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BirthweightNS 1 7.5 1 6.2 1 6.9 1 7.4 1 9.2 1 8.3 1 7.6 2 5.8 2 7.3 2 8.2 2 7.1 2 7.8 3 5.9 3 6.2 3 5.8 3 4.7 3 8.3 3 7.2 3 6.2 4 6.2 4 6.8 4 5.7 4 4.9 4 6.2 4 7.1 4 5.8 4 5.4
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7 Chapter Solutions 1. ^ satisfied n1 = 120 x1 = 30 p 1 = 0.25 ^ not satisfied n2 = 150 x2 = 62 p 2 = 0.41 95% CI for (p1 - p2) ^ ^ ^ ^ p (1 - p1) p2 (1 - p2) ^ ^ ( p1 - p2) Z1 1 + 2 n1 n2 ^ ^ Check: min(n1 p 1,n1(1- p 1))=min(120(0.25),120(0.75))=30 ^ ^ min(n2 p 2,n2(1- p 2))=min(150(0.41),150(0.59))=62 (0.25 - 0.41) + (1.960) -0.16 + 1.960 (0.056) -0.16 + 0.11 (-0.27, -0.05) 0.25(1- 0.25) 0.41(1- 0.41) + 120 150 2. a) b) c) n = 1200 x = 423 ^ = x/n = 423/1200 = 0.35 p ^ ^ p (1 - p) 0.35(1- 0.35) = = 0.014 n 1200 95% CI for p ^ ^ p (1 - p) ^ p Z1 2 n 0.35 + (1.960) (0.014) 0.35 + 0.027 (0.323, 0.377) 3. H0: Gender and rank are independent H1: Gender and rank are not independent (O - E )2 2 , df=3 = E Reject H0 if 2 > 7.81 = 0.05 88 Instructor Female Male Total 2 12 (11.44) 21 (21.56) 33 Asst. Prof. 15 (13.87) 25 (26.13) 40 Assoc. Prof. 18 (16.64) 30 (31.36) 48 Total Prof. 7 (10.05) 22 (18.95) 29 52 98 150 = (12-11.44)2/11.44 + (15-13.87)2/13.87 + (18-16.64)2/16.64 + (7-10.05)2/10.05 + (21-21.56)2/21.56 + (25-26.13)2/26.13 + (30-31.36)2/31.36 + (22-18.95)2/18.95 = 0.027 + 0.092 + 0.111 + 0.926 + 0.015 + 0.049 + 0.059 + 0.491 2 = 1.77 Do not Reject H0 since 1.77 < 7.81. We do not have significant evidence, =0.05, to show that gender and rank are not independent. 4. H0: Distribution is 10%, 20%, 40%, 20%, 10% H1: H0 is false (O - E )2 2 , df=4 = E Reject H0 if 2 > 9.49 1 12 10 0.4 2 18 20 0.2 3 50 40 2.5 4 10 20 5 = 0.05 Observed Expected (O-E)2/E 2 5 10 10 0 Total 100 100 8.1 = 8.1 Do not reject H0 since 8.1 < 9.49. We do not have significant evidence, =0.05, to show that the distribution is not 10%, 20%, 40%, 20%, 10%. 5. Plant I Plant II H0: p1 = p2 H1: p1 < p2 n1 = 250 n2 = 220 x1 = 28 x2 = 38 ^ p 1 = 0.112 ^ p 2 = 0.173 89 Z= ^ ^ p1 - p 2 ^ ^ p(1 - p) 1 + 1 n1 n 2 ^ ^ min(n1 p 1,n1(1- p 1))=min(250(0.112),250(0.888))=28 ^ ^ min(n2 p 2,n2(1- p 2))=min(220(0.173),220(0.827))=38 Check: Reject H0 if Z < -1.645 + 28 + 38 ^ x x p= 1 2= = 0.14 n1 + n 2 250 + 220 ^ ^ p1 - p 2 0.112 - 0.173 Z= = 1 1 1 1 0.14(1- 0.14) + ^ ^ p(1 - p) + 250 220 n1 n 2 = -0.061/0.032 = -1.91 Reject H0 since -1.91 < -1.645. We have significant evidence, =0.05, to show that p1 < p2, p<0.05. 6. H0: Distribution is 25%, 25%, 50% H1: H0 is false (O - E )2 2 , df=2 = E Reject H0 if 2 > 5.99 R 205 250 8.1 C 220 250 3.6 L 575 500 11.25 = 0.05 Observed Expected (O-E)2/E 2 = 22.95 Reject H0 since 22.95 > 5.99. We have significant evidence, =0.05, to show that the distribution is not 25%, 25%, 50%, p<0.005. Total 1000 1000 22.95 90 7. H0: p = 0.15 H1: p > 0.15 = 0.05 ^ - p0 p Z= p 0 (1 - p 0 ) n Check: min(np0, n(1-p0))=min(125(0.15),125(0.85))=18.75 Reject H0 if Z > 1.645 25 ^ x = p= = 0.20 n 125 ^ p - p0 0.20 0.15 Z= =0.05/0.032 = 1.57 p 0 (1 - p 0 ) 0.15(1 0.15) n 125 Do not reject H0 since 1.57 < 1.645. We do not have significant evidence, =0.05, to show that p > 0.15. 8. a) b) ^ ^ p 1- p 2 = 0.089 0.127 = -0.038 ^ ^ ( p1 - p2) Z1 Check: 2 ^ ^ p1 (1 - p1) + ^ ^ p2 (1 - p2) n1 n2 ^ ^ min(n1 p 1,n1(1- p 1))=min(1000(0.089),1000(0.911))=89 ^ ^ min(n2 p 2,n2(1- p 2))=min(1000(0.127),1000(0.873))=127 c) 0.089(1- 0.089) 0.127(1- 0.127) + 1000 1000 -0.038 + 1.960 (0.014) -0.038 + 0.027 (-0.065, -0.011) Reject H0: p1=p2 because the CI does not include 0. -0.038 + (1.960) 9. H0: Treatment and outcome are independent H1: Treatment and outcome are not independent (O - E )2 2 = , df=2 E 91 = 0.05 Reject H0 if 2 > 5.99 Control 21 (27.7) 29 (22.3) 50 E1 28 (27.7) 22 (22.3) 50 E2 34 (27.7) 16 (22.3) 50 Total 83 67 150 Improved Not Total 2 = 1.62 + 0.003 + 1.43 + 2.01 + 0.004 + 1.78 = 6.8 Reject H0 since 6.8 > 5.99. We have significant evidence, =0.05, to show that treatment and outcome are not independent, p<0.05. 