Statistical Analysis of Suicide Rates: Hypothesis Testing and Confidence Intervals, Assignments of Mathematical Statistics

Download Statistical Analysis of Suicide Rates: Hypothesis Testing and Confidence Intervals and more Assignments Mathematical Statistics in PDF only on Docsity! MT427 Problem Set 8 Chapter 9 Section 9.2 9.2.1 t = 65.2 75.5 1/ 1/ 13.9 1/ 9 1/12 p x y s n m ! ! " # # = !1.68 Since !t.05,19 = !1.7291 < t = !1.68, accept H0. 9.2.2 sp = 2 2 2 2( 1) ( 1) 3(267 ) 3(224 ) 2 4 4 X Y n s m s n m ! # ! # " # ! # ! 2 = 246.44 t = 1133.0 1013.5 1/ 1/ 246.44 1/ 4 1/ 4p x y s n m ! ! " # # = 0.69 Since !t.025,6 = !2.4469 < t = 0.69 < t.025,6 = 2.4469, accept H0. 9.2.3 sp = 2 2( 2) ( 1) 5(167.568 11(52.072) 2 6 12 2 X Yn s m s n m ! # ! # " # ! # ! = 9.39. t = 28.6 12.758 1/ 1/ 9.39 1/ 6 1/12p x y s n m ! ! " # # = 3.37 Since t = 3.37 > 2.9208 = t.005,16, reject H0. 9.2.4 sp = 2 25(15.1 ) 8(8.1 ) 6 9 2 # # ! = 11.317 t = 70.83 79.33 11.317 1/ 6 1/ 9 ! # = !1.43 Since !t.005,13 = !3.0123 < t = !1.43 < t.005,13 = 3.0123, accept H0. 9.2.5 sp = 7(7.169) 7(10.304) 8 8 2 # # ! = 2.956 t = 11.2 9.7875 2.956 1/8 1/8 ! # = 0.96 Since t = 0.96 < t.05,14 = 1.7613, accept H0. 9.2.6 sp = 2 230(1.469 ) 56(1.350 ) 31 57 2 # # ! = 1.393 t = 3.10 2.43 1.393 1/ 31 1/ 57 ! # = 2.16 Since t = 2.16 > t.025,86=1.9880, reject H0. Chapter 9 137 Chapter 9 Section 9.2 9.2.1 t = 65.2 75.5 1/ 1/ 13.9 1/ 9 1/12 p x y s n m ! ! " # # = !1.68 Since !t.05,19 = !1.7291 < t = !1.68, accept H0. 9.2.2 sp = 2 2 2 2( 1) ( 1) 3(267 ) 3(224 ) 2 4 4 X Y n s m s n m ! # ! # " # ! # ! 2 = 246.44 t = 1133.0 1013.5 1/ 1/ 246.44 1/ 4 1/ 4p x y s n m ! ! " # # = 0.69 Since !t.025,6 = !2.4469 < t = 0.69 < t.025,6 = 2.4469, accept H0. 9.2.3 sp = 2 2( 2) ( 1) 5(167.568 11(52.072) 2 6 12 2 X Yn s m s n m ! # ! # " # ! # ! = 9.39. t = 28.6 12.758 1/ 1/ 9.39 1/ 6 1/12p x y s n m ! ! " # # = 3.37 Since t = 3.37 > 2.9208 = t.005,16, reject H0. 9.2.4 sp = 2 25(15.1 ) 8(8.1 ) 6 9 2 # # ! = 11.317 t = 70.83 79.33 11.317 1/ 6 1/ 9 ! # = !1.43 Since !t.005,13 = !3.0123 < t = !1.43 < t.005,13 = 3.0123, accept H0. 9.2.5 sp = 7(7.169) 7(10.304) 8 8 2 # # ! = 2.956 t = 1.2 9.7875 2.956 1/8 1/8 ! # = 0.96 Since t = 0.96 < t.05,14 = 1.7613, accept H0. 9.2.6 sp = 2 230(1.469 ) 56(1.350 ) 31 57 2 # # ! = 1.393 t = 3.10 2.43 1.393 1/ 31 1/ 57 ! # = 2.16 Since t = 2.16 > t.025,86=1.9880, reject H0. Chapter 9 137 Chapter 9 Section 9.2 9.2.1 t = 65.2 75.5 1/ 1/ 13.9 1/ 9 1/12 p x y s n m ! ! " # # = !1.68 Since !t.05,19 = !1.7291 < t = !1.68, accept H0. 9.2.2 sp = 2 2 2 2( 1) ( 1) 3(267 ) 3(224 ) 2 4 4 X Y n s m s n m ! # ! # " # ! # ! 2 = 246.44 t = 1133.0 1013.5 1/ 1/ 246.44 1/ 4 1/ 4p x y s n m ! ! " # # = 0.69 Since !t.025,6 = !2.4469 < t = 0.69 < t.025,6 = 2.4469, accept H0. 9.2.3 sp = 2 2( 2) ( 1) 5(167.568 11(52.072) 2 6 12 2 X Yn s m s n m ! # ! # " # ! # ! = 9.39. t = 28.6 12.758 1/ 1/ 9.39 1/ 6 1/1p x y s n m ! ! " # # = 3.37 Since t = 3.37 > 2.9208 = t.005,16, reject H0. 9.2.4 sp = 2 25(15.1 ) 8(8.1 ) 6 9 2 # # ! = 1 317 t = 70.83 79.33 11.317 1/ 6 1/ 9 ! # = !1.43 Since !t.005,13 = !3.0123 < t = !1.43 < t.005,13 = 3.0123, accept H0. 9.2.5 sp = 7(7.169) 7(10.304) 8 8 2 # # ! = 2.956 t = 11.2 9.7875 2.956 1/8 1/8 ! # = 0.96 Since t = 0.96 < t.05,14 = 1.761 , accept H0. 