Solutions to Chapter 29 Special Review Exercises
1(a) This is an observational study. It is the prisoners themselves who decide whether to stay in the program or drop out.
(b) The treatment group consists of the prisoners who finish boot camp. The control group consists of those who drop out.
(c) Those who stayed the course might have been quite different to start with^×better motivated, more self-disciplined^×than those who dropped out.
(d) (i) The treatment group consists of all those who volunteer for the program, whether or not they complete it. The control group consists of those who do not volunteer.
(ii) The recidivism rate in the two groups is similar, suggesting that the program has little effect^×although it is an effective screening device. (The ones who finish are motivated and have enough self-discipline to go straight when they are released.)
Comment. If anything, this comparison is biased in favor of treatment, because those who volunteer would seem more likely to succeed in civilian life than those who do not volunteer.
(e) Most completed. The recidivism rate for the combined group is 36%; the rate for those who did not complete is 74%; for those who completed, the rate was 29%. If most dropped out, the rate for the combined group would be nearly 29%-- and it is.
A more exact answer is possible. Let x be the fraction who complete. Then 29x+74(1-x)=36, so x=0.84.
2 Not a good explanation. The comparison is between left handed ball players and right handed players. C confounder would have to be associated with left handed ness^×among the players^×and cause higher mortality rates.
This study had problems, including a small sample size.
3 False. This is like the graduate admissions study, pp.17ff. As it turns out, most of the Catholics go to the Secondary Schools, and students in those schools do rather poorly on the proficiency tests.
4 The figure is not a histogram; the class intervals are unequal: if you adjust for the lengths of the intervals, the pattern goes away. In a histogram class intervals include the left endpoint but not the right. For instance, the rectangle whose base is the interval from 25 to 30 represents the students age 25-29; those age 30 are in the next rectangle. We can start the fist rectangle at 15, and ended the last one at 75, somewhat arbitrarily. What the histogram shows is that most students are in their early 20s, with a sprinkling of precious youngsters^×and some determined oldsters.
5 False. The data are cross-sectional not longitudinal. An alternative explanation: death rates are higher among people who drink, smoke, and don^Òt eat breakfast. So, fewer of these people survive past 65 and get interviewed.
6 The difference between the 90th and 50th percentiles is bigger^×long right hand tail.
7 There is a couple where both husband and wife are about 2 years old. There is a man aged 55 married to a 5-year-old. There is a woman aged 30 married to a 5-year-old. And many other odd couples. Something is wrong.
Comment. For the motivation, see the comment on review exercise 11 in chapter 11.
8 y=0.533x+1.667
9 (a) Investigator A gets the higher correlation: B^Òs correlation is attenuated due to restriction of range. (See exercise 4 and 5 on p. 130, exercise 9 on p.144)
Comment. In the March 1993 Current Population Survey, the correlations were about 0.4 and 0.3, respectively.
(b) The "ecological" correlation^×for the state averages^×will be higher. Se section 9.4.
10 Option (iii) is right^×regression effect.
11 A first-yea GPA of 3.5 is 1 SE above average. Students with this SPA averaged about r*1=0.4 SDs above average in second year, by the regression method. Sally must have been above average on second-year GPA by about 0.4 SDs, putting her in the 66th percentile.
12 Something is wrong. The r.m.s. error has to be less than the SD.
13 Less than. The diagram is heteroscedastic, with less scatter around the regression line for women with lower educational levels (p. 192).
14 The students who scored 500 on the V-SAT averaged about 0.6*500+245=545 on the M-SAT. That is the new average. The new SD is the r.m.s. error of the regression line, which is 80 points. Now (500-545)/80=-0.56. The answer: about 70% if 25,000=0.70*25,000=17,500.
15 (a) 4/52*3/51*2/50=2/10,000.
(b) 48/52*47/51*46/50=78%
(c) 36/52*35/51*34/50=32%
(d) 100%-32%=68%
16 4/10,000
17 The net gain is like the sum of 100 draws from a box with 18 tickets marked "+$1", 18 tickets marked "-$1", and 2 tickets marked "-$0.50". (These last 2 tickets correspond to 0 and 00, where you lose half your stake.) The average of this box is -$1/38=-$0.0263. The expected net gain is -$2.63. The SD of the box^×don^Òt use the short-cut^×is $0.980. The SE for the net gain is $9.80. now $2.63/$9.80=0.27. The answer is about 40%
18 Using a telephone survey tends to exclude the homeless or recently homeless. On this basis the 3% is a little too low.
