Chapter 2 Exercise Solutions. Chapter 2

Transcription

1 Chapter 2 Exercise Solutions Chapter (a) Cumulative Class Cumulative Relative relative interval Frequency frequency frequency frequency Histogram of pindex 40 Frequency pindex Frequency Polygon of pindex 40 Frequency pindex

2 Chapter 2 Exercise Solutions (b) 40.0 % (c).7667 (d) % (e) 9 (f) % (g) (h) 2.17, because it composes almost 25 percent of the data and is the most frequently occurring value in the data set Skewed to the left (a) Cumulative Class Cumulative Relative relative interval Frequency frequency frequency frequency Histogram of sizes Frequency sizes Frequency Polygon of Sizes Frequency sizes

3 Chapter 2 Exercise Solutions (b) % (c) 152 (d) 2.52 % (e) % (f) skewed to the right (a) Cumulative Class Cumulative Relative relative interval Frequency frequency frequency frequency Histogram of BMI 10 8 Frequency bmi Frequency Polygon of BMI 10 8 Frequency bmi

4 Chapter 2 Exercise Solutions (b) % (c) % (d) % (e) Skewed to the right (f) (a) Cumulative Class Cumulative Relative relative Interval Frequency frequency frequency frequency Histogram of Selevels 10 Frequency Selevels Frequency Polygon of Selevels Frequency Selevels

5 Chapter 2 Exercise Solutions (b) symmetric (c) 6 (d) % (a) Class Relative interval Frequency frequency Histogram of Hours Frequency Hours Frequency Polygon of Hours Frequency Hours (b) Skewed to the right 5

6 Chapter 2 Exercise Solutions (a) Cumulative Class Cumulative Relative relative interval Frequency frequency frequency frequency Histogram of V Frequency V Frequency Polygon of V Frequency V (b) skewed to the left 6

7 Chapter 2 Exercise Solutions (a) Cumulative Class Cumulative Relative relative interval Frequency frequency frequency frequency Histogram of Scores 40 Frequency Scores Frequency Polygon of Scores Frequency Scores (b) Not greatly skewed 7

8 Chapter 2 Exercise Solutions (a) Stem-and-Leaf Display: AGE Stem-and-leaf of AGE N = 30 Leaf Unit = (b) Fairly symmetric (a) Stem-and-leaf display: Hospital A Stem-and-leaf of C1 N = 25 Leaf Unit = (6) Stem-and-leaf display: Hospital B Stem-and-leaf of C2 N = 25 Leaf Unit = (4) (b) Both asymmetric, A is skewed left and B is skewed right 8

9 Chapter 2 Exercise Solutions (a) Cumulative Class Cumulative Relative relative interval Frequency frequency frequency frequency Histogram of Age 40 Frequency Age Frequency Polygon of Age 40 Frequency Age (b) symmetric 9

10 Chapter 2 Exercise Solutions (a) Cumulative Class Cumulative Relative relative interval Frequency frequency frequency frequency Histogram of S/R Ratio 40 Frequency C Frequency Polygon of S/R Ratio Frequency S/R Ratio

11 Chapter 2 Exercise Solutions Stem-and-leaf of C1 N = 216 Leaf Unit = (34) (b) Skewed right (c) 10, 4.62 % (d) 196, %; 67, %, 143, % (a) Cumulative Class Cumulative Relative relative interval Frequency frequency frequency frequency Histogram of Height Frequency Height

12 Chapter 2 Exercise Solutions 30 Frequency Polygon of Height Frequency Height Stem-and-Leaf Display: Height Stem-and-leaf of Height N = 109 Leaf Unit = (27) (b) skewed to the left (d) cm or taller (e) Less than cm Mean Median Mode Range no mode 255 Variance Standard Coefficient of IQR Deviation Variation The median is a better measure of center because the data is skewed left. 12

13 Chapter 2 Exercise Solutions Mean Median Mode Range Variance Standard Coefficient of IQR Deviation Variation Boxplot of Heart rate 550 Heart rate The median is a better measure of center because of the skewed nature of the data Mean Median Mode Range , Variance Standard Coefficient of IQR Deviation Variation Either the mean or median is a good measure of center Mean Median Mode Range No mode Variance Standard Coefficient of IQR Deviation Variation Either the mean or median is a good measure of center. 13

14 Chapter 2 Exercise Solutions Mean Median Mode Range Variance Standard Coefficient of IQR Deviation Variation Boxplot of Days C The median is a better measure of center because it is not influenced by the outlier Mean Median Mode Range Variance Standard Coefficient of IQR Deviation Variation Boxplot of Months Months The median is a better measure of center because the data is skewed to the right. 14

15 Chapter 2 Exercise Solutions Mean Median Mode Range Variance Standard Coefficient of IQR Deviation Variation Boxplot of pindex pindex The median is a better measure of center because the distribution is left skewed Mean Median Mode Range Variance Standard Coefficient of IQR Deviation Variation Boxplot of Sizes Sizes The median is a better measure of center because the data is skewed to the right. 15

16 Chapter 2 Exercise Solutions Mean Median Mode Range none Variance Standard Coefficient of IQR Deviation Variation Boxplot of BMI bmi The median is a better measure of center because the distribution is right skewed Mean Median Mode Range None Variance Standard Coefficient of IQR Deviation Variation The mean or the median is a good measure of center. 16

17 Chapter 2 Exercise Solutions Mean Median Mode Range Variance Standard Coefficient of IQR Deviation Variation Boxplot of Hours hours The median is a better measure of center because the distribution is slightly right skewed Mean Median Mode Range , 54, Variance Standard Coefficient of IQR Deviation Variation The mean or the median is a good measure of center. 17

18 Chapter 2 Exercise Solutions Mean Median Mode Range , Variance Standard Coefficient of IQR Deviation Variation Boxplot of Scores Scores Either the mean or median is a good measure of center Mean Median Mode Range , 48, Variance Standard Coefficient of IQR Deviation Variation The mean or the median is a good measure of center. 18