10. a) ^ Control n1 = 50 p 1 = 21/50 = 0.42 ^ E1 n2 = 50 p 2 = 28/50 = 0.56 95% CI for (p1 - p2) ^ ^ ^ ^ p (1 - p1) p2 (1 - p2) ^ ^ ( p1 - p2) Z1 1 + 2 n1 n2 ^ ^ Check: min(n1 p 1,n1(1- p 1))=min(50(0.42),50(0.58))=21 ^ ^ min(n2 p 2,n2(1- p 2))=min(50(0.56),50(0.44))=22 (0.42 - 0.56) + (1.960) 0.42(1- 0.42) 0.56(1- 0.56) + 50 50 b) -0.14 + 1.960 (0.099) -0.14 + 0.19 (-0.33, 0.05) No, CI includes 0, proportions not significantly different. 11. a) b) c) n Z p(1 p) 1-/2 E 2 1.960 0.25 0.04 2 600.25. n = 601. 2 Z 1.960 n p(1 p) 1-/2 0.27(1- 0.27) E 0.04 (b) assuming estimate still holds. 92 2 473.2. n=474. 12. H0: Therapy and severity are independent H1: Therapy and severity are not independent = 0.05 2 (O - E ) 2 , df=2 = E Reject H0 if 2 > 5.99 Minimal Moderate Severe Total Medical 90 60 50 200 (70) (60) (70) Non50 60 90 200 Traditional (70) (60) (70) Total 140 120 140 400 2 = 5.71 + 0 + 5.71 + 5.71 + 0 + 5.71 = 22.8. Reject H0 since 22.8 > 5.99. We have significant evidence, =0.05, to show that therapy and severity are not independent, p<0.005. 13. a) n=200 ^ p Z 1-/2 ^ p = 0.60 b) ^ ^ p(1 p) n ^ ^ Check: min(n p ,n(1- p ))=min(200(0.60), 200(0.40))=80 0.60 + 0.068 (0.532, 0.668) H0: p = 0.70 H1: p 70 = 0.05 ^ p - p0 Z= p 0 (1 - p 0 ) n Check: min(np0, n(1-p0))=min(100(0.70),100(0.70))=30 Reject H0 if Z > 1.960 or if Z < -1.960 ^ p - p0 0.72 0.70 Z= =0.02/0.046 = 0.44 p 0 (1 - p 0 ) 0.70(1 0.70) n 100 Do not reject H0 since 1.960 < 0.44 < 1.960. We do not have significant evidence, =0.05, to show that p 0.70. 93 14. n Z p(1 p) 1-/2 E 2 1.960 0.25 0.04 2 600.25. n = 601. 15. H0: Gender and risk are independent H1: Gender and risk are not independent (O - E )2 2 , df=2 = E Reject H0 if 2 > 5.99 = 0.05 Male Female Total IV 24 40 64 (29.5) (34.5) Homosexual 32 18 50 (23.1) (26.9) Other 15 25 40 (18.4) (21.6) Total 71 83 154 2 = 1.03 + 0.88 + 3.43 + 2.94 + 0.62 + 0.54 = 9.44 Reject H0 since 9.44 > 5.99. We have significant evidence, =0.05, to show that gender and risk are not independent, p<0.01. 16. H0: p = 0.10 H1: p < 0.10 = 0.05 ^ p - p0 Z= p 0 (1 - p 0 ) n Check: min(np0, n(1-p0))=min(200(0.10),200(0.90))=20 Reject H0 if Z < -1.645 ^ p - p0 0.06 0.10 Z= = -0.04/0.02 = -2.0 p 0 (1 - p 0 ) 0.10(1 0.10) n 200 Reject H0 since 2.0 < -1.645. We have significant evidence, =0.05, to show that p< 0.10, p<0.05. 94 17. H0: p1 = p2 H1: p1 > p2 Z= ^ ^ p1 - p 2 ^ ^ p(1 - p) 1 n1 + 1 n2 ^ ^ min(n1 p 1,n1(1- p 1))=min(125(0.128),125(0.872))=16 ^ ^ min(n2 p 2,n2(1- p 2))=min(125(0.088),125(0.912))=11 Reject H0 if Z > 1.645 + 16 11 ^ x x p= 1 2= = 0.108 n1 + n 2 125 + 125 ^ ^ p1 - p 2 0.128 - 0.088 Z= = 1 1 1 1 ^ ^ 0.108(1- 0.108) + p(1 - p) + 125 125 n1 n 2 Check: = 0.04 / 0.039 = 1.03. Do not reject H0 since 1.03 < 1.645. We do not have significant evidence, =0.05, to show that p1 > p2. 18. a) 95% CI for (p1 - p2) ^ ^ ^ ^ p (1 - p1) p2 (1 - p2) ^ ^ ( p1 - p2) Z1 1 + 2 n1 n2 ^ 1,n1(1- p 1))=min(75(0.63),75(0.37))=28 ^ Check: min(n1 p ^ ^ min(n2 p 2,n2(1- p 2))=min(75(0.48),75(0.52))=36 (0.63 - 0.48) + (1.960) 0.15 + 1.960 (0.080) 0.15 + 0.16 (-0.01, 0.31) 0.63(1 - 0.63 ) 0.48(1 - 0.48 ) + 75 75 b) No, CI includes 0, proportions not significantly different. 