A biologist wishes to test whether the mean concentration μ, in suitable units, of a certain pollutant in a river is below the permitted level of 0.5. She measures the concentration, x, of the pollutant at 50 randomly chosen locations in the river. The results are summarised below. n = 50, Σx = 23.0, Σx² = 13.02

Mark scheme

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

1. [B1] (Est (μ) = 23/50 = 0.46)
2. [M1] For an expression of the correct form for unbiased standard deviation or variance. (Est (σ) = sqrt( (50/49) * (13.02/50 - 0.46²) ) or Est (σ²) = (50/49) * (13.02/50 - 0.46²) oe Or estimated unbiased variance = (1/49) * (13.02 - (23.0)²/50))
3. [A1] (Est (σ) = 0.22315 or Est (σ²) = 0.0497959 = 61/1225 or 0.0498)
4. [M1] Standardising with their values. ((0.46 - 0.5) / ('0.22315'/sqrt(50)))
5. [A1] (= -1.268 or -1.267 or = -1.27 (3sf))
6. [M1] For a valid comparison. (-1.268 > -1.645 or 0.102 to 0.103 > 0.05)
7. [A1] FT In context, not definite. E.g., not 'Mean concentration is not less than 0.5'. No contradictions. ([Do not reject H₀] There is insufficient evidence [at 5% level] that the mean concentration is less than 0.5.)
8. [M1] (cv - 0.5) / ('0.22315' / sqrt(50)) = -1.645
9. [A1] (cv = 0.448(1) or 0.448 (3 sf))
10. [M1] ('0.448' - 0.4) / ('0.22315' / sqrt(50)) [= 1.521 to 1.524]
11. [M1] For area consistent with their working. (1 – Φ('1.524'))
12. [A1] (= 0.0638 to 0.0642)