Standard +0.3 This is a straightforward one-tailed hypothesis test using normal approximation to binomial with continuity correction. Students must check np>5 and nq>5, set up H₀: p=0.35 vs H₁: p>0.35, apply continuity correction (49.5), calculate z-score, and compare to critical value (1.282 at 10% level). While it requires multiple steps, each is standard procedure taught explicitly in S2 with no novel problem-solving required, making it slightly easier than average.
7 Past experience shows that \(35 \%\) of the senior pupils in a large school know the regulations about bringing cars to school. The head teacher addresses this subject in an assembly, and afterwards a random sample of 120 senior pupils is selected. In this sample it is found that 50 of these pupils know the regulations. Use a suitable approximation to test, at the \(10 \%\) significance level, whether there is evidence that the proportion of senior pupils who know the regulations has increased. Justify your approximation.
Standardise with their \(np\) and \(\sqrt{npq}\), right cc. Allow both 49.5 and 50.5 and both in CR. \(z\) in range \([1.43, 1.44]\) before rounding. Comparison with 1.282, ft on \(z/p\) or \(\sqrt{120}\)
CV \(42.5 + z\sqrt{27.3}\), ignore LH, ft on \(np\), \(npq\). \(z = 1.282\) used in RH CV and compare 50. CV correct ft on \(z\), but don't worry about \(\geq\). Must round up: 49 from 49.2: A1A1A0
Reject \(H_0\)
M1
Consistent first conclusion, needs correct method and comparison. Can give M1A1 even if comparison not explicit. Allow from exact binomial
Significant evidence that proportion who know regulations has increased
A1 ft
Contextualised, needs "who know regulations" or "pupils", and "evidence"
\(np > 5\) \([= 42]\) from normal attempted
M1
From \(p = 0.35\) or 5/12, don't need 42. *or* \(n\) large or \(p\) close to 0.5 asserted
\(nq = 78 > 5\) and no others apart from \(n\) large
A1
Need 78, or 70 from 5/12, *not* \(npq\). *and* the other qualitative reason asserted
Total: 11
# Question 7:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: p = 0.35$; $H_1: p > 0.35$ | B2 | One error (e.g. $\mu$, no symbol, 2-tailed): B1, but $\bar{x}$, $t$ etc: B0. Allow $\pi$ |
| $B(120, 0.35)$ | M1 | $B(120, 0.35)$ stated or implied |
| $\approx N(42,\ 27.3)$ | M1 | $N(np,\ npq)$, their attempt at $120 \times 0.35$. $120 \times 0.35 \times 0.65$ *Not* $N(np,\ nq)$ |
| $\alpha$: $z = \dfrac{49.5 - 42}{\sqrt{27.3}} = 1.435 > 1.282$ [or $0.0757 < 0.1$] | A1 ft A1 ft | Standardise with their $np$ and $\sqrt{npq}$, right cc. Allow both 49.5 and 50.5 and both in CR. $z$ in range $[1.43, 1.44]$ before rounding. Comparison with 1.282, ft on $z/p$ or $\sqrt{120}$ |
| $\beta$: $CV = 42.5 + 1.282 \times \sqrt{27.3} = 49.198$; $z = 1.282$ and compare 50; $CR \geq 50$ or $\geq 49.2$ | A1 ft A1 A1 ft | CV $42.5 + z\sqrt{27.3}$, ignore LH, ft on $np$, $npq$. $z = 1.282$ used in RH CV and compare 50. CV correct ft on $z$, but don't worry about $\geq$. Must round up: 49 from 49.2: A1A1A0 |
| Reject $H_0$ | M1 | Consistent first conclusion, needs correct method and comparison. Can give M1A1 even if comparison not explicit. Allow from exact binomial |
| Significant evidence that proportion who know regulations has increased | A1 ft | Contextualised, needs "who know regulations" or "pupils", and "evidence" |
| $np > 5$ $[= 42]$ from normal attempted | M1 | From $p = 0.35$ or 5/12, don't need 42. *or* $n$ large or $p$ close to 0.5 asserted |
| $nq = 78 > 5$ and no others apart from $n$ large | A1 | Need 78, or 70 from 5/12, *not* $npq$. *and* the other qualitative reason asserted |
| | **Total: 11** | |
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7 Past experience shows that $35 \%$ of the senior pupils in a large school know the regulations about bringing cars to school. The head teacher addresses this subject in an assembly, and afterwards a random sample of 120 senior pupils is selected. In this sample it is found that 50 of these pupils know the regulations. Use a suitable approximation to test, at the $10 \%$ significance level, whether there is evidence that the proportion of senior pupils who know the regulations has increased. Justify your approximation.
\hfill \mbox{\textit{OCR S2 2013 Q7 [11]}}