Comment. People might not want to admit having been homeless; that also makes the 3% too low. On the other hand, if people were homeless, the y might remember that as being more recent that it really was. (Vivid experiences tend to be brought forward in memory.) That would make the 3% too high.
19 (a) Each man in the population has had sec with some number of women, perhaps 0. Adding these numbers up gives the total number of female partners for the men. There is a similarly-defined total number of male partners for the women. The two totals must be equal^×if a man has a woman as partner, that woman has the man as partner. Since the number of partners must be about the same.
Comment. Strict quality need not hold^×as laumann points out^×although the exceptions seem quite minor. For instance, American men might be more likely than women to have sex while travelling out of the country.
(b) The total of a list is related to average not the median, so laumann seems to have missed the point.
(c) Lewontin^Òs point is a good one: people in jail, and the homeless, are much more exposed to AIDs than the rest of us. The omission distorts the results in an important way.
(d) Non-respondents may well behave quite differently from respondents.
20 the number of sample families without cars is like the sum of 1500 draws from a 0-1 box. There is a ticket in the box for each of the 25,000 families in the town, marked 1 (no car) or 0(owns cars). So, the fraction of 1^Òs in the box is 0.1, and the SD of the box is (square root of 0.1*0.9)=0.3. the expected value for the sum is 1500*0.1=150. The SE is (square root of 1500)*0.3=12. The number of sample families without cars will be around 150, give or take 12 or so. Now 12 out of 1500 is 0.8%. The percentage of sample families without cas will be around 10%, give or take 0.8% or so. The chance is about 80%.
21 Option (i) is right: section 20.3.
22 The town is large, there is not much difference between drawing with or without replacement. Use the method of section 18.4. The number with phones is like the sum of 500 draws from a box with 20 tickets marked "1" and 80 marked "0". The expected number is 500*0.2=100. The SE is the square root of (500*0.2*0.8)=8.94. And 0.5/8.94=0.056. From the table, the chance is about 4%.
Comment. (i) By computer, the area under the normal curve between ± 0.056 is 4.47%.
(ii) You can also use the binomial formula (chapter 15), provide you have a computer to work out the binomial coefficient and the power. The normal approximation is very good^×4.46%.
23 (a) Bias; section 19.3
(b) A cluster sample is a probability sample, a sample of convenience isn^Òt; sections 19.4, 22.5, 23.4.
(c) Cluster samples are generally less accurate than simple random samples of the same size, but much cheaper, so they are quite cost effective; section 22.5.
24 False. This is a sample of convenience, not a simple random sample. The formula does not apply (p.424).
25 (a) (i) (b) (ii) (c) (ii) (iii)
Comment. The expected value of the sample average equals the population average. This is so even after the sample is drawn^×the expected value is sort of the average overall possible samples, not just the particular sample you happened to draw. In the frequency theory, it is a mistake to say that the expected value of the population average equals the sample average. See section 21.3 for a similar point about percentage.
26 (a) True: (488+592)/2=540.
(b) True: just work out the SE and the confidence interval from the SD of 390 and the sample size of 225.
(c) False. The SD is a large fraction of the average, if the data followed the normal curve, there would be a lot of negative distances travelled.
(d) True: pp.411, 418-19
(e) False. The population average does not have a probability histogram.
(f) False. If you double the sample size, you cut the SE for the average no by a full factor of 2, but only by the square root of 2=1.4. the confidence interval will be about 1.1.4=0.7 as wide.
Comment. The provlem is tacitly assuming that the airline used a symmetric confidence interval.
27 (a) Can^Òt be done with the information given, you have to use the half sample method (section 22.5)
(b) 0.5%
28 False. The SE measures the likely size of the chance error. It does not take bias into account. Therefore, neither foes the confidence interval.
29 Option (ii) is it. Use the half sample method (section 22.5).
30 (a) True.
