note - wiley · solutions to exercises note although i have provided complete solutions to the...

SEQ 0001 JOB WIL8513-099-007 PAGE-0001 SOLS 198-213 REVISED 23SEP02 AT 16:33 BY TF DEPTH: 62 PICAS WIDTH 43.07 PICAS

SOLUTIONS TO EXERCISES

Note

Although I have provided complete solutions to the calculation parts of the ex-ercises, I have offered only brief comments, if at all, where a commentary isrequired. This is deliberate, first because I don’t want to write the book again interms of the solutions; secondly, tutors might want to tease these answers fromthe students themselves, perhaps as part of a wider discussion.

Chapter 1

1.1

SPECIALITY 0–14

ENTMedicineSurgeryOrthopaedicsOphthalmology

12345

SPECIALITY 65 & over

MedicineSurgeryOphthalmologyOrthopaedicsENT

12345

Reasons for differences probably centre on age.

Key

N = nominal; O = ordinal; MD = metric discrete; MC = metric continuous.

1.3

Age, MC. Social class, O. No. of children, MD. Age at 1st child, MC. Age atmenarche, MC. Menopausal state, O. Age at menopause, MC. Lifetime use of oral


2 SOLUTIONS TO EXERCISES

contraceptives, N. No. of years taking oral contraceptives, MC. No. of monthsbreastfeeding, MC. Lifetime use of N, MC. Years of HRT, MC. Family history ofovarian cancer, N. Family history of breast cancer, N. Units of alcohol, MD. No. ofcigs per day, MD. Body mass index, MC.

1.4

Maternal age, MC, but given here in ordinal groups. Parity, MD. No. of cigs daily,MD. Multiple pregnancy, N. Pre-eclampsia, N. Cesarean, N.

1.5

Age, MC. Sex, N. Number of rooms in home, MD. Length of hair, O. Colour ofhair, N. Texture of hair, N. Pruritus, N. Excoriations, N. Live lice, O. Viable nits, O.

1.6

Ordinal. These are subjective judgements, which are not measured but assessed,and will probably vary from patient to patient and moment to moment. Averagecannot be computed when data is ordinal.

Chapter 22.1

PARITY FREQUENCY PERCENT FREQUENCY

012345678

56

141031001

12.515.035.025.07.52.5002.5

2.2

PARITY FREQUENCY PERCENTFREQUENCY

CUMULATIVEFREQUENCY

PERCENTCUMULATIVEFREQUENCY

012345678

56

141031001

12.515.035.025.07.52.50.00.02.5

51125353839393940

12.527.562.587.595.097.597.597.5

100.0


SOLUTIONS TO EXERCISES 3

2.3

PERCENTMORTALITY

TALLY FREQUENCY

10.0–14.915.0–19.920.0–24.925.0–29.930.0–34.9

//// //////// ///

///////

/

98531

2.4

(a) Better to have parity as the columns and diagnosis as the rows.

DIAGNOSIS PARITY TOTALS

≤ 2 > 2

BenignMalignantTotals

224

26

104

14

328

40

(b)

DIAGNOSIS PARITY

≤ 2 > 2

BenignMalignantTotals

84.615.4

100.0

71.428.6

100.0

(c) Only 15.4% of those with a parity of 2 or less had a malignant diagnosis,compared with nearly twice as many with a parity greater than 2. Low levels ofparity seem to favour a benign diagnosis.

2.5

OCP CASES(n = 106)

CONTROLS(n = 226)

YesNoTotals

3862

100

6139

100

Comment. Only 38% of those receiving a malignant diagnosis (the cases) had at sometime used OCP, whereas 61% of the controls (receiving a benign diagnosis) had usedOCP. This suggests that a woman who had used OCP is more likely to receive abenign diagnosis. This is not a contingency table. There are two distinct groups ofpatients, those with a malignant diagnosis, and those with a benign diagnosis.



Chapter 3

3.1

(a) Pie chart of hair colour of children receiving d-phenothrin:

(b) Equivalent bar chart:

3.2

Clustered percentage bar chart:



3.3


3.4


3.5

Histograms:



3.6

The distribution is not far off being symmetrical, apart from a longish tail ofhigher values, which have a maximum of about 6mmol/L. Most men had serumpotassium levels of between 3.9mmol/L and 4.6mmol/L.

