A-Level Mathematics October/November 2024 Q7(b): She takes a random sample of 100 such trees in her region and measures their heights, h…

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  Let $\mu$ be the population mean height of the trees in the scientist's region.

  1. Hypotheses The null and alternative hypotheses are: $H_0: \mu = 2.3$ $H_1: \mu > 2.3$ This is a one-tailed test at the 2% significance level.

  2. Calculate Sample Statistics Sample mean, $\bar{h} = \frac{\Sigma h}{n} = \frac{238}{100} = 2.38$

  Unbiased estimate of population variance, $s^2$: $s^2 = \frac{1}{n-1} \left( \Sigma h^2 - \frac{(\Sigma h)^2}{n} \right)$ $s^2 = \frac{1}{99} \left( 580 - \frac{238^2}{100} \right)$ $s^2 = \frac{1}{99} (580 - 566.44) = \frac{13.56}{99} \approx 0.136969...$

  Unbiased estimate of population standard deviation, $s = \sqrt{0.136969...} \approx 0.37009...$

  3. Calculate the Test Statistic Since $n=100$ is large, the Central Limit Theorem applies, and we can use a Z-test. The test statistic is calculated as: $z = \frac{\bar{h} - \mu}{s/\sqrt{n}}$ $z = \frac{2.38 - 2.3}{0.37009... / \sqrt{100}} = \frac{0.08}{0.037009...} \approx 2.1616...$ $z \approx 2.16$ (to 3 s.f.)

  4. Determine the Critical Region For a one-tailed test at the 2% significance level, we need to find the critical value $z_{crit}$ such that $P(Z > z_{crit}) = 0.02$. This means $\Phi(z_{crit}) = 1 - 0.02 = 0.98$. From the standard normal distribution tables, the critical value is $z_{crit} = 2.054$. The critical region is $z > 2.054$.

  5. Compare and Conclude Comparing the test statistic with the critical value: $2.16 > 2.054$

  The test statistic falls within the critical region, so we reject the null hypothesis, $H_0$.

  There is sufficient evidence at the 2% significance level to suggest that the mean height of the trees in the scientist's region is greater than 2.3 m. The scientist's claim is supported.

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  How the marks are awarded

  • B1 — Correctly stating both the null hypothesis ($H_0: \mu = 2.3$) and the one-tailed alternative hypothesis ($H_1: \mu > 2.3$).
  • M1 — Correctly substituting the summary statistics into the formula for the unbiased estimate of population variance, $s^2 = \frac{1}{99} ( 580 - \frac{238^2}{100} )$.
  • A1 — Correctly calculating the sample mean $\bar{h} = 2.38$ and the unbiased estimate of the standard deviation $s \approx 0.370$ (or variance $s^2 \approx 0.137$).
  • M1 — Correctly substituting the sample mean, population mean, and estimated standard error into the standardisation formula for the test statistic: $z = \frac{2.38 - 2.3}{s/\sqrt{100}}$.
  • A1 — Obtaining the correct test statistic value of 2.16 (rounded to 3 significant figures).
  • M1 — Making a valid comparison between the calculated test statistic (2.16) and the correct critical value for a one-tailed 2% test (2.054).
  • A1 — Stating a correct, non-definite conclusion in the context of the problem, consistent with rejecting H₀. For example, stating there is sufficient evidence that the mean height is greater than 2.3m.

  Common mistakes

  • Using the biased estimate of variance, $\sigma^2 = \frac{\Sigma h^2}{n} - (\frac{\Sigma h}{n})^2$, instead of the unbiased estimate $s^2 = \frac{n}{n-1}\sigma^2$.
  • Performing a two-tailed test ($H_1: \mu \neq 2.3$) and comparing with the wrong critical value (e.g., z = 2.326 for a 2% two-tailed test).
  • Using an incorrect critical value, such as the one for a 5% test (1.645) or a 1% test (2.326), instead of the required 2% one-tailed value (2.054).
  • Writing a definitive conclusion, such as 'The mean height is greater than 2.3m', instead of a statistical one like 'There is sufficient evidence to suggest...'

  Examiner tip: Always clearly define your null and alternative hypotheses before calculating anything, as this determines whether you perform a one-tailed or two-tailed test and defines your critical region.

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