## H8.dvi

8.2 It has been hypothesized that silicone breast implants cause illness. In one study it was found
that women with implants were more likely to smoke, to be heavy drinkers, to use hair dye, andto have had an abortion than were women in a comparison group who did not have implants. Usethe language of statistics to explain why this study casts doubt on the claim that implants causeillness.

Solution:In this observational study, the effect of implants on illness is confounded with the effectson illness of smoking, drinking heavily, using hair dye, and having an abortion.

8.8 Fluoridation of drinking water has long been a controversial issue in the United State. One of the
first communities to add fluoride to their water was Newburgh water supply, to begin on April 1of that year. During the month of April citizens of Newburgh complained of digestive problems,which were attributed to the fluoridation of the water. However, there had been a delay in theinstallation of the fluoridation equipment, so that fluoridation did not begin until May 2. Explainhow the placebo effect/nocebo effect is related to this example.

Solution:It appears that the digestive problems were caused by the nocebo effect. People fearedthat fluoridation of their drinking water would cause health problems and this fear led to digestiveproblems when, in fact, fluoride was not yet being added to the water.

8.16 In an experiment to compare six different fertilizers for tomatoes, 36 individually potted seedlings
are to be used, six to receive each fertilizer. The tomato plants will be grown in a greenhouse, andthe total yield of tomatoes will be observed for each plant. The experimenter has decided to use arandomized blocks design. Two possible arrangements of the blocks are shown in the accompanyingfigure. One factor that affects tomato yield is temperature, which cannot be held exactly constantthroughout the greenhouse. In fact, a temperature gradient across the bench is likely. Heat for thegreenhouse is provided by a steam pipe which runs lengthwise under one edge of the bench, and sothe side of the bench near the steam pipe is likely to be warmer.

(a) Which arrangement of blocks (I or II) is better? Why?
(b) Prepare randomized allocation of treatments to the pots within each block.

Solution: (a) II is better, because it will make each block as homogeneous as possible. Under ar-rangement II, each treatment is observed at each temperature. In contrast, arrangement I confoundsthe effects with the effect of temperature.

(b) Label the pots within each block as 1, 2,.,6. One possible allocation is as follows:
8.31 The American Heart Association designed a 12-week educational program for women with the goal
of improving their nutrition, physical activity, and knowledge of heart disease. A group of 23,171
Office hours: T 9:30-10:30 and F 11:00-12:00 at 1276 MSC

women were recruited for the program and were sent a manual that contained weekly informationregarding risk factors for cardiovascular disease, as well as how to build a support system for lifestylechange. The women were asked to return follow-up evaluations during the program in which theyreported on their physical activity, diet, and knowledge of heart disease. However, only 6,389women sent in evaluations at the end of the fourth week. Only 3,775 of the women complete theprogram and sent in their evaluations when the program ended after 12 weeks. The participantswho completed the program reported a marked increase (P-value=0.001) and a decrease in “excesscalories or fat” (P-value=0.001) compared with their levels at the beginning of the program. Isit correct to conclude, given the very small P-values, that the program is successful at promotingexercise and improved diets? Discuss.

Solution: It is not correct to conclude that the program is successful (If we define success asto help most of the people improve their physical conditions). Only a small percentage of theparticipants complete the program and sent in evaluations. It might well be that women who didnot experience success in the program dropped out, or at least did not sent in evaluations. Thosewho, for whatever reason, do well during the program are more likely to complete the program andto send in evaluations.

9.6 Trichotomy is a psychiatric illness that causes its victims to have an irresistible compulsion to pull
their hair. Twp drugs were compared as treatments for trichotillomania in a study involving 13women. Each woman took clomipramine during one time period and desipramine during anothertime period in a double-blind experiment. Scores on a trichotillmania-impairment scale, in whichhigh scores indicate greater impairment, were measured on each woman during each time period.

The average of the 13 measurements for clomipramine was 6.2; the average of the 13 measurementsfor desipramine was 4.2. A paired t test gave a value of ts = 2.47 and a two-tailed P-value of0.03. Interpret the result of the t test. That is, what does the test indicate about clomipramine,desipramine, and hair pulling?
Solution: The data provide fairly strong evidence (P=0.03) that there is difference between twotreatments. And from the two averages we know desipramine is more effective than clomipraminein reducing the compulsion to pull one’s hair.

9.38 A volunteer working at an animal shelter conducted a study of the effect of catnip on cats at the
shelter. She recorded the number of “negative interactions” each of 15 cats made in 15 minuteperiod before and after being given a teaspoon of catnip. The paired measurements were collectedon the same day within 30 minutes of one another; the data are given in the accompanying table.

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(a) Construct a 95% confidence interval for the difference in mean number of negative interaction.

(b) Construct a 95% confidence interval the wrong way, using the independent-sample method.

How does this interval differ from the one obtain in part (a)
Solution: (a) y1 − y2 = d = −1, sd = 1.2
15 = 0.3098. Using df=15-1=14, we have −1 ± 2.145 ∗ 0.3098 or (-1.66, -0.34)
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Then 95% confidence interval is −1±t0.025SEy
=−1±2.048∗0.747=−1±1.530 or (-2.530, 0.530)
Comparing with (a), we know the result in (b) is much wider than that in (a)
9.39 Refer to Exercise 9.38. Compare the before and after populations using a t test at α = 0.05. Use a
Solution: H0: The before and after means are the same (µ1 = µ2)
HA: The before and after means are different (µ1 = µ2)
s = −1/0.3098 = −3.23 With df=14, Table 4 gives t0.0005 = 4.140;
t0.005 = 2.977; thus, 0.001 < p < 0.01. We reject H0; there is strong evidence (0.001 < p < 0.01) ofa before and after difference.

(a) Compare the before and after populations using a sign test at α = 0.05. Use a nondirectionalalternative.

(b) Calculate the exact P-value for the analysis of part (a)
Solution: (a) Let p denote the probability that a before count is higher than the correspondingafter count.

N+ = 2, N− = 10, Bs = 10. Looking under nd = 12 in Table 7, we see that 0.02 < P < 0.05. Thereis sufficient evidence to conclude that the after count tends to be higher than the before count.

(b) Under H0, we know p=0.5, so both N− and N+ have binomial distribution with parametern=nd = 12, success probability p = 0.5.

Then, P-value= P(more extreme outcomes|H0 is true)=P(N− = 10, N− = 11, N− = 12, N− =2, N− = 1, N− = 0)= 2*P (N− = 10, N− = 11, N− = 12)=2*(12C100.510 ∗ 0.52 +12 C110.511 ∗ 0.51 +12C120.512 ∗ 0.50) = 2 ∗ (66 ∗ 0.510 ∗ 0.52 + 12 ∗ 0.511 ∗ 0.51 + 1 ∗ 0.512 ∗ 0.50) = 0.0386If we want to use R to get the exact P-value, the commands should be:
> pbinom(2, 12, 0.5) + 1 − pbinom(9, 12, 0.5)
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Source: http://www.stat.wisc.edu/courses/st371-ane/hw/hw08_sol.pdf

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