3.7

Suicide data:



3.8

Step chart:

3.9

(a) In both groups minimum cholesterol levels are about 4mmol/L, maximum levelsabout 11mmol/L, but the control group showed slightly higher cholesterol levelsthroughout. About half the patients had a cholesterol level of 6mmol/L and half more.

(b) Suicide data:



Ages about 26 and 33 years. Although this data is grouped, we can see that half ofthe male attempters were younger than the youngest half of the male succeeders.

Chapter 4

4.1

PSF scores are positively skewed (long tail to the right). No. The top 5% of scoreshave been omitted.

4.2

Both distributions are positively skewed, although that for succeeders is more so.

4.3

APACHE II scores are negatively skewed (long tail to the left).

4.4

Positively skewed.

Chapter 5

5.1

(a) Proportion breast-fed = 67/149 = 0.4497; percentage = 0.4497 × 100 = 44.97%.(b) Proportion bottle-fed = 93/182 = 0.5110; percentage = 0.5110 × 100 = 51.10%.

5.2

(a) Prevalence of genital chlamydia = (23/890) × 100 = 2.58%. (b) Incidence of SIDSper year = 10/10000. Incidence rate per thousand live births per year = 10/10 = 1SIDS death per 1000 live births per year.

5.3

(a) Cases and controls, modal class = II. (b) Satisfied. (c) PSF = 0.

5.4

(a) Putting the percentage mortality values in ascending order gives:

11.2 12.8 13.5 13.6 13.7 14.0 14.3 14.7 14.9 15.2 16.1 16.3 17.71 2 3 4 5 6 7 8 9 10 11 12 13

18.2 18.9 19.3 19.3 20.2 20.4 21.1 22.4 22.8 26.7 27.2 29.4 31.314 15 16 17 18 19 20 21 22 23 24 25 26

Since there are an even number of values, the median percentage mortality is theaverage of the two ‘‘middle’’ values, i.e. the average of the 13th (17.7) and 14th



(18.2) values, i.e. the 13.5th value. The median is thus = (17.7 + 18.2)/2 = 17.95%.Or you could have used the formula, median = 1⁄2 (n + 1)th value; or 1⁄2 (26 + 1) = 1⁄2× 27 = 13.5th value, as before.

(b) Men. 48 men, so the median will be the average of the 24th and 25th values, i.e.the 24.5th value. There are 13 values in the first two groups, and a further 16 in thethird group, making 29 values, so the median must be in the 35–44 years oldgroup. Women. 55 women, so the median is the value of the middle, 28th, value.There are 15 women in the first two groups, and a further 16 in the third group,making 31 values, so the median must be in the 35–44 years old group. You mightwant to repeat this exercise using the formula.

5.5

(a) mean > median; because of long tail of values to the right (positive skewness).(b) mean > median; positively skewed.

5.6

Mean percentage mortality = 18.66%, compared to median of 17.95%. Thesevalues are quite similar which suggests that the distribution might be reasonablysymmetric (which you could check with a histogram).

5.7

(a) With outliers, mean = 720.4, median = 500, standard deviation = 622.2. (b)Without outliers, mean = 610.6, median = 500, standard deviation = 319.8

5.8

25th percentile = 14.23%, 75th percentile = 21.43%. So a quarter of the ICUs have amortality of less than 14.23%, and a quarter have a mortality above 21.43%.

5.9

Breast-fed, range = 20 to 28 years; bottle-fed, range = 20 to 27 years.

5.10

Interquartile range of percentage mortality = (14.23 to 21.43)%. This means thatthe range of the middle half (50%) of the ICU percentage mortality rates is from14.23% to 21.43%.

5.11

(a) Median (IQR) pain = 51 (23.8 to 87.8). The median pain level is 51 out of amaximum of 100, so 50% of subjects had pain levels below 51 and half above 51.The interquartile range indicates that the middle 50% of pain levels lay between23.8 and 87.8. (b) The VAS does not ‘‘measure’’ the pain level using some sort ofinstrument (as would be possible with a metric variable), but asks the subject to



assess or judge it. The same patient may assess an equal level of pain differentlytomorrow, and two patients are unlikely to give an identical VAS score to equallevels of pain. The subjectiveness and potential inconsistency of such judgementsproduces ordinal data.