9.2.6 sp = 2 230(1.469 ) 56(1.350 ) 31 57 2 # # ! = 1.393 t = 3.10 2.43 1.393 1/ 31 1/ 57 ! # = 2.16 Since t = 2.16 > t.025,86=1.9880, reject H0. Chapter 9 137 9.2.15 For the data given, x = 545.45, sX = 428, and y = 241.82, sY = 183. Then t = 2 2 2 2 545.45 241.82 / / 428 /11 183 /11 X Y x y s n ! ! " # = 2.16 The degrees of freedom associated with this statistic is the greatest integer in $ % $ % $ % 2 2 2 2 2 2 2 2 2 22 2 2 2 / / (428 /11 183 /11) (428 /11) /10 (183 /11) /10/ ( 1) / /( 1) X Y X Y s n s m n n s m m # # " #! # ! = 13.5. The greatest integer is 13. Since t = 2.16 > t.05,13 = 1.7709, reject H0. 9.2.16 Decreasing the degrees of freedom also decreases the power of the test. 9.2.17 a) The sample standard deviation for the first data set is approximately 3.15; for the second, 3.29. These seem close enough to permit the use of Theorem 9.2.2. b) Intuitively, the states with th comprehensive law should have fewer deaths. However, the average for these data is 8.1, which is larger than the average of 7.0 for the states with a more limited law. Section 9.3 9.3.1 a) The critical values are F.025,25,4 and F.975,25,4. These values are not tabulated, but in this case, we can approximate them by F.025,24,4 = 0.296 and F.975,24,4 = 8.51. The observed F = 86.9/73.6 = 1.181. Since 0.296 < 1.181 < 8.51, we can accept H0 that the variances are equal. b) Yes, we can use Theorem 9.2.2, since we have no reason to doubt that the variances are equal. 9.3.2 The observed F = .2092/0.1222 = 2.935. Since F.025,9,9 = 0.248 < 2.935 < 4.03 = F.975,9,9, we can accept H0 that the variances are equal. 9.3.3 a) The critical values are F.025,19,19 and F.975,19,19. These values are not tabulated, but in this case, we can approximate them by F.025,20,20 = 0.406 and F.975,20,20 = 2.46. The observed F = 2.41/3.52 = 0.685. Since 0.406 < 0.685 < 2.46, we can accept H0 that the variances are equal. b) Since t = 2.662 > t.025,38 = 2.0244, reject H0. 9.3.4 The observed F = 3.182/5.672 = 0.315. Since F.025,9,9 = 0.248 < 0.315 < 4.03 = F.975,9,9, we can accept H0 that the variances are equal. 9.3.5 F = 0.202/0.372 = 0.292. Since F.025,9,9 = 0.248 < 0.292 < 4.03 = F.975,.9,9, accept H0. 9.3.6 The observed F = 398.75/274.52 = 1.453. Let & = 0.05. The critical values are F.025,13,11 and F.975,13,11. These values are not in Table A.4, so approximate them by F.025,12,11 = 0.301 and F.975,12,11 = 3.47. Since 0.301 < 1.453 < 3.47, accept H0 that the variances are equal. Theorem 9.2.2 is appropriate. 140 Chapter 9 9.2.15 For the data given, x = 545.45, sX = 428, and y = 241.82, sY = 183. Then t = 2 2 2 2 545.45 241.82 / / 428 /11 183 /11 X Y x y s n s m ! ! " # # 16 The degrees of freedom associated with this statistic is the greatest integer in $ % $ % $ % 2 2 2 2 2 2 2 2 2 22 2 2 2 / / (428 /11 183 /11) (4 8 /11) /10 (183 /11) /10/ ( 1) / /( 1) X Y X Y s n s m s n n s m m # # " #! # ! = 13.5. The greatest integer is 3. Since t = 2.16 > t.05,13 = 1.7709, reject H0. 