## A.4 Answer: TW 4 tutorial

### Answers for Sect. 4.2

- (No answer.)
- \(43\div (43 + 67) \times 100 = 39.1\)%.
- \(39\div (97 + 39) \times 100 = 28.7\)%.
- \(43\div 67 = 0.642\).
- \(39\div 97 = 0.402\).
- \(0.642 \div 0.402 = 1.6\).
- (No answer; see online.)

### Answers for Sect. 4.3

- Type of student:nominal (with three levels).
Number of concussions:
*ordinal*(with three levels). - \(45\div 240 = 18.75\)% of all college students in the sample have received one concussion.
- Not sure which is best; they do different things. I'd probably prefer (c) (bottom left graph).
- 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.
- 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.
- 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).
- Not given.
- Not given.
- The better table is probably row percentages: the percentages of each type of students with given number of concussions.

### Answers for Sect. 4.4

- The
*percentage*of bridges collapsing due to deterioration is*36*divided by*433*, or \(8.3\)%. - The
*odds*of a bridge collapsing due to deterioration is*36*divided by*397*, or \(0.091\). - 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*. - The percentage of bridges
*not*collapsing due to deterioration is*397*divided by*433*, or \(91.7\)%. - The odds of a bridge
*not*collapsing due to deterioration is*397*divided by*36*, or \(11.0\). - The percentage of bridge collapses due to collisions is
*82*divided by*433*, or \(18.9\)%. - The odds of a bridge collapsing due to collisions is
*83*divided by*351*, or \(0.234\). - The odds of an event cannot be larger than one:
*FALSE*. - The odds of an event cannot be smaller than one.
*FALSE*. - The following statement is true:
*c. Odds cannot be negative*

### Answer for Sect. 4.5

- \((381 + 345)/1347 = 0.53898\), or about \(53.9\)%.
- \((381 + 345)/(295 + 326) = 1.1691\), or about \(1.17\).
- \(345/326 = 1.05828\), or about \(1.06\).
- \(381/295 = 1.29153\), or about \(1.29\).
- \(1.05828 / 1.29153 = 0.81940\), or about \(0.819\).

- a \((240 + 138)/960 = 0.39375\), or \(39.4\)%.
- \((240 + 342)/960 = 0.60625\), or \(60.6\)%.
- \((240 + 240)/960 = 0.5000\), or \(50\)%.
- \((240 + 138)/(240 + 342) = 0.64948\), or about \(0.649\).
- \(240/240 = 1.000\).
- \(138/342 = 0.40351\) or about \(0.404\).
- \(0.40351/ 1.000 = 0.40351\) (i.e., the answer to f divided by the answer to e).

### Answers for Sect. 4.6.1

- Backwards.
- See Table A.1.
- \(60/320 = 0.1875\). A worker with SMND is \(0.1875\) times as likely to have worked with metal than not.
- \(33/344 = 0.096\). A worker with SMND is \(0.096\) times as likely to have worked with metal than not.

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.

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

### 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

- Observational.
- 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. - For example, the percentage of residents who detected an odour at least once a week, who lived in Area I.
- Probably column percentages.
- \(77\div 291 = 26.46\%\).