5.12

Q2, the median = 6mmol/L; Q1 = 5.5mmol/L; Q3 = 7.0mmol/L; IQR = (5.5 to7.0)mmol/L.

5.13

A boxplot:

5.14

The level of completeness of the lymphadenectomy, as measured by para-aorticradicality, in the early group of 30 patients, appears to be considerably less onaverage than that of the later group of 29 patients, a median of about 8, comparedto 20, although there is one high-valued outlier in the early group. The interquar-tile range for the early group is also smaller, IQRs of about (5 to 10) compared to(18 to 23) respectively. This is also true of the ranges, (4 to 13) and (12 to 25),respectively.

5.15

We can think of this, roughly speaking, to mean that the average distance of all ofthese cord platelet count values is 69 × 109/L from the mean value of 306 × 109/L.

5.16

SD = 5.36%. With a mean of 18.66% (Exercise 5.6), this suggests that the ICU’spercentage mortality rates are on average 5.36% away from this mean value.

5.17

For data to be Normally distributed, we need to be able to fit in three standarddeviations below the mean (and three above it). In all cases it is impossible (by along way!) to fit three SDs below the mean value without going into negativetime. This would suggest that all the distributions are positively skewed.



Chapter 6

6.1

In an observational study, the investigators do not influence in any way therecruitment, treatment or aftercare of subjects, but may simply ask questions, takemeasurements, observe events, and so on. In an experimental study, the inves-tigator takes an active role in some aspect of the study, giving a drug, changingnursing care, etc.

6.2

(a) Case–control studies are usually quicker, cheaper, and better with rare condi-tions, than cohort studies. They don’t suffer from subject fall-out over time. (b)Selection of suitable controls is often difficult. Problems with reliance on accuracyof patient recall, and medical records. Not good when exposure to risk factor israre.

6.3

By double-blinding.

6.4

Any solution to this problem will of course depend on the particular set ofrandom numbers used. My random numbers were: 2 3 1 5 (7) 5 4 (8) 5 (9)(0) 1 (8) 3 (7) 2. Since we have only six blocks we can’t use the random numbersin (). With blocks of four:

Block 1, CCTT; Block 2, CTCT; Block 3, CTTC;Block 4, TCTC; Block 5, TCCT; Block 6, TTCC

The first number is 2, so the first four subjects are allocated as block 2: C, T, C, andT. The next number is 3, so the next four subjects are allocated: C, T, T, and C.Continue this procedure until there are 20 in each group.

6.5

(a) The authors used a cross-section study of schoolchildren who were given askin-prick test of sensitivity to six common allergens (the outcome variable), todetermine atopic status, complimented by a questionnaire completed by parentsto elicit pertinent socio-economic factors (including number of siblings). Possibleconfounders identified by the researchers were family history of atopy, sex, socio-economic status, presence of pets, smoking, and age.

(b) The researchers used a double-blind RCT, with patients (aged 2–15 years)randomised to either CF or PM. To quote, ‘‘A double-dummy technique wasused: patients randomly assigned to CF also received a placebo PM, and patientsrandomly assigned to PM also received a placebo CF. Drug allocation was deter-mined by a computer-generated list of random numbers.’’ The clinical outcomevariable was the presence or absence of persistent dysentery after three days, andacceptable stool quality (satisfying a number of criteria) and no fever after 5 days.



Confounding should not be an issue in RCTs, since the randomisation process issupposed to produce two groups with identical characteristics.

(c) The researchers used a cohort design, following a group of 2185 womenbecoming pregnant and having a baby between August 1991 and May 1993. Thewomen were divided into two groups, normotensive and hypertensive. The out-come variable was defined as a birthweight below the 10th decile of expectedweight (values from reference tables). Potential confounders were parity, age,socio-economic status, ethnicity, weight and height, smoking status, and use ofaspirin.

(d) The researchers used a case–control study, in which cases were women withDown syndrome children, and controls were women selected randomly, havingchildren with no congenital abnormalities. Controls were matched only on birthyear. There were 10 controls for each case! Potential confounding factors were:maternal and paternal ages, marital status (married/unmarried), parity, alcoholconsumption (yes/no), prior fetal loss, and race (white/non-white).