19 Chapter 2 Exercise Solutions Chapter 2 Review Exercises 13. (a) Leaf Unit = (12) (b) skewed (c) 37 weeks is considered full-term, surgery is done before birth (d) x = , median = 35.00, s 2 = , s= Cumulative Class Cumulative Relative relative interval Frequency frequency frequency frequency x = 43.39, median = 42, s = 17.09, C.V. = (b) Stem-and-leaf of GFR N = 28 Leaf Unit = (7)

20 Chapter 2 Exercise Solutions (c) Boxplot of GFR GFR (d) 67.9 %, %, 100 % 16. (a) Protein Mean Median StDev Variance CoefVar Cystatin Creatinine (b) Stem-and-leaf of Cystatin N = 27 Leaf Unit = 0.10 Stem-and-leaf of Creatine N = 26 Leaf Unit = (7) (13)

21 Chapter 2 Exercise Solutions Boxplot of Cystatin 5 4 Cystatin Boxplot of Creatine Creatine (c) Creatine is more variable because it has a larger coefficient of variation. 18. Mother s Age Frequency Cumulative frequency Relative frequency Cumulative relative frequency x = 3.95, Median = 3, s 2 = , s =

22 Chapter 2 Exercise Solutions 20. (a) The sum of the squared deviations of a set of measurements about their mean is smaller than any other sum of squared deviations about any other point. (b) (c) The sum of a set of measurements is equal to the product of their mean and the number of measurements. The sum of the deviations of a set of measurements about their mean is equal to zero. 27. Variable N Mean SE Mean StDev Minimum Q1 Median Q3 Maximum C Variance =.0681, IQR =.3277, Range = , IQR/R = Boxplot of S/R Ratio S/R Ratio Variable Mean StDev Variance Q1 Median Q3 IQR Range Height Boxplot of Height Height No, the mode is approximately The data is skewed to the left. The IQR is much smaller than the range (.1/.44=.2273). 22

23 Chapter 2 Exercise Solutions 29. (a) Variable N Mean Median TrMean StDev SE Mean nutri Variable Minimum Maximum Q1 Q3 nutri Variance = , Range = 84.4, IQR = 13.1 IQR/R =.1552 Histogram of Status Frequency Status Frequency Polygon of Status Frequency Status

24 Chapter 2 Exercise Solutions Stem-and-leaf of C1 N = 107 Leaf Unit = (28) Boxplot of Status Status (d) 75.4 ± gives endpoints and There are 79 points within one standard deviation giving 79/107 =.7383 or 74% within one s.d. of the mean ± 2(13.64) gives endpoints and There 103 points within 2 s.d. s of the mean giving103/107 =.9626 or 96% within two s.d. s of the mean ± 3(13.64) gives endpoints and There are 105 points within 3 s.d. s of the mean giving 105/107 =.9813 or 98% within three s.d. s of the mean. (e) 102/107 =.9533 (f) 1/107 =

25 Chapter 3 Exercise Solutions Chapter (a) P(Woman) = 679/1024 =.6631 (b) (c) Marginal probability Using the definition of marginal probability: P(Woman) = P(Woman No Victimization) + P(Woman Partner) + P(Woman Nonpartners) + P(Woman Multiple Victimization) = 611/ / / /1024 = = The compliment rule: P(Woman) = 1- P(Man) = 1-345/1024 = =.6631 (d) 34/1024 =.0332 (e) Joint probability (f) 17/345=.0493 (g) Conditional probability (h) (345/1024) + (44/1024) (10/1024) = =.3701 (i) Addition rule (a) P(Male) = 673/1021 =.6592, P(Female) = 348/1021 =.3408, P(Split) = 569/1021 =.5573, P(Never) = 452/1021 =.4427 (b) P(Male Split) = 349/1021 =.3418, P(Male Never) = 324/1021 =.3173 P(Female Split) = 220/1021 =.2155, P(Female Never) = 128/1021 =.1254 (c) P(Split Male) = 349/673 =.5186, P(Split Female) = 220/348 =.6322 P(Never Male) = 324/673 =.4814, P(Never Female) = 128/348 =.3678 P(Male Split)=349/569 =.6134, P(Male Never) = 324/452 =.7168 P(Female Split) = 220/569 =.3866, P(Female Never) = 128/452 =.2832 (d) (e) P(Never Female)=P(Never Female)*P(Female)=.3678*.3408=.1253 A joint probability (a) P(Male Split drugs)= 349/1021 =.3418 (b) P(Male or Split drugs or both) = (673/1021) + (569/1021) (349/1021) = =.8747 (c) P(Male Split drugs) = 349/569 =.6134 (d) P(Male) = 673/1021 =

26 Chapter 3 Exercise Solutions (a) 417/2720 =.1533 (b) 196/2720 =.0721 (c) 203/2720 =.0746 (d) 676/ / /2720 = 714/2720 =.2625 (e) /2720 = = PM ( ) =.6, PS ( M) =.2, P( S M) = PS ( M)/ PM ( ) =.2/.6 = P(H) =.35, P(S H)=.86, P(S H) = P(H)*P(S H)=(.35)(.86) = (a) (b) A subject having the symptom(s) and not having the disease. A subject not having S but having the disease. (c) P(S D) = 744/775 =.96 (d) PS ( D ) = 1359/1380 =.9848 (e) PS ( D) = 21/1380 =.0152; P( D S) = (.96)(.001)/[(.96)(.001) + (.0152)(.999)] =.0595 (f) PS ( D) = 31/775 =.04; P( D S) = (.9848)(.999)/[(.9848)(.999) + (.04)(.001)] = (g) (.96)(.0001) PD ( S ) = = (.96)(.0001) + (.0152)(.9999) (.9848)(.9999) PD ( S ) = = (.9848)(.9999) + (.04)(.0001) (.96)(.01) PD ( S ) = =.3895 (.96)(.01) + (.0152)(.99) (.9848)(.99) PD ( S ) = =.9996 (.9848)(.99) + (.04)(.01) (.96)(.10) PD ( S ) = =.8753 (.96)(.10) + (.0152)(.90) (.9848)(.90) P( D S ) = =.9955 (.9848)(.90) + (.04)(.10) 26

