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527 PbAf 2006 3.48 Assignment 2 answer sheet 3.51 Using the raw counts of cases provided in the book, you can calculate the joint probabilities by dividing each cell by the total number of cases (2143). Then calculate the marginal probabilities by adding up across the rows and columns (because the joint probabilities are for mutually exclusive categories). Jury plaintiff trial win-reversd plaintiff trial affirmed/dismissed Defendant trial win-reversed Defendant trial win-affirmed/dismissed Totals 9.1% 20.0% 5.2% 34.1% 68.4% Judge Totals 3.3% 12.4% 31.2% 11.2% 8.4% 3.2% 14.0% 48.1% 31.6% 100.0% a) The sample points are all combinations of type of trial and outcome: Plaintiff win/reversed in jury trial; plaintiff win/reversed in judge trial; plaintiff win affirmed in jury; plaintiff win affirmed with judge; defendant win reversed with jury; defendant win reversed with judge; defendant win affirmed with jury; defendant win affirmed with judge. b) P(Jury)= .684 c) P( P win reversed)= .124 d) These are not mutually exclusive because some Jury decisions were won by the plaintiff then reversed (there is overlap). e) P( NOT jury tril) = 1-.684 = .316 f) P( jury OR plaintiff win reversed)= =P(jury) + P(plaintiff win reversed) P(jury AND plaintiff win reversed) =.684+.124-.091=.717 g) P( jury AND plaintiff win reversed)= .091 3.80 a) b) c) d) e) P(A)= P(fail exam)=360/1000= .360 P(B|A)= P(fail paper GIVEN failed exam)= .75 P(B| Ac)= P(failed paper GIVEN did not fail exam)= .20 P(A I B) =P(failed exam AND failed paper) P(A I B) =P(A)*P(B|A)=.360*.75=.27 Part II Here is a crosstab showing the first response to "Race" and the response to "Hispanic Origin". RACE - 1ST CHOICE * PERSON OF HISPANIC ORIGIN Crosstabulation PERSON OF HISPANIC ORIGIN 0.NO 1.YES 15285 1235 92.5% 92.7% 330 95.9% 2.0% 266 86.4% 1.6% 128 97.7% .8% 473 97.5% 2.9% 16482 92.7% 100.0% 7.5% 94.6% 14 4.1% 1.1% 42 13.6% 3.2% 3 2.3% .2% 12 2.5% .9% 1306 7.3% 100.0% RACE 1ST CHOICE 1.WHITE 2.BLACK 3.AMERICAN INDIAN/ALASKA NATIVE 4.NATIVE HAWAIIAN/OTHER API 5.ASIAN Total Count % within RACE - 1ST CHOICE % within PERSON OF HISPANIC ORIGIN Count % within RACE - 1ST CHOICE % within PERSON OF HISPANIC ORIGIN Count % within RACE - 1ST CHOICE % within PERSON OF HISPANIC ORIGIN Count % within RACE - 1ST CHOICE % within PERSON OF HISPANIC ORIGIN Count % within RACE - 1ST CHOICE % within PERSON OF HISPANIC ORIGIN Count % within RACE - 1ST CHOICE % within PERSON OF HISPANIC ORIGIN Total 16520 100.0% 92.9% 344 100.0% 1.9% 308 100.0% 1.7% 131 100.0% .7% 485 100.0% 2.7% 17788 100.0% 100.0% Overall, 7.3 percent of the WSPS sample reported being of Hispanic origin. For race, 93 percent reported their first (or only) race as white, 1.9 percent as Black, 1.7 percent as American Indian or Alaska Native, .7 as Native Hawaiian or other Asian Pacific Islander, and 2.7 as Asian. Among those who reported being of Hispanic origin, people reported being of each of the racial categories. Compared to those who were not Hispanic, Hispanics reported being White or American Indian/Alaska Native slightly more often, and being Black, Native Hawaiian or other API or Asian somewhat less often. These calculations reflect the composition of the WSPS sample, but are not adjusted by weights for the oversampling of King County and its higher proportion of people of color. Also, this table includes information only on the first racial category for each person. It does not account for the additional racial categories selected by 2.5 percent of people who named more than 1 category. Part III 1. The overall poverty rate is the sum of the rate for each group reported (e.g. each region) weighted by its population share. Let Pov stand for the event "poverty", NE for northeast, MW is for Midwest, S is for south, and W is for west. Because the region groups are mutually exclusive and exhaustive we can add the region -specific poverty rates (conditional probabilities) weighted by the proportion of people in each region (to get the "and" events--see book for the multiplicative rule). P(Pov)=P(Pov and NE) + P (Pov and MW) + P(Pov and S) + P (Pov and W) =P(Pov NE) + P (Pov MW) + P(Pov S) + P (Pov W) =P(Pov|NE) x P(NE) + P(Pov|MW) x P(MW) + P(Pov|S) x P(S) + P(Pov|W)x P(W) =(.109)(.19) + (.103)(.23) + (.138)(.36) + (.124)(.23) = .121 Thus, the overall poverty rate in 2003 was 12.1 percent. If the poverty rates for any groups or the proportion of the population in groups changes, then the overall probability will change. Similarly, for 1993 using the age groups we could calculate a poverty of rate 15 percent. Let C stand for under 18 (child), A stand for 19 -64 (adult), and O stand for 65 and above (older). P(Pov) =P(Pov and C) + P (Pov and A) + P(Pov and O) =P(Pov C) + P (Pov A) + P(Pov O) =P(Pov|C) x P(C) + P(Pov|A) x P(A) + P(Pov|O) x P(O) =(.227)(.26) + (.123)(.62) + (.122)(.12) =.150 So, the overall poverty rate in 1993 was 15 percent. The overall poverty rate fell from 15 percent in 1993 to 12.1 percent in 2003. 