95 19. H0: Center and adherence are independent H1: Center and adherence are not independent = 0.05 2 (O - E ) 2 , df=2 = E Reject H0 if 2 > 5.99 1 2 3 Total Adherent 28 30 25 83 (27.1) (32.6) (23.2) Not 21 29 17 67 Adherent (21.9) (26.4) (18.8) Total 49 59 42 150 2 = 0.03 + 0.21 + 0.14 + 0.04 + 0.26 + 0.17 = 085. Do not reject H0 since 0.85 < 5.99. We do not have significant evidence, =0.05, to show that center and adherence are not independent. 20. a) 95% CI for (p1 - p2) ^ ^ ^ ^ p (1 - p1) p2 (1 - p2) ^ ^ ( p1 - p2) Z1 1 + 2 n1 n2 ^ ^ Check: min(n1 p 1,n1(1- p 1))=min(220(0.36),220(0.64))=79 ^ ^ min(n2 p 2,n2(1- p 2))=min(190(0.50),190(0.50))=95 (0.36 - 0.50) + (1.960) 0.36(1- 0.36) 0.50(1- 0.50) + 220 190 b) -0.14 + 1.960 (0.049) -0.14 + 0.095 (-0.24 to -0.05) Yes, because CI does not include 0 (reject Ho). 21. a) ^ ^ 95% for CI p. Check: min(n p ,n(1- p ))=min(150(0.34),150(0.66))=51 ^ ^ p (1 - p) n 0.34 + (1.960) (0.039) 0.34 + 0.076 (0.264, 0.416) ^ p Z1 2 96 b) n Z p(1 p) 1-/2 E 2 1.960 0.34(1- 0.34) 0.02 2 2155.1 n = 2156. 22. 95% CI for (p1 - p2) ^ ^ ^ ^ p (1 - p1) p2 (1 - p2) ^ ^ ( p1 - p2) Z1 1 + 2 n1 n2 ^ ^ Check: min(n1 p 1,n1(1- p 1))=min(100(0.24),100(0.76))=24 ^ ^ min(n2 p 2,n2(1- p 2))=min(100(0.14),100(0.86))=14 (0.24- 0.14) + (1.960) 0.10 + 1.960 (0.055) 0.10 + 0.108 (-0.008, 0.208) 0.24(1- 0.24) 0.14(1- 0.14) + 100 100 23. a) b) c) The educational levels are different between men and women p=0.0245 as are the annual incomes p=0.0001. H0: p1 = p2 ^ ^ p1 - p 2 H1: p1 p2 Z= 1 1 ^ ^ p(1 - p) + n1 n 2 H0: p = 0.07 ^ p - p0 Z= H1: p > 0.07 p 0 (1 - p 0 ) n 24. H0: Distribution is 10%, 20%, 40%, 20% 10% H1: H0 is false (O - E )2 2 , df=5-1=4 = E Reject H0 if 2 > 9.49 = 0.05 97 Observed Expected (O-E)2/E 2 None 15 14 0.07 Minimal 25 28 0.32 Some 46 56 1.78 Moderate 36 28 2.29 Severe 18 14 1.14 Total 140 140 5.60 = 5.60 Do not Reject H0 since 5.60 < 9.49. We do not have significant evidence, =0.05, to show that the distribution is not 10%, 20%, 40%, 20% 10%. 25. H0: Medication and Pain Level are independent H1: Medication and Pain Level are not independent (O - E )2 2 , df=4 = E Reject H0 if 2 > 9.49 None 20 (15.9) 15 (19.1) 35 Minimal 35 (27.3) 25 (32.7) 60 Some 41 (39.6) 46 (47.4) 87 Moderate 15 (23.2) 36 (27.8) 51 = 0.05 New Standard Total 2 Severe 6 (10.9) 18 (13.1) 24 Total 117 140 257 = 15.39 Reject H0 since 15.39 > 9.49. We have significant evidence, medication and pain level are not independent, p<0.01. =0.05, to show that 26. H0: p1 = p2 H1: p1 p2 Z= ^ ^ p1 - p 2 ^ ^ p(1 - p) 1 + 1 n1 n 2 ^ ^ min(n1 p 1,n1(1- p 1))=min(117(0.18),117(0.82))=21 ^ ^ min(n2 p 2,n2(1- p 2))=min(140(0.39),140(0.61))=54 Check: 98 Reject H0 if Z < -1.960 or if Z > 1.960 + 21 54 ^ p = x1 x 2 = = 0.29 n1 + n2 117 + 140 ^ ^ p1 - p2 0.18 - 0.39 Z= = 1 1 1 1 0.29(1- 0.29) + ^ ^ p(1 - p) + 117 140 n1 n 2 = -0.21 / 0.057 = -3.69. Reject H0 since 3.69 < -1.960. We have significant evidence, =0.05, to show that p1 p2, p<0.001. H0: Delivery and Program are independent H1: Delivery and Program are not independent = 0.05 2 (O - E ) 2 , df = 2 = E Reject H0 if 2 > 5.99 Program 1 Program 2 Program 3 Total Preterm 34 18 12 64 (21.3) (21.3) (21.3) Term 36 52 58 146 (48.7) (48.7) (48.7) Total 70 70 70 210 2 = 7.52 + 0.52 + 4.08 + 3.30 + 0.23 + 1.79 = 17.44.. Reject H0 since 17.44 > 5.99. We have significant evidence, =0.05, to show that delivery and program are not independent, p<0.005. 27. 