27 Chapter 3 Exercise Solutions (a) Sensitivity = PT ( D ) = 38/43 =.8837; (b) Specificity = PT ( D ) = 18/28 =.6429 (c) Probability of bucket-handle tears Sensitivity = PT ( D ) =.927; Specificity = PT ( D ) =.997 ; PD ( ) = 1/ = ; PD ( ) = = PT ( D) = 1 PT ( D) = =.073 (.997)( ) Predictive Value Negative = (.997)( ) + (.073)( ) = Chapter 3 Review Exercises 3. (a) 459/2142 =.2143 (b) 319/578 =.5519 (c) 329/2142 =.1536 (d) 578/ / /2142 = =.4575 (e) 1 - (1057/2142) = (a) 1. 29/56 = /56 = /56 = ( )/56=42/56 = /29 =.5172 (b) 1. True. Each is the probability of the joint occurrence of competent and enrolled in a CPR course. 2. True. Each is the probability of the occurrence of competent or enrolled in a CPR course, or both. 3. True. The marginal probability of A is equal to the sum of the joint probabilities of A with each category of competence. 4. False. The addition rule is required. Should be PB ( C) = PB ( ) + PC ( ) PB ( C). 5. False. Training course and competence are not independent. 6. False. Training course and competence are not independent. 7. True. The training groups are mutually exclusive. 27

28 Chapter 3 Exercise Solutions 8. True. This is correct use of the multiplication rule. 9. False. Incorrect use of the multiplication rule. It should be PA ( D) = PADPD ( ) ( )or P( D A) P( A) 5. (a) 17/126 =.1349 (b) 52/126+42/126 17/126 = =.6111 (c) 42/126 =.3333 (d) 1 (52/126) =.5873 (e) 15/42 =.3571 (f) 1 (42/126) =.6667 (g) 0 (These are mutually exclusive events) (h) 17/52 = P(Over 30 Master s degree) = = 16 = (a) = = = = = (b) = = = (.7)(.7)= = = = (a) 5235/ =.0432 (b) 3102/ =.0256 (c) 2992/ =.0247 (d) 2991/3103 =.9639 (e) 2991/5235 =.5713 (f) Sensitivity = PT ( D) = P( A B) 2991/ (g) Specificity = PT ( D) = P( A B) = / =

29 Chapter 3 Exercise Solutions 10. PE ( R) =.6, P( R E) =.8, PE ( ) = PE ( R)/ P( R E) =.6/.8 = (.03)(.20) = PBC ( ) =.04, PS ( ) =.20, PBC ( S) =.03 PBC ( S) = = Symptom Disease? Y N Total Y N Total P( D S ) =.05/.8 = Yes 15. Mothers 24 years of age or younger 16. Mothers between the ages of 20 and 29, inclusive. 17. PG ( ) = 0, since A and B are mutually exclusive. A mother cannot be both under 20 and between 20 and 24 years of age. 18. A and B are mutually exclusive; A and C are not mutually exclusive; B and C are mutually exclusive. 19. (a) Plasma lipoprotein levels between 10 and 15 or greater than or equal to 30. (b) Plasma lipoprotein levels between 10 and 15 and greater than or equal to 30. (c) Plasma lipoprotein levels between 10 and 15 and less than or equal to 20. (d) Plasma lipoprotein levels between 10 and 15 or less than or equal to (a) Plasma lipoprotein levels less than 10 or grater than 15. (b) Plasma lipoprotein levels less than 30. (c). Plasma lipoprotein levels greater than

30 Chapter 3 Exercise Solutions 21. (a) Sensitivity = PT ( D ) = 214/287 =.7456 (b) Specificity = PT ( D ) = 330/1000 = (a) 22 PT ( D ) = = (b) PT ( D) =.95, PD ( ) =.031, PT ( D) = 1 PT ( D) = =.2903 (.95)(.031) PDT ( ) = =.0948 (.95)(.031) + (.2903)(1.031) (c) PD ( ) = =.969 (.7097)(.969) PDT ( ) = =.9978 (.7097)(.969) + (1.95)(.031) 23. (.95)(.002) PT ( D) = 1 specificity=1-.85=.15; PDT ( ) = =.0125 (.95)(.002) + (.15)(1.002) 30

31 Chapter 4 Exercise Solutions Chapter (a) (b) C (.76) (.24) = C (.76) (.24) + C (.76) (.24) + C (.76) (.24) = [ ] =.8915 (c) C (.76) (.24) + C (.76) (.24) + C (.76) (.24) = = (d) C (.76) (.24) + C (.76) (.24) C5(.76) (.24) + 20 C6(.76) (.24) + 20C7(.76) (.24) = = np = 20(.24) = (a) (b) (c) (d) (e) (a) C (.76) (.24) =.2536 C (.76) (.24) + C (.76) (.24) + C (.76)(.24) + C (.76) (.24) = =.3461 C (.76) (.24) + C (.76) (.24) + C (.76) (.24) = =.7330 C (.76) (.24) + C (.76) (.24) + C (.76) (.24) = =.9068 C (.76) (.24) = C = (b).5187 (c) =.4784 (d) =.2695 (e) = µ = np = = σ = = = 2 15(.32) 4.8; npq 15(.32)(.68) C =, Yes, it would be surprising to have exactly 2 indicate they 31

