Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis Chapter 7

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1 Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei Chapter 7 7. a. b. μ x = μ = μ x = μ = 0 σ 4 σ x = = =.5 n 64 σ σ x = = =.375 n 64 3 c. μ = μ μ = 0 = x x σ σ σ x x = + = + = =.65 n n d. Since n 30 and n 30, the ampling ditribution of x x i approximately normal by the Central Limit Theorem. 7.4 Aumption about the two population:. Both ampled population have relative frequency ditribution that are approximately normal.. The population variance are equal. Aumption about the two ample: The ample are randomly and independently elected from the population. 7.6 a. p = ( n ) + ( n ) (5 )0 + (5 ) = = n + n = 0 b. c. p = p = (0 ) + (0 )0 408 = (6 ).5 + (0 )..55 = = =.8 d. p = (6 ) (7 )500 85,000 = = e. p fall near the variance with the larger ample ize. Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 0

2 7.8 a. σ σ 9 6 σ x.5 x = + = + = =.5 n n b. The ampling ditribution of x x i approximately normal by the Central Limit Theorem ince n 30 and n 30. μ = μ μ = 0 x x x c. x = =. Ye, it appear that x x =. contradict the null hypothei H 0 : μ μ = 0. d. The rejection region require α/ =.05 =.05/ in each tail of the z-ditribution. From Table IV, Appendix B, z.05 =.96. The rejection region i z <.96 or z >.96. e. H 0 : μ μ = 0 H a : μ μ 0 ( x x) 0 ( ) 0 The tet tatitic i z = = σ.5 σ + n n = 4. The rejection region i z <.96 or z >.96. (Refer to part d.) Since the oberved value of the tet tatitic fall in the rejection region (z = 4. <.96), H 0 i rejected. There i ufficient evidence to indicate the difference in the population mean i not equal to 0 at α =.05. f. The form of the confidence interval i: σ σ ( x x ) ± zα / + n n For confidence coefficient.95, α =.95 =.05 and α/ =.05/ =.05. From Table IV, Appendix B, z.05 =.96. The confidence interval i: 9 6 ( ) ± ±.98 (.08, 0.) We are 95% confident that the difference in the two mean i between.08 and 0.. g. The confidence interval give more information. 0 Chapter 7

3 7.0 Some preliminary calculation: x x = = n ( x ) 654 x 8934 n 5 = = n 5 x 858 x = = = n 6 n n ( x ) 49.6 = = x = = = = ( n ) + ( n ) (5 ) (6 ) = = = p n n = a. H 0 : μ μ = 0 H a : μ μ > 0 ( x x) D0 ( ) 0.05 The tet tatitic i t = = = p + + n n 5 6 =.03 The rejection region require α =.0 in the upper tail of the t-ditribution with df = n + n = = 9. From Table VI, Appendix B, t.0 =.46. The rejection region i t >.46. Since the tet tatitic doe not fall in the rejection region (t =.03 >/.46), H 0 i not rejected. There i inufficient evidence to conclude μ μ > 0 at α =.0. b. For confidence coefficient.98, α =.0 and α/ =.0. From Table VI, Appendix B, with df = n + n = = 9, t.0 =.46. The 98% confidence interval for (μ μ ) i: ( x x ) ± t + α / p n n ( ) ± ± 4.87 (5.08, 4.84) We are 98% confident that the difference between the mean of population and the mean of population i between 5.08 and Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 03

4 7. a. Let μ = mean carat ize of diamond certified by GIA and μ = mean carat ize of diamond certified by HRD. For confidence coefficient.95, α =.05 and α/ =.05/ =.05. From Table IV, Appendix B, z.05 =.96. The 95% confidence interval i: σ σ ( x x ) ± zα / + ( ) ±.96 + n n ±.0563 (.969,.0843) b. We are 95% confident that the difference in mean carat ize between diamond certified by GIA and thoe certified by HRD i between and c. Let μ 3 = mean carat ize of diamond certified by IGI. σ σ ( x x3 ) ± zα / + ( ) ±.96 + n n ±.060 (.438,.3678) d. We are 95% confident that the difference in mean carat ize between diamond certified by GIA and thoe certified by IGI i between.438 and e. σ σ ( x x3 ) ± zα / + ( ) ±.96 + n n ±.067 (.3837,.509) f. We are 95% confident that the difference in mean carat ize between diamond certified by HRD and thoe certified by IGI i between.3837 and a. Let μ = mean core for male and μ = mean core for female. For confidence coefficient.90, α =.0 and α/ =.0/ =.05. From Table IV, Appendix B, z.05 =.645. The 90% confidence interval i: σ σ ( x x ) ± zα / + ( ) ± n n ±.45 (.6,.74) We are 90% confident that the difference in mean ervice-rating core between male and female. b. Becaue 0 fall in the 90% confidence interval, we are 90% confident that there i no difference in the mean ervice-rating core between male and female. 04 Chapter 7

