Moderate -0.3 This is a straightforward application of the normal approximation to the binomial distribution for a two-tailed hypothesis test. Students need to check conditions (np and nq both >5), apply continuity correction, standardize to find a z-score, and compare to critical values. While it requires multiple steps, each is routine and follows a standard algorithm taught explicitly in S2 with no novel problem-solving required.
2 Harry has a five-sided spinner with sectors coloured blue, green, red, yellow and black. Harry thinks the spinner may be biased. He plans to carry out a hypothesis test with the following hypotheses.
$$\begin{aligned}
& \mathrm { H } _ { 0 } : \mathrm { P } ( \text { the spinner lands on blue } ) = \frac { 1 } { 5 } \\
& \mathrm { H } _ { 1 } : \mathrm { P } ( \text { the spinner lands on blue } ) \neq \frac { 1 } { 5 }
\end{aligned}$$
Harry spins the spinner 300 times. It lands on blue on 45 spins.
Use a suitable approximation to carry out Harry's test at the \(5 \%\) significance level.
[Evidence to reject \(H_0\)] There is evidence that \(P(\text{landing on blue}) \neq \frac{1}{5}\)
A1 FT
Allow 'There is evidence that the spinner is biased.' In context, not definite, no contradictions. Condone critical values method (critical value 46.42, M1 A1 and \(45.5 < '46.42'\) M1 for valid comparison, A1 for correct conclusion). SC: \(0.0182\) unsupported: \(0.0182 < 0.025\) and there is evidence that the spinner is biased. In context, not definite B1 only
5
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $B\!\left(300, \frac{1}{5}\right) \to N(60, 48)$ | B1 | SOI |
| $\frac{45.5 - 60}{\sqrt{48}}$ | M1 | Condone with wrong or no continuity correction |
| $= -2.093$ | A1 | |
| $'2.093' > 1.96$ | M1 | Valid comparison. Note: $\phi('-2.093') (= 0.0182)$, $0.0182 < 0.025$ |
| [Evidence to reject $H_0$] There is evidence that $P(\text{landing on blue}) \neq \frac{1}{5}$ | A1 FT | Allow 'There is evidence that the spinner is biased.' In context, not definite, no contradictions. Condone critical values method (critical value 46.42, M1 A1 and $45.5 < '46.42'$ M1 for valid comparison, A1 for correct conclusion). **SC:** $0.0182$ unsupported: $0.0182 < 0.025$ and there is evidence that the spinner is biased. In context, not definite B1 only |
| | **5** | |
---
2 Harry has a five-sided spinner with sectors coloured blue, green, red, yellow and black. Harry thinks the spinner may be biased. He plans to carry out a hypothesis test with the following hypotheses.
$$\begin{aligned}
& \mathrm { H } _ { 0 } : \mathrm { P } ( \text { the spinner lands on blue } ) = \frac { 1 } { 5 } \\
& \mathrm { H } _ { 1 } : \mathrm { P } ( \text { the spinner lands on blue } ) \neq \frac { 1 } { 5 }
\end{aligned}$$
Harry spins the spinner 300 times. It lands on blue on 45 spins.\\
Use a suitable approximation to carry out Harry's test at the $5 \%$ significance level.\\
\hfill \mbox{\textit{CAIE S2 2022 Q2 [5]}}