32 Chapter 4 Exercise Solutions have been tested for HIV (a) (c) (d) (e) (f) (.81) (.19).5314 C = (b) C 2 1 (.81) (.19) = C (.81) (.19) + C (.81) (.19) = = C3(.81) (.19) = =.9931 C (.81) (.19) + C (.81) (.19) = = C (.81) (.19) = (a) PX ( = 6 n= 15, p =.75) = PX ( = 9 n= 15, p=.25) = =.0034 (b) PX ( 7 n= 15, p =.75) = P( X 8 n= 15, p=.25) =.9958 (c) PX ( 5 n= 15, p =.75) = PX ( 10 n = 15, p =.25) = =.0008 (d) P(6 X 9 n = 15, p =.75) = ([ PX ( 9) PX ( 5)] n= 15, p =.75) = PX ( 6 n= 15, p =.25) px ( 10 n = 15, p =.25) = [1 PX ( 5)] [1 PX ( 9)] = (1.8516) (1.9992) = = Number of Successes, x Probability, f(x) C (.2) (.8) =.008 C (.2) (.8) =.096 C (.2) (.8) =.384 C (.2) (.8) =.512 Total

33 Chapter 4 Exercise Solutions (a) (b) (c) (d) (a) 4 5 e 4 = 5! e 4 e 4 e 4 e 4 e 4 e = =.215 0! 1! 2! 3! 4! 5! e 4 e 4 e 4 e 4 e =.629 0! 1! 2! 3! 4! e 4 e 4 e = =.320 5! 6! 7! e.06 (.06).056 1! = or =.056 (b) =.002 (c).942 (d) = (a) e =.105 (b) =.032 (c) 7! 5 0 e 5 = 0!.007 (d) e 5 e 5 e 5 e 5 e =.440 0! 1! 2! 3! 4! (a) =.303 (b).607 (c) =.002 (d) = (a) =.086 (b) =.946 (c).463 (d) =.664 (e) P(0 z 1.43) = = P( 2.87 z 2.64) = =

34 Chapter 4 Exercise Solutions Pz ( =.55) = 0, Pz (.55) = = Pz (.55) = = Pz< ( 2.33) = Pz< ( 2.33) = P( 1.96 z 1.96) = = P( 2.58 z 2.58) = = P( 1.65 z 1.65) = = Pz= (.74) = z 1 = z 1 =

35 Chapter 4 Exercise Solutions = z 1 = P(z < 2.98) =.9986; = z 1 = ( )/2= z 1 = z 1 =

36 Chapter 4 Exercise Solutions (a) P(600 x 1000) = P z = P( 1.28 z.62) = =.6321 (b) (c) (d) Px ( > 900) = P z> = Pz ( >.14) = = Px ( < 500) = P z< = Pz ( < 1.75) = P(900 x 1100) = P z = P(.14 z 1.09) = = (a) (b) Px ( 200) = P z = Pz ( 1.20) = = Px ( < 100) = P z< = Pz ( <.8) = (c) P(100< x< 200) = P(.8< z< 1.2) = =.6730 (d) P(100< x< 200) = P < z< = P(1.2 < z< 2.2) = =.1012 (e) Px ( 200) = P z = Pz ( 1.2) = (10,000)(.1151)=

37 Chapter 4 Exercise Solutions (a) (b) (c) Px ( < 15) = P z< = Pz ( < 1.10) = Px ( > 40) = P z > = Pz ( >.71) = = P(14 x 50) = P z = P( 1.17 z.71) = =.6401 (d) (e) Px ( < 10) = P z< = Pz ( < 1.46) = P(10 x 20) = P z = P( 1.46 z.74) = = (a) (b) (c) Px ( > 50) = P z> = Pz ( >.67) = = Px ( < 30) = P z < = Pz ( < 2) = P(30 < x< 60) = P < z < = P( 2 < z< 0) =.4772 (d) Px ( > 90) = P z> = Pz ( > 2) =

38 Chapter 4 Exercise Solutions (a) (b) (c) (d) P(180< x< 200) = P < z < = P( 1 < z < 0) = Px ( > 225) = P z> = Pz ( > 1.25) = = Px ( < 150) = P z< = Pz ( < 2.5) = P(190< x< 210) = P < z < = P(.5 < z<.5) = = (a) (b) (c) (d) (e) (a) (b) P z = P( 1 z.1) = = P z > = Pz ( >.6) = = P z < = Pz ( <.6) = P z = Pz (.4) = = P z = P( 1.8 z 1.4) = = Px ( > 155) = P z > = Pz ( > 1.53) = = Px ( 100) = P z< = Pz ( 2.13) =

39 Chapter 4 Exercise Solutions (c) P(105< x< 145) = P < z < = P( 1.8 < z<.87) = =.7719 Chapter 4 Review Exercises 15. (a) C = (b) =.0949 (c) C = (d) = (a) =.0442 (b).0118 (c) = (a) =.034 (b) =.467 (c) =.923 (d).010 (e) = (a) =.180 (b).677 (c) = (a) =.4967 (b).5033 (c).1678 (d) =.0104 (e) = (a).273 (b) =.727 (c) (a) (b) Px ( > 75) = P z> = Pz ( > 1.5) = = P(55 x 75) = P z = P(.5 z 1.5) = =.6247 (c) P(50 x 70) = P z = P( 1 z 1) = =

40 Chapter 4 Exercise Solutions 22. (a) (b) (c) 23. (a) (b) (c) Px ( < 4) = P z < = Px ( > 5) = P z > = = Px ( 3) = P z = Pz ( 2.33) = Px ( < 200) = P z< = Pz ( < 3) = Px ( 650) = P z = Pz ( 1.5) = = P(350 x 675) = P z = P( 1.5 z 1.75) = = µ = np,20 = np, n = 20/ p; σ = np(1 p),16 = p(1 p) 16= 20(1 p) p 20 p =.2, n= = x µ 25. z0 = ; Pz ( z0) =.0985; =.9015 σ 70 µ Pz ( 1.29) =.9015;1.29= 70 µ = µ =