5 7.6 a. The decriptive tatitic are: Decriptive Statitic: US, Japan Variable N Mean Median TrMean StDev SE Mean US Japan Variable Minimum Maximum Q Q3 US Japan ( n ) + ( n ) (5 ).7 + (5 ).7 p = = n + n 5+ 5 =.4933 To determine if the mean annual percentage turnover for U.S. plant exceed that for Japanee plant, we tet: H 0 : μ μ = 0 H a : μ μ > 0 The tet tatitic i t = ( x x) D0 ( ) 0 = = p + + n n 5 5 The rejection region require α =.05 in the upper tail of the t-ditribution with df = n + n = = 8. From Table VI, Appendix B, t.05 =.860. The rejection region i t >.860. Since the oberved value of the tet tatitic fall in the rejection region (t = 4.46 >.860), H 0 i rejected. There i ufficient evidence to indicate the mean annual percentage turnover for U.S. plant exceed that for Japanee plant at α =.05. b. The p-value = P(t 4.456). Uing Table VI, Appendix B, with df = n + n = = 8,.005 < P(t 4.456) <.00. Since the p-value i o mall, there i evidence to reject H 0 for α >.005. c. The neceary aumption are:. Both ampled population are approximately normal.. The population variance are equal. 3. The ample are randomly and independently ampled. There i no indication that the population are not normal. Both ample variance are imilar, o there i no evidence the population variance are unequal. There i no indication the aumption are not valid. Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 05

6 7.8 Let μ = the mean relational intimacy core for participant in the CMC group and μ = the mean relational intimacy core for participant in the FTF group. Uing MINITAB, the decriptive tatitic are: Decriptive Statitic: CMC, FTF Variable N N* Mean SE Mean StDev Minimum Q Median Q3 Maximum CMC FTF Some preliminary calculation are: p ( n ) + ( n ) ( ) + ( ) = = n + n 4+ 4 = To determine if the mean relational intimacy core for participant in the CMC group i lower than the mean relational intimacy core for participant in the FTF group, we tet: H 0 : μ μ = 0 H a : μ μ < 0 The tet tatitic i ( x x ) D ( ) o t = = = = p + + n n 4 4 The rejection region require α=.0 in the lower tail of the t-ditribution with df = n + n = = 46. From Table VI, Appendix B, t The rejection region i t <.303. Since the oberved value of the tet tatitic doe not fall in the rejection region (t =.0 /.303), H 0 i not rejected. There i inufficient evidence to indicate that the mean relational intimacy core for participant in the CMC group i lower than the mean relational intimacy core for participant in the FTF group at α = a. The firt population i the et of repone for all buine tudent who have acce to lecture note and the econd population i the et of repone for all buine tudent not having acce to lecture note. 06 Chapter 7

7 b. To determine if there i a difference in the mean repone of the two group, we tet: H 0 : μ μ = 0 H a : μ μ 0 ( x x) 0 ( ) 0 The tet tatitic i z = = n n =.9 The rejection region require α/ =.0/ =.005 in each tail of the z-ditribution. From Table IV, Appendix B, z.005 =.58. The rejection region i z <.58 or z >.58. Since the oberved value of the tet tatitic doe not fall in the rejection region (z =.9 >/.58), H 0 i not rejected. There i inufficient evidence to indicate a difference in the mean repone of the two group at α =.0. c. For confidence coefficient.99, α =.0 and α/ =.0/ =.005. From Table IV, Appendix B, z.005 =.58. The confidence interval i: ( x x ) ± z.005 n + n ( ) ± ±.80 (.,.48) We are 99% confident that the difference in the mean repone between the two group i between. and.48. d. A 95% confidence interval would be maller than the 99% confidence interval. The z value ued in the 95% confidence interval i z.05 =.96 compared with the z value ued in the 99% confidence interval of z.005 = a. The bacteria count are probably normally ditributed becaue each count i the median of five meaurement from the ame pecimen. b. Let μ = mean of the bacteria count for the dicharge and μ = mean of the bacteria count uptream. Since we want to tet if the mean of the bacteria count for the dicharge exceed the mean of the count uptream, we tet: H 0 : μ μ = 0 H a : μ μ > 0 c. Uing MINITAB, the decriptive tatitic are: Decriptive Statitic: Plant, Uptream Variable N Mean Median TrMean StDev SE Mean Plant Uptream Variable Minimum Maximum Q Q3 Plant Uptream Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 07