(f) The researchers describe their study design as a ‘follow-up’’ study. Theyselected two groups of patients (and their relatives), one receiving home-basedcare in one part of a city, the other hospital-based care, in a different part of thecity. The relatives were interviewed at 10 days, one month, and one year, andgiven questionnaires to assess the burden they were experiencing, their satisfac-tion with the service, and the General Health Questionnaire). The patients wereassessed after four days, and then weekly, and given a number of psychiatricquestionnaires (Present State Examination, Morningside Rehabilitation Scale).The results from these various questionnaires constituted the outcome measures.

(g) The researchers used a randomised cross-over design. The subjects were ran-domised to either the ‘regular’’ treatment arm (two puffs of salbutamol four timesdaily), or the ‘as needed’’ treatment arm (salbutamol used as needed), each armlasting two weeks. Patients were asked to record their peak expiratory flow rate(PEFR) morning and evening before inhaler use; the number of asthma episodes;and the number of as-needed salbutamol puffs used for symptom relief.

(h) The researchers summarise their design as follows, ‘‘All new clients referredfor counselling by GPs were asked to complete a questionnaire before and aftercounselling.’’ This contained: three psychological scales to measure anxiety anddepression, self-esteem, and quality of life; and questions on levels of satisfactionwith the counselling service. GPs were also asked to complete a questionnaire ontheir level of satisfaction with the service. The prescribing of anxiolytic/hypnoticand anti-depressant drugs, and the number of referrals to other mental healthservices in practices with and without counsellors, was compared.

Chapter 7

7.1

(a) A population parameter is a defining characteristic of a population, forexample the mean age of all men dying of lung cancer in a Regional Health



Authority. The population parameter is unknown but can be estimated from arepresentative sample drawn from this population. (b) A sample will never haveexactly the same characteristics as a population because there is always the pos-sibility that those members of a population not included in the sample may insome way be different from those included. (c) Determining the parameters of atarget population is the underlying objective, but in practice this may prove to bedifficult if not impossible. The study population is the population which in prac-tice can be sampled.

Chapter 88.1

(a) (i) p(benign) = 226/332 = 0.681; (ii) p(malignant) = 106/332 = 0.319. Noticethese two probabilities sum to 1. (b) p(postmenopausal) = 200/332 = 0.602; (c) p(>3children) = 112/332 = 0.337.

8.2

(a) p(age <30) = (0.355 + 0.206 + 0.043) = 0.604. (b) p(age >29) = (0.248 + 0.148) =0.396.

8.3

(a) 0.99. (b) 0.165

8.4

(a) Men:

ALCOHOL CONSUMPTION(beverages/week)

Dead

Yes

No

Totals

< 1

195430

625

> 69

66145

211

Totals

261575

836

(i) Absolute risk of death if consuming < 1 beverage per week = 195/625 = 0.312.(ii) Absolute risk of death if consuming >69 beverages per week = 66/211 = 0.313.

(a) Women:

ALCOHOL CONSUMPTION(beverages/week)

DeadYesNoTotals

< 1

3942078

2472

> 69

119

20

Totals

3952097

2492



(i) Absolute risk of death if consuming < 1 beverage per week = 394/2472 = 0.159.(ii) Absolute risk of death if consuming >69 beverages per week = 1/20 = 0.050.Interpretation. For men there is approximately the same absolute risk of deathamong those consuming < 1 beverage per week and those consuming >69 bev-erages per week (0.312 vs. 0.313). For women the absolute risk of death if consum-ing < 1 beverage per week is about three times the absolute risk for thoseconsuming >69 beverages per week (0.159 vs. 0.050).

8.5(a) Under 35:

DOWN SYNDROME BABY

SmokedYesNoTotals

Yes112421533

No141152146625

(i) The odds that a woman having a Down syndrome baby, smoked = 112/421 =0.2660. (ii) The odds that a woman having a healthy baby, smoked = 1411/5214 =0.2706.

(b) Aged 35 or over:

DOWN SYNDROME BABY

SmokedYesNoTotals

Yes15

186201

No108611719

(i) The odds that a woman having a Down syndrome baby, smoked = 15/186 =0.0806. (ii) The odds that a woman having a healthy baby, smoked = 108/611 =0.1768.