41 Chapter 4 Exercise Solutions 26. ( )/2 = ( )/2 =.123 µ - ks 0 µ + ks µ + kσ µ =.877; Pk [ < ( µ + kσ)] = P z< = Pz ( < k) = Pz ( < 1.16) =.877; k = 1.16 σ 27. (a) (b) (c) k 100 Pz ( 2.35) = = k = k 100 Pz ( 1.23) = = k = k 100 Pz ( 2.01) = = k = (d) (1.9660)/2 =.0170; =.9830 k 100 Pz ( 2.12) = = k = µ ) Px ( 40) = P z =.0080; z= µ 2.41= 40 µ = 24.1 µ = 64.1, µ = µ 29. Pz ( z0) =.9904; z0 = = µ =

42 Chapter 4 Exercise Solutions µ Px ( 25) = P z = z = 1.62 since Pz< ( 1.62) = = µ 1.62 = 25 µ = 8.1 µ = Pz ( z0) =.0778; z0 = 1.42; 1.42= σ = 10.6 σ Px ( 50) = P z =.9772; z= 2 σ = 2σ = σ = 10 σ 33. (a) (b) (c) 34. (a) (b) (c) Could be Bernoulli if we assume each child has an equal chance of being a boy or girl; Not a Bernoulli - More than 2 possible outcomes Not a Bernoulli - Weight is a continuous variable Not a Bernoulli - Not a Yes/No outcome Not a Bernoulli - Degrees Celsius is a continuous variable Not a Bernoulli - Not a constant probability of vital signs being normal from patient to patient 42

43 Chapter 5 Exercise Solutions Chapter Sampling distribution mean: 204, Standard Error: σ 44 σ x = = = n (a) σ 43/ , 183 x = = z1 = = 2.34, z2 = = 2.16; P(170 x 195) = P( 2.34 z 2.16) = =.9747 (b) (c) (a) z = = 1.44; Px ( < 175) = Pz ( < 1.44) = z = = 1.26; Px ( > 190) = 1 Pz ( 1.26) = = σ x = 1/ 9 =.3333, z= =.90, Px ( > 6) = 1 Pz (.90) = = (b) z1 = = 2.10, z2 = P(5 x 6) = P( 2.10 z.90) = =.7980 (c) (a) z = = 1.50, Px ( < 5.2) = Pz ( < 1.50) = σ x = 454/ 50 = , z = = Px ( > 800) = 1 Pz ( 1.23) = =.1093 (b) z = =.33, Px ( < 700) = Pz ( <.33) = (c) z1 =.33, z2 = = P(700 x 850) = P(.33 z 2.01) = =

44 Chapter 5 Exercise Solutions (a) σ x = 1476/ 75 = , z = = Px ( < 2450) = Pz ( < 2.88) =.0020 (b) z = =.94, Px ( > 3100) = 1 Pz (.94) = = (c) z1 = = 2.58, z2 = = P(2500 x 3300) = P( 2.58 z 2.11) = = (d) z1 = 2.58, z2 = = P(2500 x 2900) = P( 2.58 z.23) = = (a) (b) (c) 20 σ x = = Px ( 100) = P z = Pz ( 0) = 1.5 = Px ( 110) = P z = Pz ( 2) = P(96 x 108) = P z 5 5 = P(.8 z 1.6) = =

45 Chapter 5 Exercise Solutions σ x = 16/ 64 = 2 (a) P(45 x 55) = P z 2 2 = P( 2.5 z 2.5) = =.9876 (b) (c) (d) Px ( > 53) = P z > = Pz ( > 1.5) = = Px ( < 47) = P z< = Pz ( < 1.5) = P(49 x 56) = P z 2 2 = P(.5 z 3) = = x f ( x ) µ = ( )/10= 5 x σ x (2 5) + (3 5) + 2(4 5) + 2(5 5) + (7 5) + (8 5) = = Sample x 6, 8, , 8, , 8, , 10, , 10, , 12, , 10, , 10, , 12, , 12, µ = ( )/10= 10 x σ x (8 10) + ( ) + + ( ) = =

46 Chapter 5 Exercise Solutions σ x B x = + = A ( ) Px ( B xa> 8) = P z > = Pz ( >.28) = = σ x ; M x = + = W ( ) Px ( M xw > 100) = P z> = Pz ( >.36) = = σ x 3 1 x = + = Px ( 1 x2 8) = P z = Pz ( 2.67) = = σ x 4 1 x = + = Px ( 1 x2 12) = P z = Pz ( 3) = = ( ) Px ( G xb > 10) = P z > = Pz ( > 2.20) = = z = = 1.21;.32(.68) Pz ( > 1.21) = = z = =.75;.13(.87) Pz (.75) =

47 Chapter 5 Exercise Solutions z = = 1.40;.64(.36) Pz ( 1.40) = = z = = 1.96;.64(.36) Pz ( < 1.96) = (.6)(.4) σ = =.0490 p 100 (a).65.6 P( p.65) = P z = Pz ( 1.02) = = (b).58.6 P( p.58) = P z = Pz (.41) = (c) P(.56 p.63) = P z = P(.82 z.61) = = p = 80/200 =.4;.4.35 P( p.4) = P z = Pz ( 1.48) =.0694 (.35)(.65) σ = (.095)(.905) (.049)(.951).0353 p 1 p ( ) P( p1 p2.09) = P z = Pz ( 1.25) = = σ = (.721)(.279) (.648)(.352).0438 p 2 p ( ) P( p2 p1 <.06) = P z< = Pz ( <.30) =

48 Chapter 5 Exercise Solutions σ = (.21)(.79) (.13)(.87).0475 p 1 p = P(.04 < p1 p2 <.20) = P < z< = P(.84< z < 2.53) = = Chapter 5 Review Exercises Px ( 12) = P z = Pz ( 3) = = / Px ( > 25) = P z > = Pz ( > 3.44) = = / Px ( < 24) = P z< = Pz ( < 1.50) = / σ x.7225 M x = + = W ( ) Px ( M xw > 3) = p z > = Pz ( > 1.94) = = Px ( < 12) = P z < = Pz ( < 1.91) = / Px ( > 19) = P z> = Pz ( > 1.11) = = / σ x M x = + = W ( ) Px ( M xw > 5) = p z > = Pz ( >.60) = =