28. H0: p = 0.28 H1: p > 0.28 = 0.05 ^ p - p0 Z= p 0 (1 - p 0 ) n Check: min(np0, n(1-p0))=min(200(0.28),200(0.72))=56 Reject H0 if Z > 1.645 99 = 0.04/0.032 = 1.26 p 0 (1 - p 0 ) 0.28(1 0.28) 200 n Do not reject H0 since 1.26 < 1.645. We do not have significant evidence, =0.05, to show that p > 0.28. Z= ^ p - p0 0.32 0.28 100 Solutions to SAS Problems 1. gender Frequency| Percent | Row Pct | Col Pct |assocpro|asstprof|fullprof|instruct| Total ---------+--------+--------+--------+--------+ female | 18 | 15 | 7 | 12 | 52 | 12.00 | 10.00 | 4.67 | 8.00 | 34.67 | 34.62 | 28.85 | 13.46 | 23.08 | | 37.50 | 37.50 | 24.14 | 36.36 | ---------+--------+--------+--------+--------+ male | 30 | 25 | 22 | 21 | 98 | 20.00 | 16.67 | 14.67 | 14.00 | 65.33 | 30.61 | 25.51 | 22.45 | 21.43 | | 62.50 | 62.50 | 75.86 | 63.64 | ---------+--------+--------+--------+--------+ Total 48 40 29 33 150 32.00 26.67 19.33 22.00 100.00 Statistics for Table of gender by rank Statistic DF Value Prob -----------------------------------------------------Chi-Square 3 1.7733 0.6208 Likelihood Ratio Chi-Square 3 1.8561 0.6028 Mantel-Haenszel Chi-Square 1 0.2449 0.6207 Phi Coefficient 0.1087 Contingency Coefficient 0.1081 Cramer's V 0.1087 Sample Size = 150 Do not reject Ho with p=0.6208. 2. The FREQ Procedure Table of therapy by compl therapy compl Frequency| Percent | Row Pct | Col Pct |1:Yes |2:No | Total ---------+--------+--------+ 1 | 89 | 911 | 1000 | 4.45 | 45.55 | 50.00 | 8.90 | 91.10 | | 41.20 | 51.07 | ---------+--------+--------+ 2 | 127 | 873 | 1000 | 6.35 | 43.65 | 50.00 | 12.70 | 87.30 | | 58.80 | 48.93 | ---------+--------+--------+ Total 216 1784 2000 10.80 89.20 100.00 Statistics for Table of therapy by compl Estimates of the Relative Risk (Row1/Row2) Type of Study Value 95% Confidence Limits ----------------------------------------------------------------Case-Control (Odds Ratio) 0.6716 0.5043 0.8943 Cohort (Col1 Risk) 0.7008 0.5423 0.9056 Cohort (Col2 Risk) 1.0435 1.0121 1.0759 Sample Size = 2000 The relative risk is estimated at 0.70 and the odds ratio is estimated at 0.67. The FREQ Procedure Table of gender by rank rank 101 3. The FREQ Procedure Table of therapy by severity therapy severity Frequency| Percent | Row Pct | Col Pct |minimal |moderate|severe | Total ---------+--------+--------+--------+ medical | 90 | 60 | 50 | 200 | 22.50 | 15.00 | 12.50 | 50.00 | 45.00 | 30.00 | 25.00 | | 64.29 | 50.00 | 35.71 | ---------+--------+--------+--------+ non_trad | 50 | 60 | 90 | 200 | 12.50 | 15.00 | 22.50 | 50.00 | 25.00 | 30.00 | 45.00 | | 35.71 | 50.00 | 64.29 | ---------+--------+--------+--------+ Total 140 120 140 400 35.00 30.00 35.00 100.00 Statistics for Table of therapy by severity Statistic DF Value Prob -----------------------------------------------------Chi-Square 2 22.8571 <.0001 Likelihood Ratio Chi-Square 2 23.1787 <.0001 Mantel-Haenszel Chi-Square 1 22.8000 <.0001 Phi Coefficient 0.2390 Contingency Coefficient 0.2325 Cramer's V 0.2390 Sample Size = 400 Reject Ho with p=0.0001. 4. Obs 1 2 3 alpha 0.05 0.05 0.05 beta 0.2 0.2 0.2 z_alpha2 -1.95996 -1.95996 -1.95996 z_beta -0.84162 -0.84162 -0.84162 p1 0.4 0.4 0.4 p2 0.20 0.25 0.30 power 0.8 0.8 0.8 es 0.20 0.15 0.10 n_2 82 152 356 To ensure 80% power we would need 82, 152 and 356 in each group for p2=0.20, 0.25 and 0.30, respectively. 