The masses, in kilograms, of newborn babies in country A are represented by the random variable X, with mean u and variance σ². The masses of a random sample of 500 newborn babies in this country were found and the results are summarised below. n = 500 Σχ = 1625 Σχ² = 5663.5

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [1] Est (μ) = 3.25 = 13/4 or 1625/500
2. [1] Est(σ²) = 500/499 (5663.5/500 - "3.25"²) or 1/499 (5663.5 - 500 × "3.25"²) Expression of correct form.
3. [1] = 0.766 (3 sf) or 1529/1996 Biased variance of 0.7645 scores M0A0.
4. [1] Ho: Pop mean (or μ) = ‘3.25' H1: Pop mean (or μ) ≠ ‘3.25' Not just 'mean'. FT their 3.25.
5. [1] (2.95-"3.25") / √("0.766"/60) Standardising with their values. Must have √60.
6. [1] = -2.655 Or P(X < 2.95) = 0.0039 or 0.00396 or 0.00397. SC FT their biased est(σ²), i.e. 0.7645 to give z = 2.658 A1.
7. [1] '2.655' > 2.576 or '-2.655' <-2.576 For valid comparison, e.g. 0.0039 or 0.00396 or 0.00397 < 0.005, or 0.0078 < 0.01, or 0.00792 < 0.01.
8. [1] [Reject Ho] There is evidence that (mean) mass in (country B) is different (from country A). OE. Must be in context and not definite, e.g., not ‘Mean mass is not different', No contradictions. Context needs either 'mass' or 'countries' OE. SC, Use of one-tail test. '2.655' > 2.326 or 0.0039 < 0.01 M1A0 (Max B0M1A1M1A0 3/5). Accept critical value method. Either: Xcrit=2.959 M1A1 2.95<2.959 M1A1FT with correct conclusion, or Xcrit=3.241 M1A1 3.25>3.241 M1A1FT with correct conclusion.