E.6 Answers to Lecture 6 tutorial

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E.6.1 Answers to Sect. 6.1

Answers embedded. Some answers:

  1. In order from left to right: 70; 85; 100; 115; 130.
  2. 30 points above the mean.
  3. Two standard deviations above the mean.
  4. About 2.5%.

E.6.2 Answers to Sect. 6.2

Encourage drawing diagrams! A common error: plugging numbers into calculators wrongly, and effectively computing \(z=x-(\mu/\sigma)\) rather than \(z=(x-\mu)/\sigma\).
  1. Answers vary.
  2. Answers vary.
  3. Answers vary.
  4. Answers vary. Notice that students cannot be very accurate using the 68–95–99.7 rule, which is one of the points to be made: Only a guess can be made, but using \(z\)-scores leads to more accurate answers.
  5. Answers vary.
  6. Answers vary.
  7. Two std devs from the mean is \(2\times 6.7 = 13.4\), so 95% of females aged 18 and over have a measured height between \(161.4 \pm13.4\) approximately, or from \(148.0\),cm to \(174.8\)cm.
  8. As follows:
    • \(z=(171-161.4)/6.7 = 1.43\), so the probability is \(0.9236\) or about 92%.
    • So the odds are \(92.36/(100-92.36) = 12.1\).
    • The answer is just \(1-0.9236 = 0.0764\) or about 7.6%.
    • The probability of over 171cm is \(7.64\)%.
    • So the odds are \(7.64/(100-7.64)= 0.084\).
    • \(z\)-scores are \(1.28\) and \(2.78\), so the answer is \(0.9973 - 0.8997 = 0.0976\), or about 9.8%.
    • So the odds are \(9.8/(100-9.8)=0.11\).
  9. Use the Tables: \(z=-0.84\); then using unstandardising formula, the height is \(x=\mu+(z\times\sigma) = 161.4 + (-0.84\times6.7) = 155.772\), or about 156cm.

You could have a discussion about rounding: Is it sensible to quote to the nearest millimetre, or tenth of a millimetre?

E.6.3 Answers to Sect. 6.3

Answers implied by H5P.

\[ \text{s.e.}(\hat{p}) = \sqrt{\frac{\hat{p}\times(1 - \hat{p})}{n}} \]

E.6.4 Answers to Sect. 6.4

  1. \(123/(404-123) = 123/281 = 0.44\).
  2. \(\hat{p} = 123/404 = 0.304455\).
  3. The odds: The likelihood of surviving is about 0.44 times the probability of dying (ie. it is lower). Or: For every 100 that die, about 44 survive.
  4. No: sampling variation!
  5. \(\text{s.e.}(\hat{p}) = \sqrt{0.304455 \times (1-0.304455)/404} = \sqrt{0.00052416} = 0.022894\), or about 0.023.
Students commonly forget the square root. Furthermore, the calculators may use scientific notation to display the results from some calculations (e.g. \(2.2894\times 10^{-2}\) or \(2.2894^{-2}\) as it displays on some calculators); many students will have no idea what’s going on.
  1. The values of \(\hat{p}\) will have an approximate normal distribution, with a standard deviation equal to standard error (\(0.023\)) and centred around the true proportion \(p\). Since we don’t know \(p\), we need to centre it around our best guess of \(p\), which is \(\hat{p}=0.304\). So something like this:
  2. A 95% CI: \(0.304455 \pm (2\times0.022894)\), or \(0.30446\pm0.04579\), or \(0.26\) to \(0.35\).
  3. (Answer not yet provided.)
  4. The number of surviving and non surviving both exceed 5.
  5. Larger, to get a tighter (more precise) CI than the one calculated.
  6. \(n = 1/(0.01)^2 = 10\,000\) patients.
  7. No answer given.
  8. Use a stacked or side-by-side barchart, for example. But a chart is not really needed: Just give the information in text.
  9. Odds of not surviving in ETI group is \((306/110) = 2.8\), so the OR is \((2.3/2.8) = 0.82\). The odds of a BV patient not surviving are 0.8 times as great as the odds of an ETI patient not surviving.

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