3. If poverty were statistically independent of race, then the poverty rates conditional on race should be equal. [In this data set, the white only, black onl y, Asian/Pacific Islander only, and Other Race/multiracial categories are mutually exclusive and exhaustive.] If P is poverty, W is white, B is Black, and As is Asian/Pacific Islander, and OR is other races, then by definition statistical independence is Hispanic Asian/PI Other Race P(Pov|W)=P(Pov|B)=P(P ov|As) =P(Pov|OR)=P(Pov) But because P(Pov|W) = .105 and the P(Pov|B)= .244 in 2003, there is clear evidence that race provides information about the likelihood of being poor. What kind of data would you want to further explore the cause of race differences in poverty? Could race differences in age distribution, residence, or region explain the race differences in poverty rates? Poor White Black 4. We can manipulate the conditional probability formula (and use Bayes' Theorem) to get the probability of being a child given poverty status. If C is for child (under 18), then P ( Pov I C ) P (C ) P( Pov | C ) = [multiplicative rule of probability] P(C|Pov)= P ( Pov ) P( Pov) [You could stop here and solve the equation, but I'd like to point out that the equation above equals: P ( Pov | C ) P(C ) P( Pov | C ) P(C ) + P( Pov | C c ) P(C c ) This happens to be Bayes Theorem. The denominators here and just above are equivalent.] [Do you see why?] = In 2003, we know P(Pov|C)= .176 from the chart, P(C)=.25 from the chart as well, and we calculated P(Pov) in problem 1, so =(.176 x .25)/.121 = .364 So, the probability of a being child given poverty status was about 36 percent in 2003. That is, 36 percent of the poor were children. In the last 30 years, Social Security and Medicare programs resulted in large decreases in the p robability of being poor for older people, and this makes children a larger part of the pool of poor people (can you see how this changes the formula?). Increases in single mother households between the 1960s and the 1980s pushed poverty rates for children higher. Also, the poverty rate for working-age adults in more sensitive to economic cycles than are the rates for young or old; the poverty rates for adults increase during recession making children a smaller portion of the total. 5. We want to know the proportion of poor people between 19 and 65 (A) in both 2003 and 1993--the probability of not being under 18 or over 65 given poverty status [P(A|Pov)]. We want, for 2003: P(A|Pov) = P(A and Pov)/P(Pov)=P(APov)/P(Pov) = [P(18-24 and Pov) + P((25-34 and Pov) + P((35-44 and Pov) + P((45-55 and Pov) + P((55-59 and Pov) + P((60-64 and Pov)]/P(Pov) = [P(18-24|Pov)P(Pov) + P(25-34|Pov)P(Pov) + ...]/P(Pov) = [(.165)(.10) + (.128)(.14) + (.096)(.15) + (.076)(.14) + (.082)(.06) + (.097)(.04)]/.121 =0.56 Alternatively, we know the programs won't be useful for the children and seniors, so let's find the proportion of the poor in those groups then take the complement. We want P(C or O|Pov). Since C and O are mutually exclusive categories, we can add their conditional probabilities (because they have the same denominator): P(C or O|Pov) =P(C|Pov) + P(O|Pov) We figured the P(C|Pov) in the last question as .36, similarly P(O|Pov) = P(Pov and O)/P(Pov) = P(O)P(Pov|O)/P(Pov) =(.102)(.12)/.121 =0.101 [Multiplicative Rule which gets us to Bayes Theorem again] So, P(C or O|Pov) = .36 + .10 = .46 Then, since A, Y, and O are mutually exclusive and exhaustive, take the complement: P(A|P) =1 - .46 = .54 For 1993, we can calculate the P(A|Pov) using numbers from the table and the overall poverty rate we calculated before: P(A|Pov) = P(Pov and A)/P(Pov) = P(A)P(Pov|A)/P(Pov) = (.62)(.123)/.15 = .508 Thus, in 1993 just about half and in 2003 only a little over half of the poor were likely to use labor market anti-poverty policies. Note that children will often be living with poor adults, so a program successful for adults would also change the rates for children. Also important is the fact that some programs may not raise incomes above poverty (a fixed threshold) but could increase incomes closer to that rate.

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