49 Chapter 5 Exercise Solutions P( p>.25) = P z> = Pz ( > 1.23) = =.1093 (.19)(.81) P( p>.20) = P z > = Pz ( > 1.18) = =.1190 (.17)(.83) P( p<.20) = P z< = Pz ( < 1.13) =.1292 (.23)(.77) σ = (.23)(.77) (.17)(.83).0367 p M pw =.05 (.23.17) P( pm pw <.05) = P z < = Pz ( <.27) = ! C = = 252 5!5! 22. Normally distributed 23. µ =.53, σ = (.53)(.47)/110 =.0476 p p P( p <.50) = P z< = Pz ( <.63) = At least approximately normally distributed 26. µ = 25, σ = 7/ 35 = 1.18 x x P(22 x 29) = P z = P( 2.54 z 3.39) = =

50 Chapter 5 Exercise Solutions 28. (a) n = 10 Normal n = 50 Normal n = 200 Normal (b) n = 10 Normal n = 50 Normal n = 200 Normal (c) n = 10 Not normal n = 50 Approx. normal n = 200 Approx. normal 29. (a) No, since 8(.5) = 4 and 4<5. (b) Yes, since 30(.4) = 12 and 30(.6) = 18 are both >5. (c) No, since 30(.1) = 3 is less than 5. (d) Yes, since 1000(.01) = 10 and 1000(.99)=990 are both greater than 5. (e) Yes, since 100(.9)=90 and 100(.1) = 10 are both greater than 5. (f) Yes, since 150(.05)=7.5 and 150(.95) = are both greater than 5. 50

51 Chapter 6 Exercise Solutions Chapter σ x = 10/ 49 = 1.43 (a) 90± 1.645(1.43) = (88,92) (b) 90± 1.96(1.43) = (87,93) (c) 90± 2.58(1.43) = (86,94) σ x = 3.5/ 16 =.875 (a) 5.98± 1.645(.875) = (4.54,7.42) (b) 5.98± 1.96(.875) = (4.26,7.70) (c) 5.98± 2.58(.875) = (3.72,8.24) σ x = 3/ 64 =.375 (a) 8.25± 1.645(.375) = (7.63,8.87) (b) 8.25± 1.96(.375) = (7.51,8.99) (c) 8.25± 2.58(.375) = (7.28,9.22) σ x = 15/ 100= 1.5 (a) 125± 1.645(1.5) = (123.53,127.47) (b) 125± 1.96(1.5) = (122.06,127.94) (c) 125± 2.58(1.5) = (121.13,128.87) x = , σ x = 350/ 16 = 87.5 (a) ± 1.645(87.5) = ( , ) (b) ± 1.96(87.5) = ( , ) (c) ± 2.58(87.5) = ( , ) (a) (b) (c) (d)

52 Chapter 6 Exercise Solutions (a) x = ( )/8= (b) 2 ( x i 8.875) s = = (c) 2.900/ 8= (d) 8.875± (1.0253) = (6.4506, ) (e) (1.0253) = (f) Approximately 95% of all intervals constructed in a similar manner with sample of 8 drawn from the population will contain the sample mean. (g) We are 95% confident the population mean is between and (a) (.4)( 15) = 1.549; (.1)( 15 ) =.387 (b) 3.5± (.4) = (2.64,4.36);.7± (.1) = (.49,.91) (c) (d) (e) (f) (g) Nitric oxide diffusion rates are normally distributed in the population from which the sample was drawn. Practical: We are 95% confident the mean maximal nitric oxide diffusion rate for asthmatic schoolchildren is between 2.64 and 4.36 while for control subjects it is between.49 and.91. Probabilistic: approximately 95% of intervals constructed ina similar manner with samples of size 15 drawn from the populations of asthmatic and control schoolchildren will contain the respective population means. The practical. Narrower because the t coefficient from which the interval is constructed is smaller. Wider because the t coefficient from which the interval is constructed is larger. 52

53 Chapter 6 Exercise Solutions (a) s x = 4.3/ 9= (b) 17.4± (1.433) = (12.59,22.21) (c) (1.433) = (d) That the distribution for the population of subjects with ACL-deficient knees is normally distributed s x = 12/ 16 = 3 90 % CI: 71.5± (3) = (66.2,76.8) 95 % CI: 71.5± (3) = (65.1,77.9) 99 % CI: 71.5± (3) = (62.7,80.3) x = 25.9, s= , s x = / 10 = ± (2.9943) = (19.13,32.67) The samples constitute independent simple random samples from the two populations. The two populations of free fatty acid concentrations are normally distributed, and the two population variances are equal. s = = , s = 62 11= s ( ) + 10( ) p = = , sp = = % CI: 95% CI: 99% CI: ( ) ± = ( 549.8, 340.2) ( ) ± = ( 571.3, 318.7) ( ) ± = ( 615.5, 274.5)

54 Chapter 6 Exercise Solutions The samples constitute independent simple random samples from the two populations. The two populations of prostate cancer screening knowledge scores are normally distributed, and the two population variances are equal σ x x = + = %CI: ( ) ± 1.645(.7570) = (1.95,4.45) 95 % CI: ( ) ± 1.96(.7570) = (1.72,4.68) 99 %CI ( ) ± 2.58(.7570) = (1.25,5.15) The samples constitute independent simple random samples from the two populations. The two populations of pain scores are normally distributed, and the two population variances are equal. s = = s = = σ x x = + = % CI: ( ) ± 1.645(7.1701) = ( 5.89,17.69) 95 % CI: ( ) ± 1.96(7.1701) = ( 8.15,19.95) 99 % CI: ( ) ± 2.58(7.1701) = ( 12.60,24.40) 54

