## A.5 Answers to Teaching Week 5 tutorial

### Answers to Sect. 5.2

Answers embedded.

- 70% 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 to Sect. 5.3

Answers implied by H5P.

- Five variables. ('Participants' would not be summarised, it is technically an
*identifier*and not a*variable*, as each person has a unique value). - Age; Height; Weight and quantitaive
**continuous**. - Gender (
**nominal**; two levels); GMFCS (**ordinal**; three levels) - As follows:
- 'Gender': Percentages (or number) F and M
- 'Age': Mean/median; standard deviation/IQR
- 'Height': Mean/median; standard deviation/IQR
- 'Weight': Mean/median; standard deviation/IQR
- 'GMFCS': Percentages (or numbers) in each group

- As follows:
- 'Gender': Barchart (not really needed)
- 'Age': Histogram/stemplot
- 'Height': Histogram/stemplot
- 'Weight': Histogram/stemplot
- 'GMFCS': Barchart/piechart

- As follows:
- Between Gender and Height: Boxplot
- Between Gender and GMFCS: Side-by-side or stacked bar chart.

### Answers to Sect. 5.4

- \(45\div 240=18.75\)% of all college students in the sample have received one concussion, the same for all student types.
- 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 to Sect. 5.5

Answers embedded.

- Relative frequency.
- Relative frequency.
- Subjective.
- Relative frequency.
- Classical.

### Answers to Sect. 5.6

Answers implied by H5P.

The decision-making process begin with making an assumption about the population

**parameter**.

This means we know what to**expect**from the sample**statistic**. We never know exactly what value of the statistic we will see in the sample, because of**sampling variation**. But we can have some of idea of what values are reasonable to expect.

Then we take the**sample**(that is, we make the observations).

Then we**compare**the sample statistic that we observed... to the sample statistic we expected.

If what we observe is inconsistent with what was expected, then the the assumption is**unlikely to be**true.

However, if what we observe is**consistent**with what was expected, then the the assumption is**probably**true.Step 1: Assumption about population parameter

Step 2: Expectation for sample statistic

Step 3: Observation of sample statistic

Decision: Consistent?

Conclusion A: Yes, supports assumption

Conclusion B: No, doesn't support assumption

### Answers to Sect. 5.7

Answers embedded.

- Odds
**less than**one. - Odds
**less than**one. - Odds
**greater than**one. - Odds
**greater than**one.

### A.5.1 Answers to Sect. 5.8

**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*

### A.5.2 Answers to Sect. 5.9

**1.**See Table A.1.

**2.**\(60/320 = 0.1875\). A worker with SMND is 0.1875 times as likely to have worked with metal than not.

**3.**\(33/344 = 0.096\). A worker with SMND is 0.096 times as likely to have worked with metal than not.

**4.**\(0.1875/0.096 = 1.95\); the odds are about twice as great.

**5.**\(60 \div 380 = 15.8\)%.

**6.**\(33\div 377 = 8.8\)%.

**7.**Probably not: big difference seen in a large size sample.

Worked with metals | Did not work with metals | Total | |
---|---|---|---|

SMND cases | 60 | 320 | 380 |

Controls | 33 | 344 | 377 |

Total | 93 | 664 | 757 |

### A.5.3 Answers to Sect. 5.10

**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.**\(77\div291 = 26.46\%\).

### A.5.4 Answer to Sect. 5.11

**1.**
a. \((381 + 345)/1347 = 0.53898\), or about 53.9%.
b. \((381 + 345)/(295 + 326) = 1.1691\), or about 1.17.
c. \(345/326 = 1.05828\), or about 1.06.
d. \(381/295 = 1.29153\), or about 1.29.
e. \(1.05828 / 1.29153 = 0.81940\), or about 0.819.

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