5. Obs 1 2 3 4 5 c_level 0.95 0.95 0.95 0.95 0.95 z -1.95996 -1.95996 -1.95996 -1.95996 -1.95996 p 0.34 0.34 0.34 0.34 0.50 e 0.02 0.01 0.03 0.05 0.02 n 2156 8621 958 345 2401 102 SAS Program Code options ps=55 ls=80 nodate; data in1; input gender $ rank $ count; cards; female instructor 12 female asstprof 15 female assocprof 18 female fullprof 7 male instructor 21 male asstprof 25 male assocprof 30 male fullprof 22 run; proc freq; tables gender*rank/chisq; weight count; run; data in2; input therapy compl $ count; cards; 1 1:Yes 89 1 2:No 911 2 1:Yes 127 2 2:No 873 run; proc freq; tables therapy*compl/relrisk; weight count; run; data in3; input therapy $ severity $ count; cards; medical minimal 90 medical moderate 60 medical severe 50 non_trad minimal 50 non_trad moderate 60 non_trad severe 90 run; proc freq; tables therapy*severity/chisq; weight count; run; 103 data in4; input alpha power p1 p2; z_alpha2=probit(alpha/2); beta=1-power; z_beta=probit(beta); q1=1-p1; q2=1-p2; pbar=(p1+p2)/2; qbar=1-pbar; es=abs(p2-p1); temp2_n=(sqrt(pbar*qbar*2)*z_alpha2+sqrt(p1*q1+p2*q2)*z_beta)**2/(es) **2; n_2=ceil(temp2_n); /* Input the following information (required) alpha: Level of Significance: range 0.0 to 1.0 (e.g., 0.05) power: Power: range 0.0 to 1.0 (e.g., 0.80) p1: Proportion of Successes in Group 1, and p2: Proportion of Successes in Group 2 */ cards; 0.05 0.80 0.40 0.20 0.05 0.80 0.40 0.25 0.05 0.80 0.40 0.30 run; proc print; var alpha beta z_alpha2 z_beta p1 p2 power es n_2; run; data in5; input c_level e p; z=probit((1-c_level)/2); if p=. then p=0.5; tempn=(p*(1-p))*(z/e)**2; n=ceil(tempn); /* Input the following information (required) c_level: Confidence Level: range 0.0 to 1.0 (e.g., 0.95) e: Margin of Error, and p: Proportion of Successed: range 0.0 to 1.0 NOTE: p may not be available, in which case enter . to indicate missing */ cards; 0.95 0.02 0.34 0.95 0.01 0.34 0.95 0.03 0.34 0.95 0.05 0.34 0.95 0.02 . run; proc print; var c_level z p e n; run; 104
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1 Identification of an Isolate and Pseudomonas unknown Chloe Rivera Lab 7/ TA: Aria November 12, 2007 2 Introduction Pseudomonads are bacteria found in soil that can grow on complex media as well as nonfermentable carbon sources without growth fac...
UCLA >> MIMG >> 101L (Spring, 2008)
MIMG 101L Questions LAB 1 1. What environmental and nutritional factors influence the growth of particular microbes in natural environments? Temperature, light source, carbon source, trace minerals, oxygen levels, pH, etc. 2. What is the basis of the...
UCLA >> MIMG >> 101L (Spring, 2008)
EXPERIMENT SCHEDULE - MIMG 101L - FALL 2007 WEEK 0 (9/24/07 9/28/07) TUESDAY/WEDNESDAY 1. 2. 3. 4. 1. Tutorial: Preparation of the CX41 Microscope (p 8-17) 2. Experiment 1 - (period 1) Inoculate plates. 3. Experiment 2 - (period 1) Cell morphology u...
UCLA >> MIMG >> 102 (Winter, 2007)
Hey guys and gals, here are some study questions for the final. I did not include any answers to these questions because 1) I think you will benefit more from looking up the answers yourselves instead of having them handed to you on a silver platter,...
UCLA >> MIMG >> 102 (Winter, 2007)
102 review 3 Dasgupta\'s material Basic intro to DNA viruses SV40- unenveloped, uses nucleosomes to package DNA Can cause tumors when injected at HIGH MOI into newborn rodents Method for cell entry- attach to receptor, endocytosis occurs, fusion with ...
UCLA >> MIMG >> 102 (Winter, 2007)
Herpes related review questions 1. Would a TK-/- cell line that has been stably transfected with the HSV TK gene be resistant or sensitive to acyclovir? Answer: Sensitive, because HSV TK is less specific than the cellular TK and thus will add a phosp...
UCLA >> MIMG >> 102 (Winter, 2007)
Molecular Biology Techniques Review 1. Sample question: Show that lacZ is encoded/present in a given recombinant virus (deletion mutant) What is the size of the (recombinant) insert in the viral genome? (compared to control) Southern Blot - use when ...
