A-Level Mathematics May/June 2024 Q3: A student wishes to estimate the proportion, p, of students at her college who have exa…

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  The number of students in the sample is $n=50$. The number of students with exactly one brother is $x=18$.

  The sample proportion, $p̂$, is calculated as: $p̂ = \frac{x}{n} = \frac{18}{50} = 0.36$

  The formula for an approximate confidence interval for a proportion is $p̂ \pm z \sqrt{\frac{p̂(1-p̂)}{n}}$.

  We are given that the lower limit of the confidence interval is 0.244. We can set up an equation: $p̂ - z \sqrt{\frac{p̂(1-p̂)}{n}} = 0.244$

  Substitute the known values: $0.36 - z \sqrt{\frac{0.36(1-0.36)}{50}} = 0.244$ $0.36 - z \sqrt{\frac{0.36 \times 0.64}{50}} = 0.244$ $0.36 - z \sqrt{\frac{0.2304}{50}} = 0.244$ $0.36 - z \sqrt{0.004608} = 0.244$

  Now, solve for $z$: $z \sqrt{0.004608} = 0.36 - 0.244$ $z \times 0.067882... = 0.116$ $z = \frac{0.116}{0.067882...} = 1.7088...$

  This $z$-value is the critical value for the $\alpha\%$ confidence interval. The confidence level, as a proportion, is given by the area between $-z$ and $z$ under the standard normal curve, which is $P(-z < Z < z)$. This can be calculated as $2\Phi(z) - 1$.

  Using $z = 1.7088...$: Confidence Level $= 2\Phi(1.7088...) - 1$ Confidence Level $= 2(0.95626...) - 1$ Confidence Level $= 1.91252... - 1$ Confidence Level $= 0.91252...$

  The confidence level is $\alpha\%$. Therefore, $\frac{\alpha}{100} = 0.91252...$ $\alpha = 91.252...$

  Rounding to the nearest integer, we get: $\alpha = 91$

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  How the marks are awarded

  • M1 — Awarded for setting up the correct equation for the lower limit of the confidence interval, equating it to 0.244. This involves using the sample proportion (18/50) and the standard error formula, as shown in the first step: $0.36 - z \sqrt{\frac{0.36(0.64)}{50}} = 0.244$.
  • A1 — Awarded for correctly solving the equation to find the z-value. A value of 1.709, 1.708, or a more accurate value like 1.7088... is required.
  • M1 — Awarded for using the calculated z-value to find the confidence level. This requires finding the area under the standard normal curve using $\Phi(z)$ and applying the correct method for a two-tailed interval, such as $2\Phi(z) - 1$ or $1 - 2(1 - \Phi(z))$.
  • A1 — Awarded for the final correct value of $\alpha$, rounded to the nearest integer as requested. The final answer must be 91, not 91.2 or 0.91.

  Common mistakes

  • Confusing the one-tailed and two-tailed areas when converting the z-score to a percentage. For example, calculating $\Phi(1.709) \approx 0.956$ and incorrectly stating $\alpha = 96$.
  • Making a calculation error when solving for z, for instance, by miscalculating the standard error term $\sqrt{\frac{p̂(1-p̂)}{n}}$.
  • Using an incorrect formula, such as using $n$ instead of $\sqrt{n}$ in the denominator of the standard error, or forgetting to take the square root.
  • Failing to round the final answer to the nearest integer as specified in the question, for example, leaving the answer as 91.2 or 91.3.

  Examiner tip: This question rewards the ability to work backwards from a given confidence limit to find the confidence level, a process that requires rearranging the standard formula and correctly interpreting the resulting z-value in the context of the normal distribution.

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