(b) False. You can use the variability in the data to estimate the SE of the error box, and then compute a standard error for the average, see part (c).
(c) True. The SD of the data estimates the SD of the error box, hence, the chance error in a single measurement. The SE for the average can be estimated by the square root law, as in section 24.1.
31 (a) 15, the SD of the measurements. (b) 3, the SE for the average.
32 There is a 25% chance for a chick to be c/c and have white feathers, so the chance of getting 12 chicks out of 24 with colored feathers id 0.5%. You can also use the method of section 18.4, which gives 0.37%.
33 Model: there is one ticket in the box for each household in the country, market 1 if the household experienced a burglar within the last 12 months, and 0 otherwise. There are 100 tickets in this box. The survey data are like 50,000 draws made at random from the box. (There is essentially no difference between drawing with or without replacement.) If the FBI data are accurate, the percentage of 1^Òs in the box id 3.2%; this is the null hypothesis. On the null hypothesis, the 4.9% is higher than the 3.2% just because of sample error. The alternative hypothesis: the percentage of 1^Òs in the box is bigger than 3.2%.
If the null hypothesis is right, the percentage of 1^Òs has an EV of 3.2%, and the SE is 0.08%. (The SD of the box should be computed using the 3.2% specified by the null hypothesis.) The percentage of 1^Òs in the sample is 4.9%. The difference between 4.9% and 3.2% is almost impossible to explain as chance variation: z=20.many burglaries are not reported to the police.
34 (a) True. (b) False (p.482). (c) False (p.482)
35 Parts (a-d) can be handled like example 4 on p.510; parts (e-g). like the radiation-surgery group on pp.514ff. By convention, the :observed difference" is "treatment-control."
Comment. (i) The sample size used in the calculation are at baseline; no adjustment is made for mortality.
(This would be minor.)
(ii) MRFIT was a well-designed, well-conducted study-which came up with answers contradicting the prevailing wisdom. For more cites to the literature on cholesterol, see note 7 to chapter 29.
36 the question makes sense, but cannot be answered with the information given: you have two correlated responses for each subject-the rating of lawyers and of contractors (p. 519).
37 (a) Can^Òt be done with the information give, the Current Population Survey isn^Òt a simple random sample (chapter 22).
(b) Use the method of section 28.4.
Obs Exp Obs-Exp
7 20 14.2 12.8 -7.2 7.2
21 19 21.0 19.0 0.0 0.0
13 9 11.5 10.5 1.5 -1.5
23 10 17.3 15.7 5.7 -5.7
x2@ 11.9 on 3 degree of freedom, P<1%.
Interpretation: women who are less well educated tend not to be in the labor force; women who are better educated are more likely to be in the labor force, and in professional or managerial jobs.
Note: even if Obs-Exp is 0.0 in some cells, that does not affect the formula for degree of freedom. Degree of freedom depend on the model, not on the data (pp. 533,539).
38 The quote is misleading (p. 482). The test tells you the chance of seeing a big difference, given the null hypothesis. It does not tell the chance that the null hypothesis is right, given the big difference.
39 These investigation seem to have made a number of statistical error. For one thing, they did lots of tests; data snooping makes P-values hard to interpreted (p.549):even if all their null hypotheses were right, they were almost bound to find some highly significant differences. For another thing, they seem to be thinking that P measures the size of the effect, and it doesn^Òt (p.555)/ the effect they estimated is minute: a 100-fold increase in asbestos fiber concentration only increases the risk of lung cancer, and the investigators paid no attention to this variable. The argument is weak, and there is no reason to think that asbestos in the water causes lung cancer.
Comment. The investigators found no effect for blacks or women; see note 52 to chapter 29.
40 (a) The population and the sample are the same, namely, all Dutch men of military age in the period 1963-66. The estimates and the parameters coincide, because we have date for the whole population.
(b) The statistical tests do not make much sense, because the date are for a whole population )p.558). Even if you think of these men as a sample from larger population, why is it anything like a simple random sample?
Comment. (i) The first-borns and the second-borns would generally be from different families, due to the design.
(ii) Belmont and Marolla did a first-rate piece of work, but the tests seem to be largely crermonial.