UCLA >> MIMG >> 102L (Winter, 2007)
LABORATORY WORK SCHEDULE - MIMG 102L - WINTER 2007 WEEK 1 (1/8/07 1/12/07) 1. 2. 3. 4. TUESDAY/WEDNESDAY Check in and instructions. TA Presentation & Discussion Sign-up for 102L Honors section Experiment 1 (period 1) - Practice phage plating THURSDAY...
UCLA >> MIMG >> 102L (Winter, 2007)
time log(Po/Pt) 0 0 3 0.352862 6 0.660052 9 1.948847 Time vs. log(Po/Pt) 2.5 2 y = 0.615x - 0.798 1.5 log(Po/Pt) 1 0.5 0 0 -0.5 3 6 9 Time (Minutes) ...
UCLA >> MIMG >> 102L (Winter, 2007)
Chloe Rivera MIMG 102L TA: Laura Green Lab Report 1: The Kinetics of Adsorption Purpose The purpose of this experiment was to observe the adsorption of virus to bacteria, which is the first step of the viral growth cycle. Since the temperature, compo...
UCLA >> MIMG >> 102L (Winter, 2007)
Chloe Rivera MIMG 102L TA: Laura Green Lab Report #2: One Step Growth Curve Purpose The purpose of the experiment was to observe the production of complete phage particles within a bacterial cell and observe the release of phage by cell lysis. By mea...
UCLA >> MIMG >> 102L (Winter, 2007)
Chloe Rivera MIMG 102L TA: Laura Green Lab Report #3: Host Induced Modification Purpose Most bacteria have a strain-specific DNA restriction and modification system to protect it from viral infections. The purpose of this experiment was to illustrate...
UCLA >> MIMG >> 133 (Winter, 2007)
Chloe Rivera MIMG 133 Specific Question #6 Patrick Soon-Shiong of Abraxis described a new method for diagnosing metastic cancer using realistic methods and technology. Today, in order to determine if a cancer has spread to a patient\'s lymph nodes, ...
UC Irvine >> PSB >> P9 (Spring, 2008)
Chapter 1 Introduction to Psychology Independent Reading Psychologists at Work Pg 28-30 in 3rd edition Pg 26-27 in 2nd edition Psychology at UCI Topics to be covered: Psychology vs. PSB 7A vs. P9 Psychology vs. PSB PSB studies develo...
UCLA >> MIMG >> 133 (Winter, 2007)
Course Objectives Understanding this still emerging field When it can take > a decade to bring a product to market, biotech\'s 30-year life is the blink of an eye Integrating its many facets simply Human health and the latest therapeutics Devel...
UCLA >> MIMG >> 133 (Winter, 2007)
Chloe Rivera Essays: 2) Six years ago, the pharmaceutical industry had an unchallenged monopoly in the drug market. Big Pharma\'s today are struggling against generics, the expiration of their patents, pricing pressure from the consumers, and the grow...
UCLA >> MIMG >> 133 (Winter, 2007)
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UCLA >> PHYS >> 6B (Spring, 2007)
Sample Final Exam Question 1 A spring of constant k is attached to a mass x m which can move on a frictionless surface. a- Using Newton\'s second law, find the equation of motion of the mass. b- What are the angular frequency and the frequency of the ...
UCLA >> MIMG >> 133 (Winter, 2007)
Sample Final Exam Question 1 h1 v h2 An avalanche happens on top of a mountain, at an altitude h1 = 3500 m. The inhabitants of a house at an altitude h2 = 1000 m look at the avalanche. They noticed it takes some time for the \"roar\' of this avalanch...
UCLA >> MIMG >> 133 (Winter, 2007)
Sample Mid Term Exam Ch12-14 Question 1 k m x= 0 x=A A mass m slides without friction on a horizontal table. It is attached to a spring of constant k. a- Write down Hooke\'s law for the spring and Newton\'s second law for the mass at an arbitrary lo...
UCLA >> MIMG >> 133 (Winter, 2007)
Sample Mid Term Exam Ch12-14 Question 1 A ball of mass m is attached to a string of length L and to a spring of constant k. First we ignore the spring. The ball is moved from its equilibrium position by an angle 0 and L released without initial veloc...
UCLA >> PHYS >> 6C (Spring, 2007)
Physics 6C, Fall 2006 Prof. Maha Ashour-Abdalla Page 1 of 23 1. Electromagnetic Waves Though they are omnipresent, essential, and familiar to us all, electromagnetic waves seldom elicit much thought in our everyday lives. Yet where would we be with...
UCLA >> PHYS >> 6C (Spring, 2007)
Physics 6C Physics for Life Sciences Majors: Light, Fluids, Thermodynamics, Modern Physics Lecture 1 Fall Quarter 2006 Monday, Wednesday, Friday at 12:00 - 12:50 PM - PAB Room 1425 Instructor: Phone: Maha Ashour-Abdalla (310)825-8881 Office: 3863 Sli...
UCLA >> PHYS >> 6C (Spring, 2007)
Physics 6C, Fall 2006 Problem Set 4 Page 1 of 2 Problem Set 4 Chapter 27: 3, 9, 15, 21, 31, 47, 49 Problem 27-3 Problem 27-9 Problem 27-15 Physics 6C, Fall 2006 Problem Set 4 Page 2 of 2 Problem 27-21 Problem 27-31 Problem 27-47 Problem 27-...
