A.6 Answers to Lecture 6 tutorial

A.6.1 Answers to Sect. 6.5

2. \(z=(20-0)/10 = 2\). The area or probability to the right is \(0.0228\), or about \(2.3\)%. 3. The probability the SOI exceeds 20: \(2.3\)%. So odds: \(2.3\div(100-2.3)\), or about \(0.024\). 4. \(z=(-25-0)/10 = -2.5\). The area to the left is \(0.0062\), or about \(0.6\)%. 5. \(z=(-12-0)/10 = -1.2\). The area to the right is \(0.8849\), or about \(88.5\)%. 6. The two \(z\)-scores: \((-10-0)/10 = -1\) and \((20-0)/10 = 2\). The area between these is \(0.9772 - 0.1587 = 0.8185\), or about \(81.9\)%. 7. The \(z\) score corresponds to an area of \(0.80\) to the left; from the tables, about \(z=0.84\). This corresponds to an SOI of \(0 + (0.84\times 10)\), or an SOI of about \(8.4\).

A.6.2 Answers to Sect. 6.6

1. Relational. 2. \(\hat{p} = 352/2,864=0.12291\). 3. \(\text{s.e.(}\hat{p}) = \sqrt{0.12291 \times (1-0.12291)/2864} = 0.006135\). 4. An approximate 95% CI is \(0.12291 \pm (2\times0.006135)\), or \(0.12291\pm0.01227\), or from \(0.111\) to \(0.135\). Either the ‘\(0.123\pm 0.012\)’ form or the ‘\(0.111\) to \(0.135\)’ form is fine; percentages or proportions are fine (but the calculations must done with the proportions, not the percentages). 5. We need the number of boys who are late maturers and who are not late maturers to both be greater than 5. This is true, so the calculations are valid. 6. Smaller; the current sample size estimates \(p\) to within \(1.2\%\), and less accuracy needs fewer in the sample. 7. \(n = 1/(0.02)^2 = 2500\) boys.