8 ( n ) + ( n ) (6 )3.9 + (6 ).355 p = = n + n 6+ 6 = 7.86 The tet tatitic i ( x x) 0 ( ) 0.53 t = = 7.86 p + + n n 6 6 = No α level wa given, o we will ue α =.05. The rejection region require α =.05 in the upper tail of the t-ditribution with df = n + n = = 0. From Table VI, Appendix B, t.05 =.8. The rejection region i t >.8. Since the oberved value of the tet tatitic doe not fall in the rejection region (t =.53 >/.8), H 0 i not rejected. There i inufficient evidence to indicate the mean bacteria count for the dicharge exceed the mean of the count uptream at α =.05. d. We mut aume:. The mean count per pecimen for each location i normally ditributed.. The variance of the ditribution are equal. 3. Independent and random ample were elected from each population. 7.4 a. We cannot make inference about the difference between the mean alarie of male and female accounting/finance/banking profeional becaue no tandard deviation are provided. b. To determine if the mean alary for male i ignificantly greater than that for female, we tet: H 0 : μ μ = 0 H a : μ μ > 0 The rejection region require α =.05 in the upper tail of the z-ditribution. From Table IV, Appendix B, z.05 =.645. To make thing eaier, we will aume that the tandard deviation for the group are the ame. The tet tatitic i ( ) ( ) x x D o 69,484 5,0 0 7,836 47, z = = = = σ ( ) σ σ σ + σ + n n 08 Chapter 7

9 In order to reject H 0 thi tet tatitic mut fall in the rejection region, or be greater than.645. Solving for σ we get: 47, , z = >.645 σ < = 86, σ.645 Thu, to reject H 0 the average of the two tandard deviation ha to be le than $86, c. Ye. In fact, reaonable value for the tandard deviation will be around $5,000. which i much maller than the required $86, d. Thee data were collected from voluntary ubject who reponded to a Web-baed urvey. Thu, thi i not a random ample, but a elf-elected ample. Generally, ubject who repond to urvey tend to have very trong opinion, which may not be the ame a the population in general. Thu, the reult from thi elf-elected ample may not reflect the reult from the population in general. 7.6 a. Pair Difference d = d = n d di = nd 6 = n d di i= di nd n i= nd i= d = () 34 6 = 5 b. μ d = μ μ c. For confidence coefficient.95, α =.05 and α/ =.05. From Table VI, Appendix B, with df = n D = 6 = 5, t.05 =.57. The confidence interval i: d d ± tα / =.57 ±.484 (.56, 3.484) n 6 d Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 09

10 d. H 0 : μ d = 0 H a : μ d 0 The tet tatitic i t = d t = = = 3.46 n / 6 d d The rejection region require α/ =.05/ =.05 in each tail of the t-ditribution with df = n D = 6 = 5. From Table VI, Appendix B, t.05 =.57. The rejection region i t <.57 or t >.57. Since the oberved value of the tet tatitic fall in the rejection region (3.46 >.57), H 0 i rejected. There i ufficient evidence to indicate that the mean difference i different from 0 at α = a. H 0 : μ μ = 0 H a : μ μ < 0 The rejection region require α =.0 in the lower tail of the z-ditribution. From Table IV, Appendix B, z.0 =.8. The rejection region i z <.8. b. H 0 : μ μ = 0 H a : μ μ < 0 The tet tatitic i d z = = = 4.7. d n 38 d The rejection region i z <.8 (Refer to part a.) Since the oberved value of the tet tatitic fall in the rejection region (z = 4.7 <.8), H 0 i rejected. There i ufficient evidence to indicate μ μ < 0 at α =.0. c. Since the ample ize of the number of pair i greater than 30, we do not need to aume that the population of difference i normal. The ampling ditribution of d i approximately normal by the Central Limit Theorem. We mut aume that the difference are randomly elected. d. For confidence coefficient.90, α =.0 and α/ =.0/ =.05. From Table IV, Appendix B, z.05 =.645. The 90% confidence interval i: d d ± z ± ±.3 ( 4.73,.77) n 38 d e. The confidence interval provide more information ince it give an interval of poible value for the difference between the population mean. 0 Chapter 7