UCLA >> PHYS >> 6C (Spring, 2007)
Instructor\'s Manual Volume 1 to accompany Principles of Physics Third edition by Raymond A. Serway and John W. Jewett, Jr. Ralph V. McGrew Broome Community College Harcourt College Publishers Fort Worth Philadelphia Toronto San Diego Montreal New ...
UCLA >> PHYS >> 6C (Spring, 2007)
CHAPTER 4 ANSWERS TO QUESTIONS Q4.1 (a) ma = R + mg (b) ma = T - mg (c) ma = f - R Q4.2 When the bus starts moving, the mass of Claudette is accelerated by the force of the back of the seat on her body. Clark is standing, however, and the only...
UCLA >> BIO CHEM >> 153A (Spring, 2007)
Seat # _ Last initial _ Chemistry and Biochemistry 153A Fall 2004 Midterm Examination _KEY_ Print your full name (last name first) Question # 1 2 3 4 5 6 7 TOTAL Value 28 26 21 25 21 15 16 152 INSTRUCTIONS READ EACH QUESTION CAREFULLY! SHOW YOUR CALC...
UCLA >> BIO CHEM >> 153A (Spring, 2007)
Chemistry 153A Practice Final Exam KEY First letter of your last name _ Question # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 TOTAL TOTAL COURSE SCORE FINAL COURSE GRADE Value 25 5 10 10 7 I have read the instructions below. 14 18 15 18 20 20 16 25 17 15 __...
UCLA >> BIO CHEM >> 153A (Spring, 2007)
Seat # _ Last initial _ Chemistry and Biochemistry 153A Practice Midterm Examination KEY _KEY_ Print your full name (last name first) Question # 1 2 3 4 5 6 7 8 TOTAL Value 23 23 20 10 11 25 10 30 152 __ Signature INSTRUCTIONS READ EACH QUESTION CARE...
UCLA >> BIO CHEM >> 153A (Spring, 2007)
Seat # _ Last initial _ Chemistry and Biochemistry 153A Practice Midterm Examination B -KEY _KEY_ Print your full name (last name first) Question # 1 2 3 4 5 6 TOTAL Value 28 21 32 28 15 28 152 INSTRUCTIONS READ EACH QUESTION CAREFULLY! SHOW YOUR CAL...
UCLA >> BIO CHEM >> 153A (Spring, 2007)
Seat # _ Last initial _ Chemistry and Biochemistry 153A Winter 2005 Midterm Examination -KEY _KEY_ Print your full name (last name first) Question # 1 2 3 4 5 6 TOTAL Value 28 21 32 28 15 28 152 INSTRUCTIONS READ EACH QUESTION CAREFULLY! SHOW YOUR CA...
UCLA >> BIO CHEM >> 153A (Spring, 2007)
Name _KEY_ 5 3. (32 ) a. (1) One factor in \"buffering capacity\" (i.e., effectiveness of a buffer) is the proximity of the buffer\'s pKa to the target pH. Why is the total buffer concentration also important to \"buffering capacity\"? [35 words] Higher...
Alabama Huntsville >> PHY >> 112 (Spring, 2007)
Inclined Plane Date of Experiment: September 7, 2007 Date of Submission: September 14, 2007 1 Purpose: This lab deals the gravity, vectors, and forces. Part 1 of the lab deals with calibration of the force sensor. Part 2 deals with the measurement ...
Alabama Huntsville >> PHY >> 112 (Spring, 2007)
Simple Pendulum Date of Experiment: August 31, 2007 Date of Submission: September 7, 2007 1 Purpose: The purpose of this experiment is to study the motion of a simple pendulum; which includes its amplitude and frequency and the relationships betwee...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Fundamentals of Software Engineering Agenda Class 6 Container Classes Container Classes The Bag Key Concepts UAH CPE 212 This Time Container Classes What are they? Structures which hold things BUT also Abstract Data Types / Use...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Fundamentals of Software Engineering Introductions Introductions and Administrative details Development Concepts C+ Review Key Concepts UAH CPE 212 David Skipper Introductions Part Time Instructor Use email Office hours De...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Fundamentals of Software Engineering Agenda Class 7 Derived Classes 2 Derived Classes 2 Key Concepts UAH CPE 212 Today Last Time Derived Classes 1 This Time Derived Classes 2 UAH CPE 212 Review Derived Classes Defin...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Fundamentals of Software Engineering Development Environments Windows Unix Makefiles Key Concepts UAH CPE 212 Tools and Techniques Separate declaration: These are NOT directly compiled Function prototypes and class prototype...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Fundamentals of Software Engineering Classes and Objects Class and Object Concepts Classroom Example Key Concepts UAH CPE 212 This Time Two parts to programming Develop classes (user defined types) Use the classes (types) in ...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Fundamentals of Software Engineering Agenda Class 10 Linked Lists Linked Lists Key Concepts UAH CPE 212 Today Last Time Lists This Time Linked Lists UAH CPE 212 Linked Lists Non-Contiguous (Usually) Linear Structures...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Fundamentals of Software Engineering Agenda Class 11 Linked Lists 2 Linked Lists Application Key Concepts UAH CPE 212 Today Last Time First Fundamental Structure, Linked Lists This Time Bagged Again UAH CPE 212 Linked List...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Fundamentals of Software Engineering Agenda Class 10 List ADT The List Key Concepts UAH CPE 212 Last Time Templates Function Templates Class Templates Capture Behavior Syntax Issues UAH CPE 212 This Time Example: The...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Fundamentals of Software Engineering Agenda Class 5 Classes and Operator Overloads More on Classes Operator Overloading Key Concepts UAH CPE 212 Last Time Concept of a class Common attributes Basic class definitions and util...