CAIE S2 2013 June — Question 2 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2013
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeOne-tailed hypothesis test (upper tail, H₁: p > p₀)
DifficultyModerate -0.3 This is a straightforward one-tailed binomial hypothesis test with clear parameters (n=20, x=19, p=0.82, α=0.10). Students must state the assumption of independence, set up H₀: p=0.82 vs H₁: p>0.82, and calculate P(X≥19). The calculation is routine and the question structure is standard for S2, making it slightly easier than average but still requiring proper statistical methodology.
Spec2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail5.05b Unbiased estimates: of population mean and variance

2 A hockey player found that she scored a goal on \(82 \%\) of her penalty shots. After attending a coaching course, she scored a goal on 19 out of 20 penalty shots. Making an assumption that should be stated, test at the 10\% significance level whether she has improved.

Question 2:
AnswerMarks Guidance
Assume shots independent OR prob of scoring constantB1 In context
\(H_0: P(\text{score}) = 0.82\); \(H_1: P(\text{score}) > 0.82\)B1 Both, allow \(p\)
\(20 \times 0.82^{19} \times 0.18 + 0.82^{20} = 0.102\) (3 sf); No evidence that improvedM1, A1, B1f [5] For use of \(\text{Bin}(20, 0.82)\) and either \(P(19)\) and/or \(P(20)\) attempted; valid comparison with 0.05 if \(H_1\ p \neq 0.82\); correct conclusion ft numerical errors in 0.102 only
Normal approx'n note: B1 B1 (\(\mu = 16.4\) acceptable), then \(CR = 1.222\) from \(\frac{18.5 - 20 \times 0.82}{\sqrt{20 \times 0.82 \times (1-0.82)}}\); comp \(z = 1.282\); No evidence that improved SC 1; Same scheme for proportions
## Question 2:

| Assume shots independent OR prob of scoring constant | B1 | In context |
| $H_0: P(\text{score}) = 0.82$; $H_1: P(\text{score}) > 0.82$ | B1 | Both, allow $p$ |
| $20 \times 0.82^{19} \times 0.18 + 0.82^{20} = 0.102$ (3 sf); No evidence that improved | M1, A1, B1f [5] | For use of $\text{Bin}(20, 0.82)$ and either $P(19)$ and/or $P(20)$ attempted; valid comparison with 0.05 if $H_1\ p \neq 0.82$; correct conclusion ft numerical errors in 0.102 only |

Normal approx'n note: B1 B1 ($\mu = 16.4$ acceptable), then $CR = 1.222$ from $\frac{18.5 - 20 \times 0.82}{\sqrt{20 \times 0.82 \times (1-0.82)}}$; comp $z = 1.282$; No evidence that improved SC 1; Same scheme for proportions

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2 A hockey player found that she scored a goal on $82 \%$ of her penalty shots. After attending a coaching course, she scored a goal on 19 out of 20 penalty shots. Making an assumption that should be stated, test at the 10\% significance level whether she has improved.

\hfill \mbox{\textit{CAIE S2 2013 Q2 [5]}}