11 7.30 a. Let μ = the mean alary of technology profeional in 003 and μ = the mean alary of technology profeional in 005. Let μ d = μ - μ. b. To determine if the mean alary of technology profeional at all U.S. metropolitan area ha increaed between 003 and 005, we tet: H 0 : μ μ = 0 H 0 : μ d = 0 OR H a : μ μ < 0 H a : μ d < 0 Metro Area 003 Salary ($ thouand) 005 Salary ($ thouand) Difference ( ) Silicon Valley New York Wahington, D.C Lo Angele Denver Boton Atlanta Chicago Philadelphia San Diego Seattle Dalla-Ft. Worth Detroit n d di 5.7 c. d = = =.977 n 3 d d n d n d di d ( 5.7) i nd = = 3 =.989 n 3 d d d = =.989 = d. The tet tatitic i d μo t = = =.978 n d d e. The rejection region require α =.0 in the lower tail of the t-ditribution with df = n d = 3 = 4. From Table VI, Appendix B, t.0 =.345. The rejection region i t <.345. Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei

12 f. Since the oberved value of the tet tatitic fall in the rejection region (t =.978 <.345), H 0 i rejected. There i ufficient evidence to indicate the mean alary of technology profeional at all U.S. metropolitan area ha increaed between 003 and 005 at α =.0. g. In order for the inference to be valid, we mut aume that the population of difference i normal and that we have a random ample. Uing MINITAB, the hitogram of the difference i: Hitogram of Diff Frequency Diff The graph i fairly mound-haped although it i omewhat kewed to the right. Since there are only 3 obervation, thi graph i cloe enough to being mound-haped to indicate the normal aumption i reaonable. 7.3 a. The data hould be analyzed a a paired difference experiment becaue each actor who won an Academy Award wa paired with another actor with imilar characteritic who did not win the award. b. Let μ = mean life expectancy of Academy Award winner and μ = mean life expectancy of non-academy Award winner. To compare the mean life expectancie of Academy Award winner and non-winner, we tet: H 0 : μ μ = μ d = 0 H a : μ d 0 c. Since the p-value wa o mall, there i ufficient evidence to indicate the mean life expectancie of the Academy Award winner and non-winner are different for any value of α >.003. Since the ample mean life expectancy of Academy Award winner i greater than that for non-winner, we can conclude that Academy Award winner have a longer mean life expectancy than non-winner. Chapter 7

13 7.34 a. Let μ = mean driver chet injury rating and μ = mean paenger chet injury rating. Becaue the data are paired, we are intereted in μ μ = μ d, the difference in mean chet injury rating between driver and paenger. b. The data were collected a matched pair and thu, mut be analyzed a matched pair. Two rating are obtained for each car the driver chet injury rating and the paenger chet injury rating. c. Uing MINITAB, the decriptive tatitic are: Decriptive Statitic: DrivCht, PaCht, diff Variable N Mean Median TrMean StDev SE Mean DrivCht PaCht diff Variable Minimum Maximum Q Q3 DrivCht PaCht diff For confidence coefficient.99, α =.0 and α/ =.0/ =.005. From Table IV, Appendix B, z.005 =.58. The 99% confidence interval i: d 5.57 d ± z ± ±.438 (.999, 0.877) n 98 d d. We are 99% confidence that the difference between the mean chet injury rating of driver and front-eat paenger i between.999 and Since 0 i in the confidence interval, there i no evidence that the true mean driver chet injury rating exceed the true mean paenger chet injury rating. e. Since the ample ize i large, the ampling ditribution of d i approximately normal by the Central Limit Theorem. We mut aume that the difference are randomly elected a. Let μ C = mean relational intimacy core for the CMC group on the firt meeting and μ C3 = mean relational intimacy core for the CMC group on the third meeting. Let μ Cd = difference in mean relational intimacy core between the firt and third meeting for the CMC group. To determine if the mean relational intimacy core will increae between the firt and third meeting, we tet: H o : μ Cd = 0 H a : μ Cd < 0 b. The reearcher ued the paired t-tet becaue the ame individual participated in each of the three meeting eion. Thu, the ample would not be independent. c. Since the p-value i o mall (p =.003), H 0 would be rejected. There i ufficient evidence to indicate that the mean relational intimacy core for participant in the CMC group increaed from the firt to the third meeting for any value of α >.003. Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 3