55 Chapter 6 Exercise Solutions The samples constitute independent simple random samples from the two populations. The two populations of pain scores are normally distributed, and the two population variances are equal. x = 18.83, s = 4.36, x = 22.21, s = 3.36 F F M M (4.36) + 13(3.36) sp = = , sp = = % CI: 95 % CI: 90 % CI: ( ) ± = ( 6.48,.28) ( ) ± = ( 7.14,.38) ( ) ± = ( 8.53,1.77) The samples constitute independent simple random samples from the two populations. The two populations of doses of methadone are normally distributed, and the two population variances are equal (156) + 7(316) sp = = , sp = = % CI: 95 % CI: 99 % CI: ( ) ± = (64.1,479.9) ( ) ± = (19.2,524.8) ( ) ± = ( 77.5,621.5)

56 Chapter 6 Exercise Solutions (1.05) + 8(1.01) = = s p sx.4557 M x = + = F % CI: ( ) ± (.4557) = (1.42,3.00) 95 % CI: ( ) ± (.4557) = (1.26,3.16) 99 % CI: ( ) ± (.4557) = (.91,3.51) (1.5) + 11(2.0) s p = = sx x = + = % CI: ( ) ± (.7217) = (2.1,4.5) 95 % CI: ( ) ± (.7217) = (1.8,4.8) 99 % CI: ( ) ± (.7217) = (1.3,5.3) (5) + 13(6) = = s p sx x = + = % CI: (26 21) ± (2.2512) = (1.1,8.9) 95% CI: (26 21) ± (2.2512) = (.3,9.7) 99% CI: (26 21) ± (2.2512) = ( 1.3,11.3) 56

57 Chapter 6 Exercise Solutions sx x = + = (4/20)(1.7291) + (100/24)(1.7139) % CI: t 1 α /2 = = = (4/20) + (100/24) (7 36) ± ( ) = ( 32.58, 25.42) (4/20)(2.093) + (100/24)(2.0687) % CI: t 1 α /2 = = = (4/20) + (100/24) (7 36) ± ( ) = ( 33.32, 24.67) (4/20)(2.8609) + (100/24)(2.8073) % CI: t 1 α /2 = = = (4/20) + (100/24) (7 36) ± ( ) = ( 34.87, 23.13) x = 14.72, s = 6.20, n = 18, x = 20.41, s = 6.81, n = 14 N N N S S S (6.20) + 13(6.81) sp = = , s= = % CI: 95 % CI: 99 % CI: ( ) ± = ( 9.60, 1.78) ( ) ± = ( 10.40,.98) ( ) ± = ( 12.03,.65) Yes, it does appear likely mean cadmium level is higher among smokers than nonsmokers because the 90 and 05% confident intervals do not contain 0. 57

58 Chapter 6 Exercise Solutions Point estimate: p = 63/472 =.1335, Reliability coefficient: 1.96 Standard error: (.1335)(.8665) 472 =.0157 CI:.1335± 1.96(.0157) = (.1027,.1643) Practical: We are 95% confident the true proportion of ventilator-associated pneumonia is between.1027 and Probabilistic: Approximately 95% of intervals constructed in a similar manner with samples of size 472 drawn from the populations of ventilated patients will contain the respective population proportion Point estimate: p = 118/202 =.5842, Reliability coefficient: Standard error: (.5842)(.4158) 202 =.0347 CI:.5842± 1.645(.0347) = (.5271,.6413) Point estimate: p = 75/136 =.5515, Reliability coefficient: 2.58 Standard error: (.5515)(.4485) 136 =.0426 CI:.5515± 2.58(.0426) = (.4415,.6615) Point estimate: p = 88/125 =.7040, Reliability coefficient: 1.96 Standard error: (.704)(.296) 125 =.0408 CI:.704± 1.96(.0408) = (.624,.784) Point estimates: p = 114/637 =.1790, p = 57/510 =.1118 Reliability coefficient: Standard error: (.1790)(.8210) (.1118)(.8882) + = CI: ( ) ± 1.96(.0206) = (.0268,.1076) 58

59 Chapter 6 Exercise Solutions Practical: We are 95% confident the true difference in proportions between the abused or neglected group and the control group is between.0268 and Probabilistic: Approximately 95% of intervals constructed in a similar manner with samples of size 637 and 510 drawn from the two populations will contain the difference in the respective population proportions Point estimates: p = 109/138 =.7899, p = 120/136 =.8824 Reliability coefficient: Standard error: (.7899)(.2101) (.8824)(.1176) + = CI: ( ) ± 1.96(.0443) = (.1793,.0057) Point estimates: p = 16/49 =.3265, p = 12/51 =.2353 Reliability coefficient: Standard error: (.3265)(.6735) (.2353)(.7647) + = CI: ( ) ± 1.96(.0895) = (.0842,.2666) Point estimates: p = 24/50 =.48, p = 50/75 = Reliability coefficient: Standard error: (.48)(.52) (.6667)(.3333) + = CI: ( ) ± 1.645(.0892) = (.3334,.0400) (2.58) (1) n = = (1.96) (1) n = =

60 Chapter 6 Exercise Solutions (1.96) (15) n = = ( 2.5) 2 2 (1.645) (60) n = = (1.96) (90) n = = (1.96) (.20)(.80) n = = (.03) 2 (1.96) (.5)(.5) n = = (.03) (1.96) (.5)(.5) n = = (.04) 2 (2.58) (.5)(.5) n = = (.04) (1.96) (.5)(.5) n = = (.05) 2 (1.96) (.25)(.75) n = = (.05) (1.96) (.20)(.80) n = = (.03) 2 s χ.975 χ.025 = , = , = ( ) 9( ) 2 2 < σ < < σ < s χ.975 χ.025 = = , = ( ) 11( ) 2 2 < σ < < σ <