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Fundamentals of Software Engineering Agenda Class 5 Pointers and Dynamic Memory Pointers and Dynamic memory Key Concepts UAH CPE 212 Help Resources First Read Text Ask questions in class email Next Schedule extra instruct...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Fundamentals of Software Engineering Agenda Class 14 Queues 2 Priority Queues Key Concepts UAH CPE 212 Today Last Time Queue Concept Queue Implementations Queue Application This Time Priority Queue UAH CPE 212 Priority ...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Purpose Project 1 Understand constructing your own class. Background With this project you must design your own class and manipulate the class for the calculations and selection of the winner. Tasks Perform programming assignme...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Project 2 Purpose Understand ADTs. Measuring performance. Background The class lectures cover a number of ADTs and implementations. This projects gives the student the opportunity to create their own ADT class. Tasks Develop a q...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Fundamentals of Software Engineering Agenda Class 13 Queues Queues Key Concepts UAH CPE 212 Today Last Time Stacks LIFO, push, pop This Time Queue Concept Queue Implementations Queue Application UAH CPE 212 Queues Abst...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Fundamentals of Software Engineering Agenda Class 15 Recursion Recursion Key Concepts UAH CPE 212 Today Last Time Priority Queues This Time Recursion UAH CPE 212 Recursion Procedure calls itself Recursion: Divide and C...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
The University of Alabama in Huntsville ECE Department Spring 2008 CPE 212 Fundamentals of Software Engineering Revision A INSTRUCTOR INFORMATION: Time: M/W 3:55 5:15 p.m. Location: EB 207 Name: Dr. David Skipper Email: skipper@ece.uah.edu Office: TH...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 Fundamentals of Software Engineering Agenda Class 12 Stacks Stacks Key Concepts UAH CPE 212 Today Last Time Linked Lists This Time Stacks UAH CPE 212 Stacks Abstract Concept Can have heterogeneous or homogeneous stack...
Alabama Huntsville >> CPE >> 212 (Spring, 2007)
UAH CPE 212 #include <climits> #include <ctime> #include <iostream> using namespace std; Timer int main() { const int Extra=1000; /Machine is too fast, add extra work int MyInts[USHRT_MAX]; /Static array of ints bool DisplayIt=false; /Show filled a...
UC Irvine >> PSB >> P9 (Spring, 2008)
Biology and Behavior Ch 2: How does the brain work? Independent reading The neuron and the nuerotransmitters 39-44 3rd edition 37-41 2nd edition The Human Nervous System What parts of the brain are important to psychologists? Topics: Spinal ...
UCSB >> PSYCH >> 1 (Spring, 2008)
1. Chapter 13 Social Psychology a. Prisoner\'s dilemma i. In your own words choose between a cooperative act and a competitive act that benefits yourself but not others ii. Example narc and go free, friend gets 20 years, both narc and get 5 years e...
UC Irvine >> PSB >> P9 (Spring, 2008)
Chapter 3 Sensation & Perception Independent Reading Unusual Perceptual Experiences Page 105 in 3rd edition Sensation Senses pick up sensory stimuli (visual, auditorial, and other info and transmit it to the brain. Topics to be covered: Touc...
UC Irvine >> PSB >> P9 (Spring, 2008)
Chapter 4 Consciousness: Part I Independent Reading What\'s up with the different states of consciousness? What is Consciousness? Page 116-117 (3rd edition) Note: this section does NOT EXIST in the 2 nd edition! Borrow from a friend Read in ou...
UCSB >> PSYCH >> 1 (Spring, 2008)
Psychology 1 Introduction to Psychology Winter Quarter, 2008 Reading Guide for Chapters 13.1-13.3, 13.5, 9: pp. 476-508, 518-527, 332-359 Week 9 * This sheet is pretty comprehensive, but you\'ll notice some terms missing as you read. I chose not to...
UCSB >> PSYCH >> 1 (Spring, 2008)
Psychology 1 Introduction to Psychology Reading Guide for Chapters 8.1-8.2 and 5 pp. 284 315, 163 199 Week 7 * This sheet is pretty comprehensive, but you\'ll notice some terms missing as you read. I chose not to emphasize these topics however you...
UCSB >> PSYCH >> 1 (Spring, 2008)
Psychology 1 Introduction to Psychology Winter Quarter, 2008 Reading Guide for Chapters 14 pp. 528564 Week 10 This sheet is pretty comprehensive, but you\'ll notice some terms missing as you read. I chose not to emphasize these topics however you a...
UCSB >> PSYCH >> 1 (Spring, 2008)
Psychology 1 Introduction to Psychology Winter Quarter, 2008 Reading Guide for Chapters 5.1, 12.1-2, 14.3 pp. 150-162, 414-425, 434-464M, 509-517 1. Week 8 2. Chapter 11.3 Sexual Motivation a. Kinsey Survey i. Main findings in your own words ...
Clemson >> ACCT >> 312 (Spring, 2008)
Intangible Assets Result from legal or contractual rights, do not have physical substance Accounting for Intangibles Recorded at cost Those that are amortized are reported on a company\'s balance sheet at their book value (cost accumulated amortizat...
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