ScanHW9 - Chapter 8 Solutions Solutions to Section 8.1 8.1(a Matched pairs(b Independent samples(c Matched pairs(d Independent samples 8.2(a(b(c(d

ScanHW9 - Chapter 8 Solutions Solutions to Section 8.1...

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Chapter 8 Solutions Solutions to Section 8.1 8.1 (a) Matched pairs. (b) Independent samples. (c) Matched pairs. (d) Independent samples. 8.2 (a) Experimental. (b) Observational. (c) Experimental. (d) Experimental. 8.3 (a) Matched pairs. (b) Matched pairs. (c) Independent samples. (d) Independent samples. 8.4 Observational. Observational. Observational. Experimental. Solutions to Section 8.2 8.5 (a) The clouds in each group are not matched to one another on some characteristic. (b) Q-Q Plot 3000 ~ 2500 ]1 c: 'OJ 2000 a: "0 ::> .Q U 1500 "0 ., "0 al 1000 en c: ::> i 500 - :; 0 0 0 500 1000 1500 2000 2500 3000 Ordered Seeded Cloud Rainfalls -117 -
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Seeded cloud rainfalls tend to be larger than unseeded cloud rainfalls. 8.6 (a) The ball bearings from each line are not matched on any characteristic. (b) Q-Q Plot of Ball Bearing Measurements "I ~ 1.8 -g 1.6 8 1.41 Gl en ,g l!! 1.2 S Gl E 1.01 as i5 '0 e .81 Q) "E 0 .6 .6 .8 1.0 1.2 1.4 1.6 1.8 2.0 Ordered Diameters for First Une The second production line tends to have larger diameter ball bearings. 8.7 (a) The two different types of eyes are or belong to the same person. (b) Scatterplot of Corneal Thicknesses =1 · · · as 480 •• E 8 ill 440, ~ > I .<:I 420 '0 .! 1 ~ 400j ~ 380.1 380 400 420 440 480 480 500 Eye Not Affected by Glaucoma The pairs tend to lie pretty close to the 45 degree line through the origin. Eyes with glaucoma do not appear to have thicker corneas than unaffected eyes. 8.8 (a) The two viruses are appli~d to the same leaf, which makes each leaf a block. - 118-
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25 - , (21 - 1)4.5 + (21 - 1)2.0 21 + 21 - 2 = 1.803. (WI + W2)2 wi/(nl-1) + wV(n2 -1) (0.463 + 0.309)2 (0.463)2/(21 - 1) + (0.309)2/(21 - 1) = 34.845 ~ 35. = v = C\l 15 '" 2 :> 10 >. .&J "'C ! 5 f 0 I- 0 5 10 15 20 25 30 35 s= Then a 95% CI for jJ.I - jJ.2 is given by: _ _ s~ s~ )4.5 2.0 x - y ± tva/2 - + - = 8 - 6.5 ± 2.032 x -21 + -21 = [0.392,2.608]. , nl n2 . s~ 4.5 s~ 2.0 WI = nl = 21 = 0.463 and W2 = n2 = 21 = 0.309, the degrees of freedom are 20 Scatterplot of Tobacco Leaf Lesions 35,----------------~ - 119- Treated by Virus 1 The results are similar, indicating that the assumption of equal variances is reasonable. 30 The pairs tend to lie belo~ the 45 degree line through the origin. Virus 1 tends to produce more lesions than virus 2. Then a 95% CI for jJ.I - jJ.2 is given by: X - fi ± tn1+n2-2 a/2sV 1 + 1 = 8 - 6.5 ± 2.021 x 1.803 [222 = [0.376,2.625]. , nl n2 V 21 Since the CI does not contain 0, we reject Ho and conclude that there is a significant difference between the two filters. (b) Since (b) 8.10 Solutions to Section 8.3 8.9 (a) The pooled SD is (nl - 1)s~ + (n2 - 1)s~ nl +n2 - 2
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-120 - Since the 01 does not contain 0, we reject Ho and conclude that there is a significant difference between the average dopamine levels of the two groups of patients. Since It I < tS,.025 = 2.571, we do. not reject Ho at a = 0.05. This is the opposite conclusion as that obtained from assuming equal variances. - . (10 -1)(0.0079)2 + (5 - 1)(0.0132)2 10 + 5 - 2 = 0.0099. (15 - 1)(0.0047)2 + (10 - 1)(0.0051)2 _ 0 15 + 10 _ 2 - O. 049. x - y 12.0026 - 12.0143 === = ---======= = -1.817. 5. 5. ./ (0.0079)2 + (0.0132)2 n1 + n2 V 10 5 (WI + W2)2 wV(n1 - 1) + wV(n2 - 1) ((0.0025)2 + (0.0059)2)2 (0.0025)4/(10 - 1) + (0.0059)4/(5 - 1) = 5.481 ~ 5. 1/ = t= (n1 - 1)8~ + (n2 - l)s~ '---~---~- n1 +n2 -2 - s= s= The test statistic is (n1 - l)s~ + (n2 - 1)8~ _ n1 +n2 - 2 Then the test statistic is WI = s~ = (0.0079)2 = (0.0025)2 and W2 = s~ = (0.0132)2 = (0.0059)2, n1 10 n2 5 the degrees of freedom are Then a 95% 01 for ILl - IL2 is given by: ~1 95% 01: = x - y ± tn1 +n2-2 0./2S - + - , n1 n2 {I 1 = 0.0164 - 0.0243.
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  • Spring '08
  • SAWYER
  • Statistical hypothesis testing, Statistical significance

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