14 d. Let μ F = mean relational intimacy core for the FTF group on the firt meeting and μ F3 = mean relational intimacy core for the FTF group on the third meeting. Let μ Fd = difference in mean relational intimacy core between the firt and third meeting for the FTF group. To determine if the mean relational intimacy core will change between the firt and third meeting, we tet: H 0 : μ Fd = 0 H a : μ Fd 0 e. Since the p-value i not mall (p =.39), H 0 would be not be rejected. There i inufficient evidence to indicate that the mean relational intimacy core for participant in the FTF group changed from the firt to the third meeting for any value of α < Uing MINITAB, the decriptive tatitic are: Decriptive Statitic: Method, Method, Diff Variable N N* Mean SE Mean StDev Minimum Q Median Q3 Maximum Method Method Diff To determine if the mean tranition error for method differ from the mean tranition error for method, we tet: H 0 : μ μ = 0 H 0 : μ d = 0 OR H a : μ μ 0 H a : μ d 0 The tet tatitic i d μo t = = = 0.5 n d d The rejection region require α/ =.0/ =.05 in each tail of the t-ditribution with df = n d = 0 = 9. From Table VI, Appendix B, t.05 =.833. The rejection region i t <.833 or t >.833. Since the oberved value of the tet tatitic doe not fall in the rejection region (t = 0.5 >/.833), H0 i not rejected. There i inufficient evidence to indicate the mean tranition error for method differ from the mean tranition error for method at α = Uing MINITAB, the decriptive tatitic are: Decriptive Statitic: HMETER, HSTATIC, Diff Variable N N* Mean SE Mean StDev Minimum Q HMETER HSTATIC Diff Variable Median Q3 Maximum HMETER HSTATIC Diff Chapter 7

15 For confidence coefficient.95, α =.05 and α/ =.05/ =.05. From Table VI, Appendix B, with df = n d = 40 = 39, t The 95% confidence interval i: d d ± t ± ± n 40 ( , ) We are 95% confident that the true difference in mean denity meaurement between the two method i between and Since the abolute value of thi interval i completely le than the deired maximum difference of.00, the winery hould chooe the alternative method of meauring wine denity. 7.4 a. The rejection region require α =.0 in the lower tail of the z-ditribution. From Table IV, Appendix B, z.0 =.33. The rejection region i z <.33. b. The rejection region require α =.05 in the lower tail of the z-ditribution. From Table IV, Appendix B, z.05 =.96. The rejection region i z <.96. c. The rejection region require α =.05 in the lower tail of the z-ditribution. From Table IV, Appendix B, z.05 =.645. The rejection region i z <.645. d. The rejection region require α =.0 in the lower tail of the z-ditribution. From Table IV, Appendix B, z.0 =.8. The rejection region i z < For confidence coefficient.95, α =.95 =.05 and α/ =.05/ =.05. From Table IV, Appendix B, z.05 =.96. The 95% confidence interval for p p i approximately: pˆˆ q pˆˆ q ˆ ˆ ± + (.65.58) ±.96 a. ( p p ) zα / n n (.003,.37).65(.65).58(.58) pˆˆ q pˆˆ q ˆ ˆ ± + (.3.5).96 b. ( p p ) zα / n n.06 ±.086 (.06,.46).3(.3).5(.5) pˆˆ q pˆˆ q ˆ ˆ ± + (.46.6) ±.96 c. ( p p ) zα / n n.5 ±.3 (.8,.09).46(.46).6(.6) npˆ + npˆ 55(.7) + 65(.6) ˆp = = = n + n =.65 ˆq = ˆp =.65 =.35 H 0 : p p = 0 H a : p p > 0 Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 5

16 ( pˆ pˆ) 0 (.7.6) 0. The tet tatitic i z = = = pq ˆˆ +.65(.35) + n n =.4 The rejection region require α =.05 in the upper tail of the z-ditribution. From Table IV, Appendix B, z.05 =.645. The rejection region i z >.645. Since the oberved value of the tet tatitic doe not fall in the rejection region (z =.4 >/.645), H 0 i not rejected. There i inufficient evidence to indicate the proportion from population i greater than that for population at α = a. Let p = proportion of men who prefer to keep track of appointment in their head and p = proportion of women who prefer to keep track of appointment in their head. To determine if the proportion of men who prefer to keep track of appointment in their head i greater than that of women, we tet: H 0 : p p = 0 H a : p p > 0 npˆ + npˆ 500(.56) + 500(.46) pˆ = = =.5 and qˆ = pˆ =.5 =.49 n + n b. ( pˆ pˆ) 0 (.56.46) 0 The tet tatitic i z = = pq ˆˆ +.5(.49) + n n = 3.6 c. The rejection region require α =.0 in the upper tail of the z ditribution. From Table IV, Appendix B, z.0 =.33. The rejection region i z >.33. d. The p-value i p = P(z 3.6).5.5 = 0. e. Since the oberved value of the tet tatitic fall in the rejection region (z = 3.6 >.33), H 0 i rejected. There i ufficient evidence to indicate the proportion of men who prefer to keep track of appointment in their head i greater than that of women at α = a. Let p = proportion of cutomer returning the printed urvey and p = proportion of cutomer returning the electronic urvey. Some preliminary calculation are: x 6 x 55 = = = pˆ = = =.374 n 44 pˆ.44 n 63 For confidence coefficient.90, α =.0 and α/ =.0/ =.05. From Table IV, Appendix B, z.05 =.645. The 90% confidence interval i: 6 Chapter 7

