## A.4 Answer: TW 4 tutorial

### Answers for Sect. 4.2

1. (No answer.)
2. $$43\div (43 + 67) \times 100 = 39.1$$%.
3. $$39\div (97 + 39) \times 100 = 28.7$$%.
4. $$43\div 67 = 0.642$$.
5. $$39\div 97 = 0.402$$.
6. $$0.642 \div 0.402 = 1.6$$.
7. (No answer; see online.)

### Answers for Sect. 4.3

1. Type of student:nominal (with three levels). Number of concussions: ordinal (with three levels).
2. $$45\div 240 = 18.75$$% of all college students in the sample have received one concussion.
3. Not sure which is best; they do different things. I'd probably prefer (c) (bottom left graph).
4. Odds (non-athlete receiving 2+ concussions) is $$21\div(81 - 21) = 21\div 60 = 0.35$$. Non-athletes are $$0.35$$ times as likely to receive two or more concussions than fewer than two. Or: For every $$100$$ students with fewer than two concussions, there are $$35$$ with two or more.
5. Odds (soccer player receiving 2+ concussions) is $$13\div(63 - 13) = 13\div 50 = 0.26$$. Soccer players are $$0.26$$ times more likely to receive two or more concussions than fewer than two concussions.
6. OR: $$0.35 \div 0.26$$, or about $$1.35$$. The odds of a non-athlete receiving two or more concussions is about $$1.35$$ times the odds of a soccer player receiving two or more concussions (that is, smaller odds for soccer players).
7. Not given.
8. Not given.
9. The better table is probably row percentages: the percentages of each type of students with given number of concussions.

### Answers for Sect. 4.4

1. The percentage of bridges collapsing due to deterioration is 36 divided by 433, or $$8.3$$%.
2. The odds of a bridge collapsing due to deterioration is 36 divided by 397, or $$0.091$$.
3. The odds of a bridge collapsing due to deterioration is $$0.091$$; this means: b. For every $$100$$ bridges that do not collapse due to deterioration, about $$9.1$$ bridges do collapse.
4. The percentage of bridges not collapsing due to deterioration is 397 divided by 433, or $$91.7$$%.
5. The odds of a bridge not collapsing due to deterioration is 397 divided by 36, or $$11.0$$.
6. The percentage of bridge collapses due to collisions is 82 divided by 433, or $$18.9$$%.
7. The odds of a bridge collapsing due to collisions is 83 divided by 351, or $$0.234$$.
8. The odds of an event cannot be larger than one: FALSE.
9. The odds of an event cannot be smaller than one. FALSE.
10. The following statement is true: c. Odds cannot be negative

### Answer for Sect. 4.5

1. $$(381 + 345)/1347 = 0.53898$$, or about $$53.9$$%.
2. $$(381 + 345)/(295 + 326) = 1.1691$$, or about $$1.17$$.
3. $$345/326 = 1.05828$$, or about $$1.06$$.
4. $$381/295 = 1.29153$$, or about $$1.29$$.
5. $$1.05828 / 1.29153 = 0.81940$$, or about $$0.819$$.
1. a $$(240 + 138)/960 = 0.39375$$, or $$39.4$$%.
1. $$(240 + 342)/960 = 0.60625$$, or $$60.6$$%.
2. $$(240 + 240)/960 = 0.5000$$, or $$50$$%.
3. $$(240 + 138)/(240 + 342) = 0.64948$$, or about $$0.649$$.
4. $$240/240 = 1.000$$.
5. $$138/342 = 0.40351$$ or about $$0.404$$.
6. $$0.40351/ 1.000 = 0.40351$$ (i.e., the answer to f divided by the answer to e).

### Answers for Sect. 4.6.1

1. Backwards.
2. See Table A.1.
3. $$60/320 = 0.1875$$. A worker with SMND is $$0.1875$$ times as likely to have worked with metal than not.
4. $$33/344 = 0.096$$. A worker with SMND is $$0.096$$ times as likely to have worked with metal than not.
TABLE A.1: SMND cases, and whether they had worked with metals
Worked with metals Did not work with metals Total
SMND cases 60 320 380
Controls 33 344 377
Total 93 664 757

### Answers for Sect. 4.6.2

Answers embedded.

1. $$70$$%, as given directly from the table.
2. $$281/(404 - 281) - 2.28$$.
3. For patients treated with ETI, about $$278$$ die for every $$100$$ that survive.
4. $$2.78/2.28 = 1.22$$.

### Answers for Sect. 4.6.3

Answers embedded.

Odds less than one. Odds less than one. Odds greater than one. Odds greater than one.

### Answers for Sect. 4.6.4

1. Observational.
2. The percentage of residents in each area that detected the smell at the given frequency. So, for example, the first number in the first column would become $$48\div 97\times 100 = 49.5$$. This means that $$49.5$$% of the sample that lived in Area I detected the odour at least once a week.
3. For example, the percentage of residents who detected an odour at least once a week, who lived in Area I.
4. Probably column percentages.
5. $$77\div 291 = 26.46\%$$.