## A.6 Answers to Teaching Week 6 tutorial

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.

A person with an IQ of 85 has an IQ that is 15 points below the mean.
This is equivalent to 1 standard deviation(s) below the mean IQ.
Using the 68--95--99.7 rule, this means that the proportion of the population with an IQ less than 85 is about 32%. In addition, the proportion of the population with an IQ above 85 is about 68%.

4. Answers vary. You probably cannot be very accurate using the 68--95--99.7 rule.
6. Two standard deviations 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.
7. 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 $$9.8/(100-9.8)=0.11$$.
8. 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.

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

• $$\hat{p}$$ is the sample proportion.
• $$n$$ is the sample size.
• "s.e." stands for "standard error".

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.
A definition can be found in the Glossary. Essentially, each sample is likely to produce a different value for the sample proportion, $$\hat{p}$$ (the estimate of the population proportion, $$p$$), and that is what we mean by “sampling variation”.

6. 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:

7. A 95% CI: $$0.304455 \pm (2\times0.022894)$$, or $$0.30446\pm0.04579$$, or $$0.26$$ to $$0.35$$.

8. One way of writing communicating: “The population proportion of patients surviving after BVM treatment has a 95% chance of lying between 26% and 35%.” This is not strictly correct, but acceptable and very, very commonly used (as explained in the textbook).

9. The number of surviving and non surviving both exceed 5.

10. Larger, to get a tighter (more precise) CI than the one calculated.

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

12. See table below.

13. Use a stacked or side-by-side barchart, for example. But a chart is not really needed: Just give the information in text.

14. 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.

15. A greater sample size would give a more precise estimate. But rather than a greater sample size (which would still be helpful), probably more important is to consider other relevant issues that have not been discussed so far: Relative costs; ease of use; confounding variables; potential side effects; etc.

Method Survived Did not survive Total
BVM 123 281 404
ETI then BVM 110 306 416

### A.6.1 Answers to Sect. 6.6

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 $$8.4$$. 8. The $$z$$-score is 0.385 from Table B.3, remembering that the area to the left would be 0.650 (draw a diagram!). So, the $$z$$-score is 0.385, so that the SOI values is $$x = \mu + (z\times\sigma) = 0 + (0.385\times 10) = 3.85$$.

### A.6.2 Answers to Sect. 6.7

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.

### A.6.3 Answers to Sect. 6.8

1. $$\sqrt{0.70\times(1 - 0.70)/25} = \sqrt{0.084} = 0.2898275$$, or about 0.2898. 2. $$\sqrt{0.25\times(1 - 0.25)/100} = \sqrt{0.001875} = 0.04330127$$, or about 0.04330. 3. 0.08724964, or about 0.08725. 4. 0.0534479, or about 0.05345.

Note: Students commonly forget to take the square root.

Note: If you calculator gives an answer something like 1.875 E-03 or similar, it is using scientific notation. It means $$1.875\times 10^{-3}$$, or $$0.001875$$.