CAIE S2 2003 June — Question 2 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2003
SessionJune
Marks5
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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 clearly stated hypotheses (p=0.6 vs p>0.6), small n=12, and standard 10% significance level. It requires only routine application of the binomial test procedure with no conceptual complications, making it slightly easier than average but still requiring proper statistical method.
Spec5.05b Unbiased estimates: of population mean and variance

2 Before attending a basketball course, a player found that \(60 \%\) of his shots made a score. After attending the course the player claimed he had improved. In his next game he tried 12 shots and scored in 10 of them. Assuming shots to be independent, test this claim at the \(10 \%\) significance level.

AnswerMarks Guidance
\(H_0: p = 0.6\)\(B1\) For correct \(H_0\) and \(H_1\)
\(H_1: p > 0.6\)\(B1\)
\(P(X = 10) = {}_{12}C_{10}0.6^{10}0.4^2 + {}_{12}C_{11}0.6^{11}0.4^1 + 0.6^{12} = 0.0834\)\(M1^*\) For one Bin term (\(n = 12, p = 0.6\))
\(M1^*dep\)For attempt \(X = 10, 11, 12\) or equiv.
\(A1\)For correct answer (or correct individual terms and digits showing 0.1)
Reject \(H_0\), i.e. accept claim at 10% level\(B1ft\) For correct conclusion
S.R. Use of Normal scores 4/5 max
AnswerMarks Guidance
\(z = \frac{9.5 - 7.2}{\sqrt{2.88}} = 1.3552\)\(M1\) Use of \(N(7.2, 2.88)\) or \(N(0.6, 0.24/12)\) and standardising with or without cc
(or equiv. Using \(N(0.6, 0.24/12)\)) For correct answer or \(1.3552\) and \(1.282\) seen
\(\Pr(>9.5) = 1 - 0.9123 = 0.0877\)\(A1\) For correct answer or \(1.3552\) and \(1.282\) seen
Reject \(H_0\), i.e. accept claim at 10% level\(B1ft\) For correct conclusion
$H_0: p = 0.6$ | $B1$ | For correct $H_0$ and $H_1$
$H_1: p > 0.6$ | $B1$ |

$P(X = 10) = {}_{12}C_{10}0.6^{10}0.4^2 + {}_{12}C_{11}0.6^{11}0.4^1 + 0.6^{12} = 0.0834$ | $M1^*$ | For one Bin term ($n = 12, p = 0.6$)
| $M1^*dep$ | For attempt $X = 10, 11, 12$ or equiv.
| $A1$ | For correct answer (or correct individual terms and digits showing 0.1)

Reject $H_0$, i.e. accept claim at 10% level | $B1ft$ | For correct conclusion

**S.R. Use of Normal scores 4/5 max**

$z = \frac{9.5 - 7.2}{\sqrt{2.88}} = 1.3552$ | $M1$ | Use of $N(7.2, 2.88)$ or $N(0.6, 0.24/12)$ and standardising with or without cc
(or equiv. Using $N(0.6, 0.24/12)$) | | For correct answer or $1.3552$ and $1.282$ seen

$\Pr(>9.5) = 1 - 0.9123 = 0.0877$ | $A1$ | For correct answer or $1.3552$ and $1.282$ seen
Reject $H_0$, i.e. accept claim at 10% level | $B1ft$ | For correct conclusion
2 Before attending a basketball course, a player found that $60 \%$ of his shots made a score. After attending the course the player claimed he had improved. In his next game he tried 12 shots and scored in 10 of them. Assuming shots to be independent, test this claim at the $10 \%$ significance level.

\hfill \mbox{\textit{CAIE S2 2003 Q2 [5]}}