61 Chapter 6 Exercise Solutions (1,000,000) 19(1,000,000) 2 < σ < ,307.86< σ < 1,878,027.08, < σ < (1.03) 29(1.03) 2 < σ < < σ < 2.28,.75< σ < (2.25) 24(2.25) 2 < σ < < σ < 4.35, 1.17 < σ < s 15(2.1796) 15(2.1796) = , < σ < < σ < 5.221, < σ < s 19( ) 19( ) = , < σ < < σ < s = , s = , n = 5, n = Incomplete Complete Incomplete Complete s 2 Incomplete 2 Complete s = = < σ σ σ <.1131 < < 2 2 Incomplete Incomplete 2 2 Complete σcomplete s = 8.1 = 65.61, s = 6 = 36, n = 33, n = 32, S NS S NS s s 2 S 2 NS = = σ σ < <.71< < σ S S 2 2 NS σ NS 61

62 Chapter 6 Exercise Solutions /10 σ 12/10 σ < <.49 < < σ 1/ σ 2 15/8 σ 15/8 σ < <.78< < σ 1/ σ 2 35,000/20,000 σ 35,000/20,000 σ < <.90 < < σ 1/ σ 2 148/105 σ 148/105 σ < <.39< < σ 1/3.37 s 2 2 N U σ 2 s = , s = , = = s U 2 N σ σ < < 5.13< < σ 1/3.77 Chapter 6 Review Exercises σ2 13. x =79.87, s 2 = , s = 5.3; 79.87± ( 5.30/ 15 ) = (76.93,82.81) 14. p n ( ) =.35, = 140,.35± 1.96 (.35)(.65)/140 = (.27,.43) 15. p = 21/70 =.30,.30± 1.96 (.3)(.7)/70 = (.19,.41) 16 (a) (b) 2 (1.96) (.3)(.7) n = = (.05) (1.96) (.3)(.7) 2 2 (.05) (1499) + (1.96) (.3)(.7) n = = p = 44/220 =.20, p = 150/280 = (.54)(.46) (.20)(.80) (.54.20) ± = (.26,.42)

63 Chapter 6 Exercise Solutions (20)(1.95) + (15)(1.70) = = s p ( ) ± = (.19,2.29) (.9)(.1) p = 180/200 =.90,.90± = (.87,.93) (125 95) ± = (17.06,42.94) x = 19.23, s 2 = ; 19.23± )/10 = (16.01,22.45) 22. x s x s 2 2 A = , A = , B = , B = ( ) + 11( ) = = s p ( ) ± = (.0753,.4199) x s x s 2 2 A = 6.125, A = , B = 5.806, B = ( ) + 15(5.9980) = = s p ( ) ± = ( 2.18,2.82) ( ) ± = ( 246.4, 3.6) ± 1.96 = (362.73,507.27) 34 (.293)(.707) (.173)(.827) ( ) ± = (.026,.214)

64 Chapter 6 Exercise Solutions 27. (.5897)(.4103) p = 23/39 =.5897,.5897± 1.96 = (.4353,.7441) x s x s 2 2 M = , M = , H = , H = ( ) ± = (5.903,11.305) s = (.0582) 10 =.1840, s = (.1641) 10 =.5189 CP C Assume equal variances for the two groups, (.1840) + 9(.5189) = = s p ( ) ± = (.416,.188) Assume unequal variances for the two groups ω = 8.2 /19 = , ω = 23 /17 = , t = , t = (2.8784) (2.9208) t.995 = = ( ) ± = ( 40.57, 6.23) Level of confidence decreases. The interval would have no width. The level of confidence would be zero. 32. The interval would have to cover the entire spectrum of possible answers. The interval would be too large to be useful in any way. 33. Since the sample size is 32, the reliability coefficient should be z. The precision of the estimate equals the margin of error, which is The target population are people staying in hospitals with and without delirium. The sampled population was the 204 subjects with delirium and the 118 without delirium in the study. 35. All drivers aged 55 and older. Drivers 55 and older participating in the vision study. 64

65 Chapter 6 Exercise Solutions 36. The target population is babies born of HIV-1 infected mothers. The sampled population is the babies in the sample from Cameroon. 37. x =.3197, s =.2486, n= ± 1.96 = (.2865,.3529) 216 We use z since n > x = , s=.0760, n= ± = (1.7398,1.7638)

66 Chapter 7 Exercise Solutions Chapter H : µ 75, H : µ < 75, z = = A ( 14.6/ 76 ) Reject H 0 since < -2.33, p =.0051 < H : µ 60, H : µ < 60, x = 54.94, s= , t = = A ( / 16 ) Reject H 0 since < ,.01 < p < H : µ 9, H : µ > 9, t= =.76 0 A ( 7.3/ 18 ) Fail to reject H 0 since.76 < 1.333, p > H : µ 4, H : µ > 4, t = = A ( 2/ 25 ) Reject H 0 since 2.00 > ,.025 < p <.05 We assume the data comes from a normally distributed population H : µ 30, H : µ < 30, z= = A ( 11/ 49 ) Yes, Reject H 0 since z = < , p < H : µ 6, H : µ > 6, t = = A (.6/ 9) Reject H 0 since 2.50 > ,.01 < p < H : µ 80, H : µ < 80, t = = A ( 10/ 25 ) Fail to reject H 0 since t = > ,.05 < p <.10 66

67 Chapter 7 Exercise Solutions H : µ 2000, H : µ < 2000, z= = A ( 210/ 500 ) Fail to reject H 0 since z = > , p = H : µ 25; H : µ > 25; z= = A ( 6.5/ 100 ) Reject H 0 since z = 3.08 > 1.645, p = H : µ 70, H : µ > 70, t = = A ( 12/ 16 ) Fail to reject H 0 since t = 1.33 < , p > H : µ 10, H : µ > 10, z= = A ( 3/ 16 ) Reject H 0 since z = 4.0 >1.645, p < H : µ = 12, H : µ 12, t = = A ( 1.282/ 15 ) Reject H 0 since t = > , p <.005(2) = H : µ = 110, H : µ 110, t = = A ( / 20 ) Fail to Reject H 0 since t =.1271 < , p >.1(2) = H : µ 165, H : µ < 165, t = = A ( / 12 ) Fail to Reject H 0 since t = > , p > H0 : µ 30, HA: µ < 30, z= = / 50 Reject H 0 since z = > 1.645, p <