Interpretation. Among the under 35 mothers there is little difference in the oddsfor Down syndrome between smoking and non-smoking mothers (0.2660 vs.0.2706). Among mothers ≥ 35, the odds for Down syndrome among smokingmothers is about a half the odds for non-smoking mothers (0.0806 vs. 0.1768).

8.6(a) Age < 35. (i) p = 0.2660/(1 + 0.2660) = 0.2101; (ii) p = 0.2706/(1 + 0.2706) = 0.2130.(b) Age ≥ 35. (i) p = 0.0806/(1 + 0.806) = 0.0746; (ii) p = 0.1768/(1 + 0.1768) = 0.1502.

8.7(a) Men. Risk ratio of death among those drinking >69 beverages per week com-pared to those drinking < 1 beverage per week = 0.313/0.312 = 1.003. Women. Riskratio = 0.050/0.159 = 0.314.Interpretation. For men a risk ratio very close to 1 implies that there is noincreased or decreased risk of death among those drinking < 1 compared to those



drinking >69 beverages per week. For women, the risk of death among the heavydrinkers appears to be only about a third the risk for light (or none) drinkers. Butsmall numbers in sample not reliable.

8.8

(a) Mothers < 35. Odds ratio for a woman with a Down syndrome baby havingsmoked, compared to a woman with a healthy baby = 0.2660/0.2706 = 0.9830. (b)Mothers ≥ 35. Odds ratio = 0.0806/0.1768 = 0.4558.Interpretation. In younger mothers, the odds ratio close to 1 (0.9830) implies thatsmoking neither increases or decreases the odds for Down syndrome. In oldermothers, the odds ratio of 0.4558 implies that mothers who smoked during preg-nancy have under half the odds of having a Down syndrome baby compared tonon-smoking mothers.

8.9Cohort study:

PERIODONTITIS

Death from CHDYesNoTotals

Yes151

16351786

No92

34503542

Totals243

50855328

Absolute risk of dying from CHD with periodontitis = 151/1786 = 0.084. Absoluterisk of dying from CHD with no dental disease = 92/3542 = 0.026. So risk reduc-tion = 0.084 – 0.026 = 0.058. Therefore NNT = 1/0.058 = 17.2, i.e. 18 people.

Chapter 9

9.1

The smaller the SE of the sample mean, the more precise the estimate of thepopulation mean. In this case the sample mean vitamin E intake of 6.30mg (non-cases), has a standard error of 0.05mg, so we can be 95% confident that thepopulation mean vitamin E intake (non-cases) is no further than 2 SE from thismean, i.e. within ± 0.01mg. The largest SE, 5.06mg, and therefore the least preciseestimate of the population mean, is that for vitamin C (cases).

9.2

(a) Cases. Sample mean age = 61.6y , sample SD = 10.9y, n = 106. Thus SE(X) =10.9/√106 = 1.059. The 95% confidence interval is therefore: (61.6 ± 2 × 1.059), or(59.582 to 63.718) years. (b) Controls. Sample mean age = 51.0y, sample SD = 8.5y,n = 226. Thus SE(X) = 8.5/√226 = 0.565. The 95% confidence interval is therefore:(51.0 ± 2 × 0.565), or (49.870 to 52.13) years. The fact that the two CIs don’t overlapmeans that we can be 95% confident that the two population mean ages aresignificantly different.



9.3

For the integrated care group, over 12 months the sample mean number of admis-sions is 0.15. The 95% confidence interval means we can be 95% confident that theinterval from 0.11 to 0.19 will contain the population mean number of visits forthe population of which this is a representative sample. For the conventional caregroup the sample mean number of visits is lower, 0.11, and the 95% confidenceinterval means we can be 95% confident that the interval from 0.08 to 0.15 willcontain the population mean number of visits.

9.4

0.290(1 – 0.290)p = 0.290, SE(p) = –––––––––––––– = 0.030. 95% CI is:√ 226

(0.290 – 2 × 0.030 to 0.290 + 2 × 0.030 = (0.230 to 0.350)

So we can be 95% confident that the interval from 0.230 to 0.350 (or 23.0% to35.0%), will contain the population proportion of women who are pre-menopausal.

9.5

Percentage abstaining in three groups was 31% (Manual group), 35% (CBTgroup), and 17% (Waiting list group). The CBT group appear to get the bestresults. However, the fact that these very wide CIs (caused by too-small samplesizes) all overlap, means that we cannot conclude that there is a statisticallysignificant difference in the proportions who are no longer binge eating across thethree groups.

