Ординатура / Офтальмология / Английские материалы / Principles Of Medical Statistics_Feinstein_2002

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Answers to Exercises

The following “official” answers are for Exercises that have an odd second number. Answers to the even-numbered Exercises are contained in an Instructor’s Manual.

Chapter 1

1.1.4. (Architecture)

1.3.3. (Stochastic)

1.5.1. (A descriptive summary, expressed in nonquantitative terms.)

1.7. 5. (A decision about classifying the raw data. If you cited 6 as the answer, you can also be justified, although data processing usually refers to the conversion of sick into a coded category, such as “1 in column 37 of card 6,” rather than the data acquisition process that converts raw data, such as “Apgar ≤ 6,” into a designated category, such as “sick.” Because the distinctions are somewhat blurred, either answer is acceptable.)

1.9. 5, if you think the research was done to demonstrate the quality of the raw data; 4, if you think the research has an architectural role with respect to biased observations. Either answer is acceptable. 1.11. 4. (Architecture)

Chapter 2

2.1.

2.1.1.Age, height, and each blood pressure measurement are dimensional. Sex is binary. Diagnosis is nominal. Treatment is nominal, if different agents are used; dimensional, if different dosages of the same agent are given in the same schedule; ordinal, if the dosages of the same agent can be ranked but cannot be arranged in exact dimensions because of the different schedules.

2.1.2.To analyze blood pressure responses, some of the following changes might be examined:

Ultimate Effect: Increment of before and after values. Immediate Effect: Increment of before and during values.

Therapeutic Trend: Rating of rise, stable, fall, up-and-down, down-and-up, etc. for trend observed in the three consecutive BP values

Ultimate Success: Deciding if the blood pressure after treatment is normal or abnormal.

2.3.Examples

“How do you feel today compared with yesterday?”

“What has happened to your pain since you received the medication?” “What sort of change has occurred in your child’s temperature?” “How much heavier have you gotten since your last birthday?”

2.5. The TNM index expressions are composite “profiles” that contain three single-state variables. They are nominal and cannot be readily ranked because they do not have an aggregated expression. In the TNM staging system, each TNM profile index is assigned to an ordinal stage such as I, II, III, … . If desired, the TNM indexes could be ordinally ranked according to their locations in the staging system.

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magnitudes of differences in actual scores. The three cited percentiles (84, 98, and 99.9) are roughly about l, 2, and 3 standard deviations above the mean in a Gaussian distribution, so that the actual magnitudes of scores are about equidistant.

4.7.4.False. In this “middle” zone of the distribution, where data are usually most abundant, a small change in actual score can produce large changes in percentiles.

4.7.5.Uncertain. The overall percentile scores for each student last year represent a comparative ranking among all students who took the test last year. The percentile rating averaged for different students will not be a meaningful number. On the other hand, if percentiles (rather than actual test scores) are the only information available to the dean, he has no other option if he wants to compare this year’s results.

4.9.As Joe’s lawyer, you argue that “grading on a curve” is an abominable way to assess competence. If you give the exam exclusively to an assembled collection of the most superb omphalologists in the country, some of them will nevertheless fail if “graded on a curve.” Why should Joe, whose actual score of 72% is usually accepted as passing in most events, be failed because of the performance of other people? As lawyer for the Board, you argue that it is impossible to determine an absolute passing score for the complex examination. Accordingly, the passing score is determined from a point on the curve of scores obtained by a reference group of Joe’s peers. He failed because his performance was worse than that of other people with presumably similar backgrounds and training. (This approach has been used, and successfully defended despite legal attacks, by almost all certification Boards in the U.S.)

Chapter 5

5.1. It seems peculiar to assemble a group of people who have been deliberately chosen for all being “healthy,” and then to exclude 5% of them arbitrarily as being outside the “range of normal.” Why not include the full range of data in the healthy, i.e. “normal,” people?

5.3. Individual answers.

5.5.

The value 28 occupies ranks 41–44. The quartile coefficient of variation will be (28 17)/(28 +

17) = 11/45 = .24.