17 pq ˆˆ pq ˆˆ.44(.586).374(.66) pˆ pˆ ± z + ± ( ).05 ( ).645 n n.04 ±.05 (.0,.09) We are 90% confidence that the difference in the repone rate for the two type of urvey i between.0 and.09. b. Since the value.05 fall in the 90% confidence interval, it i not an unuual value. Thu, there i no evidence that the difference in repone rate i different from.05. The reearcher would be able to make thi inference. 7.5 a. Let p = proportion of manager and profeional who are male and p = proportion of part-time MBA tudent who are male. To ee if the ample are ufficiently large: pq pq ˆˆ (.95)(0.5) pˆ ± 3σ pˆ ± 3 pˆ ± 3.95± 3 6 pˆ n n.95 ±.05 (.90,.00) pq pq ˆ ˆ (.689)(.3) pˆ ± 3σ pˆ ± 3 pˆ ± 3.95± 3 09 pˆ n n.689 ±.33 (.556,.8) Since both interval are contained within the interval (0, ), the normal approximation will be adequate. Firt, we calculate the overall etimate of the common proportion under H 0. pˆ npˆ + npˆ 6(.95) + 09(.689) n + n = = =.845 To determine if the population of manager and profeional conit of more male than the part-time MBA population, we tet: H 0 : p = p H a : p > p ( pˆ pˆ) 0 ( ) 0 The tet tatitic i z = = pq ˆˆ +.845(.55) + n 6 09 n = 5.8 The rejection region require α =.05 in the upper tail of the z-ditribution. From Table IV, Appendix B, z.05 =.645. The rejection region i z >.645. Since the oberved value of the tet tatitic fall in the rejection (z = 5.8 >.645), H 0 i rejected. There i ufficient evidence to indicate that population of manager and profeional conit of more male than the part-time MBA population at α =.05. Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 7

18 b. We had to aume:. Both ample were randomly elected. Both ample ize are ufficiently large. c. Firt, we calculate the overall etimate of the common proportion under H 0. pˆ npˆ + npˆ 6(.9) + 09(.534) n + n = = =.760 To determine if the population of manager and profeional conit of more married individual than the part-time MBA population, we tet: H 0 : p = p H a : p > p ( pˆ pˆ) 0 (.9.534) 0 The tet tatitic i z = = pq ˆˆ +.760(.40) + n 6 09 n = 7.4 The rejection region require α =.0 in the upper tail of the z-ditribution. From Table IV, Appendix B, z.0 =.33. The rejection region i z >.33. Since the oberved value of the tet tatitic fall in the rejection (z = 7.4 >.33), H 0 i rejected. There i ufficient evidence to indicate that population of manager and profeional conit of more married individual than the part-time MBA population at α =.0. d. We had to aume:. Both ample were randomly elected. Both ample ize are ufficiently large Let p = accuracy rate for module with correct code and p = accuracy rate for module with defective code. Some preliminary calculation are: x 400 x 0 p ˆ = = =.89 p ˆ = = =. 408 n 449 n 49 8 Chapter 7

19 For confidence coefficient.99, α =.0 and α/ =.0/ =.005. From Table IV, Appendix B, z.005 =.58. The 99% confidence interval i: ˆˆ ˆˆ.89(.09).408(.59) ˆ ˆ ± pq + pq ± ( p p ) z.005 ( ).58 n n.483 ±.85 (.98,.668) We are 99% confident that the difference in accuracy rate between module with correct code and module with defective code i between.98 and a. Let p = proportion of all children who recognize Joe Camel. ˆp = x = n =.735 ˆq = ˆp =.735 =.65 To ee if the ample i ufficiently large: ˆp ± 3σ p ˆ ˆp ± 3 pq n ˆp ± ˆˆ 3 pq n.735 ±.735(.65) ±.45 (.590,.880) Since the interval lie within the interval (0, ), the normal approximation will be adequate. For confidence coefficient.95, α =.05 and α/ =.05/ =.05. From Table IV, Appendix B, z.05 =.96. The 95% confidence interval i: ˆp ± z.05 pq ˆˆ n.735 ± (.65) ±.095 (.640,.830) We are 95% confident that the proportion of all children who recognize Joe Camel i between.640 and.830. b. Let p = proportion of children under the age of 6 who recognize Joe Camel and p = proportion of children age 6 and over who recognize Joe Camel. x 5 ˆp = n = 8 =.536 ˆq = ˆp =.536 =.464 x 46 ˆp = n = 55 =.836 ˆq = ˆp =.836 =.64 Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 9