9.6

For all three time periods the median differences in pain levels are reasonablysimilar (38, 31, and 35), as are the 95% confidence intervals, which all overlap,indicating no statistically significant difference between the two groups at anytime period.

Chapter 10

10.1

Three of the confidence intervals include 0, so there is no statistically significantdifference in population mean infant weights between non-smoking and smokingmothers. The confidence interval for the difference in the mean weight of non-smoking mothers and mothers smoking 1–9 cigarettes per day, (–118 to –10)g, forboys, does not include 0, so this difference in population mean weights is statis-tically significant.

10.2

That for the radius which has the narrowest confidence interval.



10.3

Because overlapping confidence intervals imply that the difference is not statis-tically significant.

10.4

The difference in sample median alcohol intakes is 5.4g. The 95% confidenceinterval of (1.2 to 9.9)g, does not include 0, so we can be 95% confident that thepopulation difference in median alcohol intake is statistically significant and liessomewhere between 1.2g and 9.9g.

Chapter 11

11.1

For gingivitis, the confidence intervals for both CHD and mortality contain 1, sodifference in risk compared to no disease is not statistically significant. For perio-dontitis neither confidence interval includes 1, so the difference in risk is statis-tically significant. For no teeth, the confidence interval for CHD includes 1, so notstatistically significant; but for mortality, the confidence interval does not include1, so the difference in risk compared to no disease is statistically significant.

11.2

(a) Age and sex are notorious as confounders of many other variables, and adjust-ment for them is nearly always advisable. (b) With no exercise taken as thereferent state, the odds ratios for all three age groups are less than 1, implying thatexercise at any age reduces the odds for a stroke. However, only exercise takenbetween 15 and 40 has a statistically significant effect, since the confidence inter-val for the 40–55 year-old group, (0.3 to 1.5), includes 1. Note by the way that a 25year-old and a 40 year-old individual could each be allocated to either one of twogroups. The groups are not well defined.

11.3

The following risk factors are statistically significant for increasing the risk ofthromboembolic events: being aged ≤ 19; having any parity other than 1; smoking≥ 10 cigarettes per day; having multiple pregnancy; having pre-eclampsia; havinga cesarean. The latter two appear to increase the risk the most.

Chapter 12

12.1

Add up probabilities, starting at top of column, until sum equals or exceedssignificance level, α. (a) α = 0.1; 41 tails or fewer. (b) α = 0.01; 34 tails (or fewer).



12.2

Since both p-values (0.25 and 0.32) exceed 0.05, there is no statistically significantdifference in the two means.

12.3

Mean age, mean age at menopause, and mean body mass index are statisticallysignificant, since their p-values are all less than 0.05. The other four variables showno statistically significant difference since their p-values are all greater than 0.05.

12.4

(b) Because β would be unacceptably large.

12.5

(a) (i) n = (2 × 122/102) × 10.5 = 31; (ii) n = (2 × 122/102) × 14.9 = 43; (iii) n = (2 ×122/102) × 11.7 = 34. (b) (i) n = [(0.4 × 0.6 + 0.20 × 0.80)/0.202] × 10.5 = 105; (ii) n =[(0.4 × 0.6 + 0.20 × 0.80)/0.202] × 14.9 = 149; (iii) n = [(0.4 × 0.6 + 0.20 × 0.80)/0.202]× 11.7 = 117.

12.6

Pa = 0.70 and Pb = 0.80, so (Pb – Pa) = –0.10. Therefore:(a) n = [(0.70 × 0.30 + 0.80 × 0.20)/–0.102] × 7.8 = 289; (b) n = [(0.70 × 0.30 + 0.80 ×0.20)/–0.102] × 14.9 = 551.

Chapter 13

13.1

There are only two statistically significant risk factors, both of which show higherrisks for the alteplase patients (i.e. RR < 1)); CAPG, RR = 0.884, p-value = 0.049, seetable footnote for meaning of CAPG; and a Killip classification >1; RR = 0.991,p-value = 0.026. Anaphylaxis is a complication which is almost statistically signifi-cant (RR = 0.376, p-value = 0.052, and we might want to consider it so.