5.7.Individual answers.

Chapter 6

6.1.

6.1.1.Since each tossed coin can emerge as a head or tail, there are four possibilities: HH, HT, TH, and TT. The chance is 1/4 for getting two heads or two tails. The chance is 2/4 = 1/2 for

getting a head and a tail.

6.1.2.Each of the two dice has six sides marked 1, 2,..., 6. When tossed, the pair can have 36

possible outcomes, ranging from 1-1 to 6-6. A value of 7 occurs on six occasions:1 – 6, 2–5, 3–4, 4–3, 5–2, and 6–1. Thus, the probability of a 7 is 6/36 = 1/6.

6.1.3. Common things occur commonly, but uncommon things can also occur if given enough opportunity. A pair of consecutive 7’s can be expected about once in 36 pairs of tosses. If you stay at the dice table long enough to observe 72 or more tosses, two of them could yield consecutive 7’s under ordinary circumstances. If the consecutive pairs of 7’s appear with substantially greater frequency than .03, you might suspect that the dice are “loaded”.

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Using the parametric result (.05) for the coefficient of potential variation would in general produce a more cautious decision than the jackknife results. In this instance, if we accept the parametric coefficient of .05 as indicating stability, the mean might be regarded as stable.

7.3. The laboratory is trying to indicate a 95% interval for the range of the observed data, but the phrase “95% confidence interval” refers to dispersion around an estimate of the mean, not to dispersion of the data. The correct phrase should be simply “range of normal” or “customary inner 95% range.” The word confidence should be used only in reference to the location of a mean (or proportion or other central index of a group).

Chapter 8

8.1. The value of .40 is relatively unstable since the standard error is (.40 )(.60 ) ⁄30 = .089 and .089/.40 = .22. To answer the policy question, one approach is to put a 95% Gaussian confidence interval around the observed proportion. The interval would be .40 ± 1.96 × .089 = .40 ± .175; it extends from .225 to

.575. Because the confidence interval does not include the hypothesized value of .10, the result is probably not compatible with the 10% policy. A second approach is to do a one-sample Z test for the hypothesized parameter of .10. The calculation would be Z = (.40 – .10)/ (.40 )(.60 ) ⁄30 = .30/.089 = 3.37, for which 2P ð .001.

An entirely different approach is to jackknife 12/30 into 12/29 = .41 or 11/29 = .38. This range of maximum difference is .41 – .38 = .03, which is proportionately .03/.40 = .075. Because the ratio exceeds

.05, it immediately shows that the original proportion (.40) is unstable. The jackknife approach, however, does not answer the policy question.

8.3. With an 80% failure rate, the usual rate of success is 20% or .20. The chance of getting three consecutive successes is .2 × .2 × .2 = .008. Unless you think that the clinician is a wizard or that his patients have unusually good prognoses, this low probability seems hard to believe.

8.5.

8.5.1.Since 95% confidence intervals are regularly accepted, use Zα = Z.05 = 1.96. If the proportion of dementia is unknown, use p = q = .5. With a 1% tolerance, e = .01. Then solve n ≥ (.5)(.5)(1.96)2/(.01)2 to get n ≥ 9604.

8.5.2.With .20 as the estimated proportion, solve n ≥ (.20)(.80)(1.96)2/(.01)2 to get n ≥ 6147.

8.5.3.When the commissioner is distressed by the large sample sizes, offer to raise the tolerance

level to 5%. The sample size will then be n ≥ (.20) (.80)(1.96)2/(.05)2, which reduces the sample to n ≥ 246, which is within the specified limit of 300.

8.7.Individual answers.

Chapter 9

Chapter 10

10.1. Like beauty, the concept of what is cogent is in the eye of the beholder. For the investigators, the most important issue seems to have been myocardial infarction. If this is your choice, do you also want to focus on all myocardial infarctions, or just the associated deaths? Would you prefer to focus, instead, on total deaths? And how about strokes, which seemed more common in the aspirin group?

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