9.2.16 Decreasing the degrees of freedom also decreases the power of the test. 9.2.17 a) The sample standard deviation for the first data set is approximately 3.15; for the second, 3.29 These seem close enough to permit the use of Theorem 9.2.2. b) Intuitively, the states with the comprehensive law should have fewer deaths. However, the average for these data is 8.1, which is larger than the average of 7.0 for the states with a more limited law. Section 9.3 9.3.1 a) The critical values are F.025,25,4 and F.975,25,4. These valu s are not tabulated, but in this case, w can approximate them by F.025,24,4 = 0.296 and F.975,24,4 = 8.51. The observed F = 86.9/73.6 = 1.18 . Since 0.296 < 1.181 < 8.51, we can accept H0 that the variances are equal. b) Yes, we can use Theorem 9.2.2, since we have no reason to doubt that the variances are equal. 9.3.2 The observed F = 0.2092/0.1222 = 2.935. Since F.025,9,9 = 0.248 < 2.935 < 4.03 = F.975,9,9, we can accept H0 that the variances are equal. 9.3.3 a) The critical values are F.025,19,19 and F.975,19,19. These values are not tabulated, but in this case, we can approximate them by F.025,20,20 = 0.406 and F.975,20,20 = 2.46. The observed F 2.41/3.52 = 0.685. Since 0.406 < 0.685 < 2.46, we can accept H0 that the variances are equal. b) Since t = 2.662 > t.025,38 = 2.0244, reject H0. 9.3.4 The observed F = 3.182/5.672 = 0.315. Since F.025,9,9 = 0.248 < 0.315 < 4.03 = F.975,9,9, we can accept H0 that the variances are equal. 9.3.5 F = 0.202/0.372 = 0.292. Since F.025,9,9 = 0.248 < 0.292 < 4.03 = F.975,.9,9, accept H0. 9.3.6 The observed F = 398.75/274.52 = 1.453. Let & = 0.05. The critical values are F.025,13,11 and F.975,13,11. These values are not in Table A.4, so approximate them by F.025,12,11 = 0.301 and F.975,12,11 = 3.47. Since 0.301 < 1.453 < 3.47, accept H0 that the variances are equal. Theorem 9.2.2 is appropriate. 140 Chapter 9 9.3.7 Let ! = 0.05. F = 65.25/227.77 = 0.286. Since 0.208 = F.025,8,5 < 0.286 < 6.76 = F.975,8,5, accept H0. Thus, Theorem 9.2.2 is appropriate. 9.3.8 For these data, 2 X s = 56.86 and 2 Y s = 66.5. The observed F = 66.5/56.86 = 1.170. Since F.025,8,8 = 0.226 < 1.170 < 4.43 = F.975, 8, 8, we can accept H0 that the variances are equal. Thus, Theorem 9.2.2 can be us d, s it has the hypothesis that the variances are equal. 9.3.9 If 2 2 X Y 2" " "# # , the maximum likelihood estimator for "2 is 2 2 2 1 1 1 ˆ ( ) ( ) n m i i i i x x y y n m " # # $ % # & ' &( )' * + , , . Then 2 2 2 1 1 1( ) / 2 ( ) ( ) ˆ2 2 1 ˆ( ) ˆ2 n m i i i i n m x x y y L e "- ." # # $ %' ( )& & ' &( ) * + , ,$ %# ( ) * + = ( ) / 2 ( ) / 2 2 1 ˆ2 n m n m e ." ' & '$ % ( ) * + If 2 2 X Y " "/ the maximum likelihood estimators for 2 2andX Y" " are 2 2 1 1 ˆ ( n X i i )x x n " # # &, , and 2 2 1 1 ˆ ( ) m Y i i y y m " # # &, . Then 2 2 2 2 1 1 1 1/ 2 / 2 ( ) ( ) ˆ ˆ2 2 2 2 1 1ˆ( ) ˆ ˆ2 2 n m i i X Yi i n m x x y y X Y L e e " " ." ." # # $ % $ ( ) (& & & &( ) ( * + * , ,$ % $ % 0 # ( ) ( ) * + * + % ) ) + = / 2 / 2 / 2 / 2 2 2 1 1 ˆ ˆ2 2 n m m n X Y e e ." ." & &$ % $ % ( ) ( ) * + * + The ratio 1 = 2 / 2 2 / 2 2 ( ) / 2 ˆ ˆ ˆ( ) ( )( ) ˆ ˆ( )( ) n m X Y n m L L " "- " ' # 0 , which equates to the expression given in the statement of the question. 9.3.l0 Since 2X and 2Y are known, the maximum likelihood estimator uses 2X instead of x and 2Y instead of y . For the GLRT, 1 is as in Question 9.3.9 with those substitutions. Section 9.4 9.4.1 55 40 ˆ 200 200 x y p n m ' ' # # ' ' = 0.2375 z = ˆ ˆ ˆ ˆ(1 ) (1 ) x y n m p p p p n m & & & ' = 55 40 200 200 0.2375(0.7625) 0.2375(0.7625) 200 200 & ' = 1.76 Since &1.96 < z = 1.76 < 1.96 = z.025, accept H0. Chapter 9 141 1