusion), each step follows a routine procedure taught in S2 with no novel problem-solving required. The calculations are straightforward using tables, making it slightly easier than average for A-level statistics. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \leq 1) = (0.75)^{20} + 20(0.75)^{19}(0.25) = 0.0243\) | M1 A1 | Attempt correct expression |
| \(P(X = 2) = (0.75)^{20} + 20(0.75)^{19}(0.25) + {^{20}C_2}(0.75)^{18}(0.25)^2 = 0.0913\) or \(0.0912\) | M1 A1 | Attempt correct expression OR Find P(2); = 0.0669 or 0.0670 |
| Critical region is 0 or 1 pkt contain gift or < 2 pkts contain gift oe | A1 [5] | dep M1M1 & their P(X ≤ 1) < 0.05 < their P(X ≤ 2); (S.R. Use of Normal: N(5.3.75²) used B1 −1.645=(x + 0.5 − 5)/√3.75 M1 x < 1.31 A1 (3/5)) |
| Answer | Marks | Guidance |
|---|---|---|
| P(Type II error) = 0.0243 (3 sfs) | B1ft [1] | ft their P(X ≤ 1) dep < 0.05 ft Normal |
| Answer | Marks | Guidance |
|---|---|---|
| 2 is outside rej reg; No evidence to reject claim | M1 A1ft [2] | or P(X = 2) > 0.05; No contradictions |
**(i)**
$P(X \leq 1) = (0.75)^{20} + 20(0.75)^{19}(0.25) = 0.0243$ | M1 A1 | Attempt correct expression
$P(X = 2) = (0.75)^{20} + 20(0.75)^{19}(0.25) + {^{20}C_2}(0.75)^{18}(0.25)^2 = 0.0913$ or $0.0912$ | M1 A1 | Attempt correct expression OR Find P(2); = 0.0669 or 0.0670
Critical region is 0 or 1 pkt contain gift or < 2 pkts contain gift oe | A1 [5] | dep M1M1 & their P(X ≤ 1) < 0.05 < their P(X ≤ 2); (S.R. Use of Normal: N(5.3.75²) used B1 −1.645=(x + 0.5 − 5)/√3.75 M1 x < 1.31 A1 (3/5))
**(ii)**
P(Type II error) = 0.0243 (3 sfs) | B1ft [1] | ft their P(X ≤ 1) dep < 0.05 ft Normal
**(iii)**
2 is outside rej reg; No evidence to reject claim | M1 A1ft [2] | or P(X = 2) > 0.05; No contradictions4 A cereal manufacturer claims that $25 \%$ of cereal packets contain a free gift. Lola suspects that the true proportion is less than $25 \%$. In order to test the manufacturer's claim at the $5 \%$ significance level, she checks a random sample of 20 packets.\\
(i) Find the critical region for the test.\\
(ii) Hence find the probability of a Type I error.
Lola finds that 2 packets in her sample contain a free gift.\\
(iii) State, with a reason, the conclusion she should draw.
\hfill \mbox{\textit{CAIE S2 2012 Q4 [8]}}