## E.6 Answers to Lecture 6 tutorial

### E.6.1 Answers to Sect. 6.1

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.

### 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 $$z=(x-\mu)/\sigma$$.
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.
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

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

### 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 $$0.35$$.
6. $$n = 1/(0.01)^2 = 10\,000$$ patients.
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.