A.10 Answer: TW 10 tutorial

Answers for Sect. 10.3

Some answers embedded.

  1. See Table A.3.

  2. Using artificial limb: \(49/16 = 3.0625\). Not using artificial limb: \(21/19 = 1.105263\). The OR is \(3.0625/1.105263 = 2.771\); that is, the odds of being alive after five years is almost three times higher for those using an artificial limb compared to those who do not. See Table A.4.

  3. The CI for the OR is from \(1.198\) to \(6.411\).

  4. Chi-squared: \(5.836\); like \(z = \sqrt{5.836/1} = 2.42\): large; \(P\) about \(0.016\).

    The sample provides evidence to suggest that the odds of dying within five years is not the same between having a wearing an artificial limb and the five-year mortality rate in the population (\(\text{chi-square} = 5.836\); \(\text{df} = 1\); \(P=0.016\); OR: \(2.771\) and \(95\)% CI from \(1.198\) to \(6.411\)).

TABLE A.3: Five-year mortality for artifical limb users
Alive Dead Total
Used art. limb 49 16 65
Did not use art. limb 21 19 40
Total 70 35 105
TABLE A.4: Five-year mortality and use of an artificial limb: Numerical summary
Percentage alive after 5 years Odds alive after 5 years Sample size
Use artificial limb 75.4 3.06 65
Did not use artifical limb 52.5 1.11 40
Diff in percentages: 22.9 OR: 2.771 626

Answers for Sect. 10.4

1. Two-sample \(t\)-test. 2. A \(\chi^2\) test. 3. A paired \(t\)-test. 4. A two-sample \(t\)-test. 5. None of the other options are correct: requires regression or correlation.

Answers for Sect. 10.5

\(n = (2\times 7.145\div 0.5)^2 = 816.8\), so use guesses from \(817\) students.

Answers for Sect. 10.6.1

  1. Odds HIE: \(263/151 = 1.742\). HIE is \(1.74\) times more likely to be for GM foods than against.
  2. Odds LIE: \(258\div222 = 1.162\). LIE is \(1.16\) times more likely to be for GM foods than against.
  3. \(\text{OR}(\text{HIE in favour})\div\text{OR}(\text{LIE in favour}) = 1.742/ 1.162 = 1.5\) (\(1.499\) in table). The odds of HIE being for GM food \(1.5\) times the odds that a LIE for GM foods.
  4. From the sample, we estimate the OR in the population to be between \(1.145\) to \(1.961\). (Loosely, though technically incorrect: the true OR is likely to be between \(1.145\) and \(1.961\).) Importantly, this interval does not include \(1\).
  5. \(\chi^2 = 8.74\) and \(P = 0.003\); strong evidence of a difference.

Answers for Sect. 10.6.2

1. Incorrect is the one about means. 2. \(113/626 = 18.05\%\). 4. \(18.05\)% of \(47\) is \(8.48\). See Table A.5. 5. \(25\div 88 = 0.284\). 6. \(22\div 491 = 0.0448\). 7. The OR is \(0.284\div 0.0448 = 6.3\). This OR means that the odds of having Hep. C is \(6.3\) times greater for students who have a tattoo, compared to those who do not have a tattoo.

TABLE A.5: Five-year mortality for artifical limb users
Had Hep. C Did not have Hep. C Total
Had tattoo 25 88 113
Did not have tattoo 22 491 513
Total 47 579 626

Answers for Sect. 10.6.3

  1. The third. The first is about samples; the second about means. \(H_0\): \(\text{Pop. odds having byss.}_{\text{Smokers}} = \text{Pop. odds having byss.}_{\text{Non-Smokers}}\) or \(\text{OR} = 1\); \(H_1\): \(\text{Pop. odds having byss.}_{\text{Smokers}} > \text{Pop. odds having byss.}_{\text{Non-Smokers}}\) or \(\text{OR} > 1\) for the OR suitably defined.
  2. Smokers: \(\hat{p} = 125/3,189 = 0.0392\); Non-smokers: \(\hat{p} = 40/2,230 = 0.0179\).
  3. For smokers: \(\text{Odds having byssinosis} = 125/3,064 = 0.0408\). This means: smokers are \(0.041\) times as likely to have byssinosis than not (or, inverting the ratio, \(24.5\) times as likely not to have byssinosis than have it). For non-smokers: \(\text{odds} = 40/2,190 = 0.01826\). Non-smokers are \(0.018\) times as likely to have byssinosis than not (or, inverting the ratio, \(54.8\) times as likely not to have byssinosis than have it).
  4. The OR is \(0.04080/0.01826 = 2.2\), so the odds of a smoker having byssinosis are \(2.2\) times the odds of a non-smoker having byssinosis.
  5. OR not one could be due to chance or because of a real difference in the population, due to smoking and/or other reasons.
  6. The sample provides strong evidence (one-tailed \(P < 0.001)\); \(\text{df} = 1\); \(\text{chi-square} = 20.092\); OR: \(2.234\) (\(95\)% CI: \(1.6\) to \(3.2\))) that the population proportion of smokers with byssinosis (\(\hat{p} = 0.0392\)) is greater than the population proportion of non-smokers with byssinosis (\(\hat{p} = 0.0179\)).
  7. Valid, since expected counts all exceed five (which they do: no SPSS warnings given for example).
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