A-Level Mathematics May/June 2024 Q6(b): Later the inspector carries out a similar test at the 2.5% significance level, using th…

1 answer

• ▲0
  accepted ✓

  A Type II error occurs when the null hypothesis, $H_0$, is not rejected even though it is false. This means the sample mean, $\bar{X}$, falls into the acceptance region of the original test.

  First, we must find the acceptance region for the test on $H_0: \mu = 510$ against $H_1: \mu < 510$ at the 2.5% significance level.

  The population distribution is $X \sim N(510, 10^2)$. The sampling distribution of the mean is $\bar{X} \sim N\left(510, \frac{10^2}{120}\right)$.

  For a 2.5% significance level in a one-tailed test, the critical z-value is found from $\Phi(z) = 0.025$, which gives $z = -1.96$.

  We find the critical value for the sample mean, $\bar{x}_{crit}$: $\frac{\bar{x}_{crit} - 510}{\frac{10}{\sqrt{120}}} = -1.96$ $\bar{x}_{crit} = 510 - 1.96 \times \frac{10}{\sqrt{120}}$ $\bar{x}_{crit} = 510 - 1.7892... = 508.2107... \approx 508.21$ The null hypothesis is not rejected if $\bar{X} > 508.21$. This is the acceptance region.

  Now, we find the probability of a Type II error. This is the probability that $\bar{X}$ falls into the acceptance region, given that the mean is now actually $\mu = 506$.

  We need to calculate $P(\bar{X} > 508.21 | \mu = 506)$.

  Under this new condition, the sampling distribution is $\bar{X} \sim N\left(506, \frac{10^2}{120}\right)$.

  Standardise using the new mean: $P\left(Z > \frac{508.2107... - 506}{\frac{10}{\sqrt{120}}}\right)$ $= P(Z > \frac{2.2107...}{0.91287...})$ $= P(Z > 2.4216...)$ This is equivalent to $1 - \Phi(2.422)$. $= 1 - 0.99228$ $= 0.00772$

  The probability of a Type II error is 0.0077 (to 2 s.f.).

  ---

  How the marks are awarded

  • M1 — For correctly standardising to find the critical value of the test. The calculation (cv - 510) / (10 / sqrt(120)) = -1.96 uses the null hypothesis mean (510), the correct standard error, and the correct z-value for a one-tailed 2.5% test.
  • A1 — For correctly calculating the critical value for the sample mean as 508.21.
  • M1 — For standardising again to find the Type II error probability. The expression (508.2107... - 506) / (10 / sqrt(120)) correctly uses the calculated critical value, the new actual mean (506), and the standard error.
  • M1 — For identifying the correct area for the probability calculation. The expression P(Z > 2.4216...) or 1 - Φ(2.422) shows understanding that a Type II error corresponds to the sample mean being in the acceptance region (greater than the critical value).
  • A1 — For the final correct probability of 0.00772, which is within the acceptable range of 0.0077 to 0.0080.

  Common mistakes

  • Using the z-value for a two-tailed test (z = -1.96 and 1.96) or the wrong significance level (e.g., z = -1.645 for 5%), leading to an incorrect critical value.
  • Using the original mean (μ = 510) instead of the new actual mean (μ = 506) when calculating the final probability for the Type II error.
  • Calculating the wrong tail area for the final probability, for example finding Φ(2.422) instead of 1 - Φ(2.422), resulting in an answer close to 1.
  • Forgetting to divide the standard deviation by the square root of n, i.e., using σ = 10 instead of the standard error σ/√n = 10/√120 in the standardisation steps.

  Examiner tip: A Type II error calculation is a two-stage process: first, define the acceptance region based on the null hypothesis, then calculate the probability of landing in that region using the alternative hypothesis's true mean.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers

Your answer

Sign in to answer this question.