13.2

In the model with the seven variables shown, all are statistically significant exceptpassive smoking from husband, and at work. With only the first five variablesincluded, plus passive smoking from husband and/or at work, makes this lastvariable statistically significant (p-value = 0.049).



Chapter 1414.1Expected values:

MOTHER SMOKED

Yes No TOTALS

Apgar < 7 { YesNo

Totals

3.676.33

10

7.3312.6720

111930

14.2

The test statistic = √{(8 – 3.67)2/3.67 + (3 – 7.33)2/7.33 + (2 – 6.33)2 /6.33 + (17 –12.67)2/12.67} = √12.109 = 3.480. Since we have a 2 × 2 table, the critical chi-squaredvalue which must be exceeded to reject the null hypothesis is 3.85. The test statisticvalue of 3.480 does not exceed this value so the evidence is not strong enough for usto reject the null hypothesis of equal proportions of smokers in both Apgar groups.

14.3

The null hypothesis of equal proportions is equivalent to a null hypothesis ofindependent variables. Since we have rejected the former we have also rejectedthe latter, so these variables are independent.

14.4

(i) No trend across categories of social class, p-value = 0.094; (ii) statisticallysignificant trend across the two categories (yes/no) of oral contraceptive use,p-value = 0.000; (iii) no trend across categories of alcohol consumption, p-value =0.927; (iv) no trend across categories of cigarette consumption, p-value = 0.383.

Chapter 15

15.1

The association seems to be strong and positive, and linear-ish.



15.2

Overall, the association seems to be strong and positive.

15.3

The association appears to be strong and positive but does not appear to be linear.

15.4

It’s positive and seemingly linear, but the spread is only moderately concentrated.Actual value of r is 0.50.

15.5

(a) None. (b) 0.896 for mothers less than two years from birth date. (c) 0.632 formothers where the baby concerned was ≥ 3rd born.

15.6

All correlation values are statistically significant (all p-values < 0.05), with thestrongest that between physical symptom scores as measure by the STAS and thestaff-POS (r = 0.80).

Chapter 16

16.1

Yes. No.

16.2

Contingency table:

Observer 1 Totals

<16 ≥ 16

Observer 2

Totals

<16≥16

50

5

29

11

79

16

(a) Observed proportional agreement = (5 + 9)/16 = 0.875.



(b) Expected values are as follows:

Observer 1 Totals

<16 ≥ 16

Observer 2

Totals

<16≥16

2.192.81

5

4.816.19

11

79

16

Expected agreement = (2.19 + 6.19)/16 = 0.524. So kappa = (0.875 – 0.524)/(1 – 0.524) = 0.734. Chance adjusted agreement is good.

Chapter 17

17.1

(a) Percentage mortality = 46.886 – 0.620E, a decrease of 0.620%. (b) Peak = 1.336 +0.833 × Initial, 5.5. (c) BMI = –17.86 + 0.43HIP, increased by 0.43kg/m2.

17.2

Mean BMI = 41.902kg/m2.

17.3

(a) Severity of disability; mental disorders; respiratory system disorders; numbersof residents in private residential homes (all p-values < 0.05). (b) (i) natural log ofutilisation time increases by 0.006, or 1.006 minutes (taking antilog). (ii) increase of0.043 in natural log, or 1.044 minutes. (c) About 11% (see R2 in table footnote).

17.4

(a) Age; age squared; family history of hypertension; calcium intake. (b) We canbe 95% confident that the population regression parameter on Age is between 0.28and 0.64. (c) The blood lead model (largest age coefficient value).

Chapter 18

18.1

(a) Because only two values for the dependent variable. Better to group variablefirst and plot proportions in each group. (b) Yes, confidence interval for odds ratioof (1.08 to 1.14) does not include 1. (c) P(y=1) = e(–6.4672 + 0.10231×Age). (d)(i) 0.1343.(ii) 0.2707. (e) 0.8657, 0.7299. Odds ratio = 0.4182. A woman aged 45 has onlyabout 41% the odds of a malignant diagnosis as a woman aged 50. (f) The antilogeof 0.10231 equals 1.108 (rounded to 1.11 by Minitab). (g) 10 × 0.10231 = 1.0231;antiloge = 2.78. In other words an increase in age of 10 years increases the oddsratio by 2.78.

note - wiley · solutions to exercises note although i have provided complete solutions to the...

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