A-Level Mathematics May/June 2024 Q6(a): The masses of cereal boxes filled by a certain machine have mean 510 grams. An adjustme…

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  Let $\mu$ be the population mean mass of cereal boxes in grams after the adjustment.

  1. Hypotheses: Null hypothesis, $H_0: \mu = 510$ Alternative hypothesis, $H_1: \mu < 510$

  This is a one-tailed test at the 2.5% significance level.

  2. Test Statistic: The population standard deviation is known ($\sigma = 10$) and the sample size is large ($n=120$), so we use a z-test. The distribution of the sample mean, $\bar{X}$, is approximately normal by the Central Limit Theorem. Under $H_0$, $\bar{X} \sim N(510, \frac{10^2}{120})$.

  The test statistic is calculated as: $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ $z = \frac{508 - 510}{\frac{10}{\sqrt{120}}}$ $z = \frac{-2}{0.91287...}$ $z = -2.19089... \approx -2.191$ (to 3 d.p.)

  3. Critical Value and Comparison: For a one-tailed test at the 2.5% significance level, the critical value is found from the standard normal distribution. Critical value = $\Phi^{-1}(0.025) = -1.96$

  We compare the test statistic with the critical value: $-2.191 < -1.96$

  The test statistic falls within the rejection region.

  4. Conclusion: Since the test statistic is in the rejection region, we reject the null hypothesis, $H_0$. There is sufficient evidence at the 2.5% significance level to suggest that the mean mass of cereal boxes filled by the machine has decreased.

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  How the marks are awarded

  • B1 — Correctly stating both the null hypothesis ($H_0: \mu = 510$) and the one-tailed alternative hypothesis ($H_1: \mu < 510$). Using the symbol 'μ' is required.
  • M1 — Correctly applying the standardisation formula for the test statistic, ensuring the standard deviation is divided by the square root of the sample size: $\frac{10}{\sqrt{120}}$.
  • A1 — Obtaining the correct value for the test statistic, $z = -2.191$ or $-2.190$.
  • M1 — Making a valid comparison between the calculated test statistic and the correct critical value. Here, comparing $-2.191$ with the one-tailed 2.5% critical value of $-1.96$.
  • A1FT — Stating a correct conclusion in context, based on the comparison. This includes rejecting H₀ and stating there is sufficient evidence that the mean mass has 'decreased', avoiding definitive language.

  Common mistakes

  • Incorrectly setting up a two-tailed test ($H_1: \mu \neq 510$), which leads to using the wrong critical value (e.g., $\pm 2.241$) and losing marks for the hypothesis and conclusion.
  • Forgetting to divide the standard deviation by the square root of the sample size, i.e., calculating $z = \frac{508 - 510}{10}$, which gives an incorrect test statistic.
  • Making an error with the inequality when comparing negative numbers, for example, incorrectly stating that $-2.191 > -1.96$, which leads to the wrong conclusion.
  • Writing a definitive conclusion, such as 'The mean mass has decreased', instead of a nuanced one like 'There is sufficient evidence to suggest the mean mass has decreased'.

  Examiner tip: Always identify whether a hypothesis test is one-tailed (e.g., 'decreased', 'increased') or two-tailed (e.g., 'changed') from the question's wording, as this determines both the alternative hypothesis and the critical region.

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