Moderate -0.8 This is a straightforward one-tailed binomial hypothesis test with clearly stated hypotheses and significance level. Students need to calculate P(X ≤ 25) under H₀: p = 1/6, n = 200 and compare to 10%. The setup is explicit with no ambiguity, making it easier than average—essentially a direct application of the standard procedure with normal approximation or tables.
1 A six-sided die shows a six on 25 throws out of 200 throws. Test at the \(10 \%\) significance level the null hypothesis: P (throwing a six) \(= \frac { 1 } { 6 }\), against the alternative hypothesis: P (throwing a six) \(< \frac { 1 } { 6 }\).
Evidence to reject \(H_0\); There is some evidence that \(p < \frac{1}{6}\), or e.g. It is likely that \(p < \frac{1}{6}\)
A1ft [5]
No contradictions
## Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $B(200, \frac{1}{6}) \rightarrow N(\frac{100}{3}, \frac{250}{9})$ | B1 | seen or implied |
| $\frac{25.5 - \frac{100}{3}}{\sqrt{\frac{250}{9}}}$ | M1 | allow with wrong or no continuity correction |
| $= -1.486$ | A1 | Accept alternative correct methods |
| comp '1.486' with 1.282 | M1 | or comp ('1.486') with 0.1 |
| Evidence to reject $H_0$; There is some evidence that $p < \frac{1}{6}$, or e.g. It is likely that $p < \frac{1}{6}$ | A1ft [5] | No contradictions |
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1 A six-sided die shows a six on 25 throws out of 200 throws. Test at the $10 \%$ significance level the null hypothesis: P (throwing a six) $= \frac { 1 } { 6 }$, against the alternative hypothesis: P (throwing a six) $< \frac { 1 } { 6 }$.
\hfill \mbox{\textit{CAIE S2 2016 Q1 [5]}}