All Problems are from: Myra L. Samuels and Jeffrey A. Witmer,Statistics for the Life Sciences, 3rd edition, Prentice-Hall (2003)12.612.1212.1512.22a: From Exercise 12.6, s_Y|X = sqrt[SS(resid)/df] = sqrt[16.7812/(12-2)] = 1.3 The standard error of the slope is12.27a: r = 82.8977/sqrt[(28465.7)(.363708)] = .8147Stat 13 Final Exam Review Problem Solutionshttp://www.stat.ucla.edu/~dinov/courses_students.htmlAll Problems are from: Myra L. Samuels and Jeffrey A. Witmer, Statistics for the Life Sciences, 3rd edition, Prentice-Hall (2003)Chapter 10 – 2 Tests10.4: H0: Timing of births is random (Pr(weekend) = 2/7)HA: Timing of births is not random (Pr(weekend) not= 2/7). Weekend WeekdayObserved 216 716Expected 266.29 665.71Difference -50.29 +50.29Chi-Square = Sum of (O-E)2/E = (-50.29)2/266.29 + (+50.29)2/665.71 = 13.3With df = 1, Table 8 gives http://socr.stat.ucla.edu/Applets.dir/OnlineResources.html#Tables.0001 < P-value < .001. There is sufficient evidence to conclude that the timing of births is not random. 10.5 Let WF and DF denote white and dark feathers; let SC and LC denote small and large comb.H0: The model is correct; that is, Pr(WF,SC) = 9/16, Pr(WF,LC) = 3/16, Pr(DF,SC)=3/16, Pr(DF,LC)=1/16.HA: The model is incorrect; that is, Probabilities are not as specified by H0. OBS EXPWF,SC 111 106.875WF,LC 37 35.625DF,SC 34 35.625DF,LC 8 11.875Chi-square = Sum of (O-E)2/E = 1.55, with df = 3, P-value > .20 since Table 8 (http://socr.stat.ucla.edu/Applets.dir/OnlineResources.html#Tables) gives chi-square_{.20} = 4.64. We do not reject H0. There is little or no evidence (P-value > .20) to conclude that the model is incorrect; the evidence is consistent with the Mendelian model. 10.6a: n = 1000 OBS EXP DIFFBOY 520 500 20GIRL 480 500 -20Chi-square = 1.6. With df = 1, Table 9 (http://socr.stat.ucla.edu/Applets.dir/OnlineResources.html#Tables) shows P-value > .2010.6b: n = 5000 OBS EXP DIFFBOY 2600 2500 100GIRL 2400 2500 -100Chi-sq = 8. With df = 1, Table 9 (http://socr.stat.ucla.edu/Applets.dir/OnlineResources.html#Tables) shows 0.001 < P-value < 0.01.10.6c: n = 10000 OBS EXP DIFFBOY 5200 5000 200GIRL 4800 5000 -200Chi-sq = 16. With df = 1, Table 9 http://socr.stat.ucla.edu/Applets.dir/OnlineResources.html#Tables shows . P-value < .0001.10.11: The hypotheses areH0: The men are guessing (Pr(correct) = 1/3) Ha: The men have some ability to detect their partners (Pr(correct) > 1/3)Observed ExpectedCorrect 18 12Wrong 18 24 Total 36 36Chi-Square statistic = 4.5. With df = 1, Table 9 http://socr.stat.ucla.edu/Applets.dir/OnlineResources.html#Tablesgives .01 < P-value < .025 and we reject H0. Note that no alpha level was specified, but a P-valueless than 0.025 is generally considered to be small. 10.17: table is striped all red alive 65 (70.31) 23 (17.69) TOTAL = 88 dead 98 (92.69) 18 (23.31) TOTAL = 116 TOTAL = 163 TOTAL = 41 TOTAL = 204 Null is that there is no difference in the survival rates for the two types, and alternative is that the mimic form (all red) survives more than the striped kind. Test stat is chi-sq = [(65 - 70.31)2/70.31] + [(98 - 92.69)2/92.69] + [(23 - 17.69)2/17.69] + [(18 - 23.31)2/23.31] = 0.40 + 0.30 + 1.59 + 1.21 = 3.50 Again, since alternative is one-tailed, we half to get p-values: 0.025 < P-value < 0.05. Since P-value < , we conclude that the mimic form of P. cinereus seem to survive more successfully that the red-striped. (df = 1) 10.22a: H0: E. coli had no effect on tumor incidences.Ha: E. coli increased tumor incidences. H0: p1 = p2 Ha: p2 > p1 = .05Df = 1Germ-free E. coliTumors 19 (21.34) 8 (5.66) 27No tumors 30 (27.66) 5 (7.34) 35Total 49 13 62chi-sq = 2.17chi-sq_.20 = 1.64 and chi-sq_.10 = 2.71Multiply by half because Ha is directional: therefore, .10 > P > .05We do not reject H0.There is insufficient evidence (.10 > P > .05) to conclude that E. coli increases the number of tumors in mice.10.22b: If the percentages stay the same but the sample sizes double, then the O (Observed) and E (Expected) values double. Also (O-E) doubles, which means that (O-E)2 is four times larger. But when divided by a doubled E, we get that (O-E)2 / E is doubled. So the Chi-square statistic is doubled. Then H0 is rejected because .01 < P-value < .025.Similarly, if the samples were to triple, then the Chi-square statistic would triple. Then .005 < P-value < .01 and, of course, H0 is rejected.This makes sense. If you toss a coin 4 times and get 3 (75%) heads, that is not unusual (z = 1). But if you tossed a coin 100 times and got 75 (75%) heads, then that would be very unusual (z = 5). 10.35: p1 = Pr{HP / MP} and p2 = Pr{HP / MA}. Null is that p1 = p2, and alternative is that p1 and p2 differ. From table 8, http://socr.stat.ucla.edu/Applets.dir/OnlineResources.html#Tables0.001 < P-value < 0.01, and, since P-value < alpha, reject the null and conclude that there is an association (dependence) between the species. Data suggests repulsion. 47.3 % = p1-hat < p2-hat = 70.8%. (df = 1) 10.37a:Pr {Yes|A} : 111/513 = 0.21637 = 21.637% Pr {Yes|B} : 74/515 = 0.1437 = 14.37%10.37b: Pr {A|Yes} : 111/185 = 0.60 = 60%Pr {A|No} : 402/843 = 0.4767 = 47.67%10.73: Let p denote the probability that the uninfected mouse in a cage becomes dominant.H0: Infection has no effect on development of dominant behavior (p = 1/3)HA: Infection tends to inhibit development of dominant behavior (p > 1/3)Uninfected mouseDominant NotDominant15(10) 15(20)Chi-square statistic = 3.75. With df = 1, we get 0.025 = 0.05/2 < P-value < 0.10/2 = 0.05 and we reject Ho. There is sufficient evidence (0.025 < P-value < 0.05) to conclude that infection tends toinhibit development of dominant behavior. 10.87: The hypotheses are H0: Type of treatment does not affect survivalHA: Type of treatment affects survival table is Zidovudine Didanosine Both Total Died 17 (11.29) 7 (11.50) 10(11.21) 34 Survived 259(264.71) 274(269.50)
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