A.5 Answers to Teaching Week 5 tutorial

Answers to Sect. 5.2

Answers embedded.

  1. 70% 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 to Sect. 5.3

Answers implied by H5P.

  1. Five variables. ('Participants' would not be summarised, it is technically an identifier and not a variable, as each person has a unique value).
  2. Age; Height; Weight and quantitaive continuous.
  3. Gender (nominal; two levels); GMFCS (ordinal; three levels)
  4. 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
  5. As follows:
    • 'Gender': Barchart (not really needed)
    • 'Age': Histogram/stemplot
    • 'Height': Histogram/stemplot
    • 'Weight': Histogram/stemplot
    • 'GMFCS': Barchart/piechart
  6. As follows:
    • Between Gender and Height: Boxplot
    • Between Gender and GMFCS: Side-by-side or stacked bar chart.

Answers to Sect. 5.4

  1. \(45\div 240=18.75\)% of all college students in the sample have received one concussion, the same for all student types.
  2. Not sure which is best; they do different things. I'd probably prefer (c) (bottom left graph).
  3. 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.
  4. 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.
  5. 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).
  6. Not given.
  7. Not given.
  8. 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.

  1. Relative frequency.
  2. Relative frequency.
  3. Subjective.
  4. Relative frequency.
  5. Classical.

Answers to Sect. 5.6

Answers implied by H5P.

  1. 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.

  2. 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.

  1. Odds less than one.
  2. Odds less than one.
  3. Odds greater than one.
  4. 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.
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

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).