20 To ee if the ample are ufficiently large: ˆp ± ˆp ± 3σ p ˆ 3σ p ˆ ˆp ± 3 p q n ˆp ± 3 p q n ˆˆ ˆp ± 3 p q n.536 \± 3.536(.464) ±.83 (.53,.89) ˆ ˆ ˆp ± 3 p q.836(.64).836 ± 3 n ±.50 (.686,.986) Since both interval lie within the interval (0, ), the normal approximation will be adequate. To determine if the recognition of Joe Camel increae with age, we tet: H 0 : p p = 0 H a : p p < 0 ( pˆ pˆ) 0 ( ) 0 The tet tatitic i z = = pq ˆˆ +.735(.65) + n 8 55 n =.93 The rejection region require α =.05 in the lower tail of the z-ditribution. From Table IV, Appendix B, z.05 =.645. The rejection region i z <.645. Since the oberved value of the tet tatitic fall in the rejection region (z =.93 <.645), H 0 i rejected. There i ufficient evidence to indicate that the recognition of Joe Camel increae with age at α = a. For confidence coefficient.95, α =.95 =.05 and α/ =.05/ =.05. From Table IV, Appendix B, z.05 =.96. ( z ) ( ) α / σ σ n = n = ME + (.96) (5 + 7 ) = = b. If the range of each population i 40, we would etimate σ by: σ 60/4 = 5 For confidence coefficient.99, α =.99 =.0 and α/ =.0/ =.005. From Table IV, Appendix B, z.005 =.58. ( z ) ( ) α / σ σ n = n = ME + (.58) (5 + 5 ) = = Chapter 7

21 c. For confidence coefficient.9, α =.9 =. and α/ =./ =.05. From Table IV, Appendix B, z.05 =.645. For a width of, the bound i.5. ( z ) ( ) α / σ σ n = n = ME + (.645) ( ) = = Firt, find the ample ize needed for width 5, or margin of error.5. For confidence coefficient.9, α =.9 =. and α/ =./ =.05. From Table IV, Appendix B, z.05 =.645. ( z ) ( ) α / σ σ n = n = ME + (.645) (0 + 0 ) = = Thu, the neceary ample ize from each population i 87. Therefore, ufficient fund have not been allocated to meet the pecification ince n = n = 00 are large enough ample. 7.6 For confidence coefficient.95, α =.05 and α/ =.05/ =.05. From Table IV, Appendix B, z.05 =.96. ( ) ( z ) α / σ σ n = n = + + = ( ME).5.96 ( ) = We would need to ample 7 pecimen from each location For confidence coefficient.90, α =.90 =.0 and α/ =.0/ =.05. From Table IV, Appendix B, z.05 =.645. Since no information i given about the value of p and p, we will be conervative and ue.5 for both. A width of.04 mean the bound i.04/ =.0. n = n = ( z ) ( ) α ( p q + p q ).645.5(.5) +.5(.5) / ( ME).0 = = 3,38.5 3, a. For confidence coefficient.80, α =.80 =.0 and α/ =.0/ =.0. From Table IV, Appendix B, z.0 =.8. Since we have no prior information about the proportion, we ue p = p =.5 to get a conervative etimate. For a width of.06, the margin of error i.03. n = n = ( z ) ( ) α ( pq + p q ) (.8).5(.5) +.5(.5) / = ME.03 = b. For confidence coefficient.90, α =.90 =.0 and α/ =.0/ =.05. From Table IV, Appendix B, z.05 =.645. Uing the formula for the ample ize needed to etimate a proportion from Chapter 7, n ( z ) pq ( ).645.5(.5) α / = = = = ME No, the ample ize from part a i not large enough. Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei

22 7.68 For confidence coefficient.95, α =.95 =.05 and α/ =.05. From Table IV, Appendix B, z.05 =.96. ( ) ( z ) α / σ σ n = n = +.96 ( ) = = ( ME) a. With ν = and ν = 30, P(F 5.39) =.0 (Table XI, Appendix B) b. With ν = 4 and ν = 0, P(F.74) =.05 (Table IX, Appendix B) Thu, P(F <.74) = P(F.74) =.05 =.95. c. With ν = 7 and ν =, P(F 36.8) =.05 (Table VIII, Appendix B) Thu, P(F < 36.8) = P(F 36.8) =.05 =.95. d. With ν = 40 and ν = 40, P(F >.) =.0 (Table XI, Appendix B) 7.7 To tet H 0 : σ = σ againt H a : σ σ, the rejection region i F > F α/ with ν = 0 and ν =. a. α =.0, α/ =.0 Reject H 0 if F > F.0 =.9 (Table VIII, Appendix B) b. α =.0, α/ =.05 Reject H 0 if F > F.05 =.75 (Table IX, Appendix B) c. α =.05, α/ =.05 Reject H 0 if F > F.05 = 3.37 (Table X, Appendix B) d. α =.0, α/ =.0 Reject H 0 if F > F.0 = 4.30 (Table XI, Appendix B) 7.74 a. To determine if a difference exit between the population variance, we tet: H 0 : H a : σ σ = σ σ The tet tatitic i F = 8.75 = = Chapter 7

23 The rejection region require α/ =.0/ =.05 in the upper tail of the F-ditribution with ν = n = 7 = 6 and ν = n = =. From Table IX, Appendix B, F The rejection region i F >.60. Since the oberved value of the tet tatitic doe not fall in the rejection region (F =.6 >/.60), H 0 i not rejected. There i inufficient evidence to indicate a difference between the population variance. b. The p-value i P(F.6). From Table VIII and IX, with ν = 6 and ν =, (.05) < P(F.6) < (.0).0 < P(F.6) <.0 There i no evidence to reject H 0 for α Let σ = variance of carat ize for diamond certified by GIA, diamond certified by HRD, and σ = variance of carat ize for σ = variance of carat ize for diamond certified by IGI. 3 a. To determine if the variation in carat ize differ for diamond certified by GIA and diamond certified by HRD, we tet: H 0 : H a : σ = σ σ σ The tet tatitic i Larger ample variance.456 F = = = =.799 Smaller ample variance.83 The rejection region require α/ =.05/ =.05 in the upper tail of the F-ditribution with ν = n = 5 = 50 and ν = n = 79 = 78. From Table X, Appendix B, F The rejection region i F >.43. Since the oberved value of the tet tatitic fall in the rejection region (F =.799 >.43), H 0 i rejected. There i ufficient evidence to indicate the variation in carat ize differ for diamond certified by GIA and thoe certified by HRD at α =.05. b. To determine if the variation in carat ize differ for diamond certified by GIA and diamond certified by IGI, we tet: H 0 : H a : σ = σ σ σ 3 3 The tet tatitic i Larger ample variance.456 F = = = =.89 Smaller ample variance.63 3 The rejection region require α/ =.05/ =.05 in the upper tail of the F-ditribution with ν = n = 5 = 50 and ν = n 3 = 78 = 77. From Table X, Appendix B, F The rejection region i F >.43. Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 3

24 Since the oberved value of the tet tatitic doe not fall in the rejection region (F =.89 >/.43), H 0 i not rejected. There i inufficient evidence to indicate the variation in carat ize differ for diamond certified by GIA and thoe certified by IGI at α =.05. c. To determine if the variation in carat ize differ for diamond certified by HRD and diamond certified by IGI, we tet: H 0 : H a : σ = σ σ σ 3 3 The tet tatitic i Larger ample variance.63 F = = = =.396 Smaller ample variance.83 3 The rejection region require α/ =.05/ =.05 in the upper tail of the F-ditribution with ν = n 3 = 78 = 77 and ν = n = 79 = 78. From Table X, Appendix B, F The rejection region i F >.67. Since the oberved value of the tet tatitic doe not fall in the rejection region (F =.396 >/.67), H 0 i not rejected. There i inufficient evidence to indicate the variation in carat ize differ for diamond certified by HRD and thoe certified by IGI at α =.05. d. We will look at the 4 method for determining if the data are normal. Firt, we will look at hitogram of the data. Uing MINITAB, the hitogram of the carat ize for the 3 certification bodie are: Percent 0 0 Percent GIA HRD Percent IGI From the hitogram, none of the data appear to be mound-haped. It appear that none of the data et are normal. 4 Chapter 7

25 Next, we look at the interval x ±, x ±, x ± 3. If the proportion of obervation falling in each interval are approximately.68,.95, and.00, then the data are approximately normal. Uing MINITAB, the ummary tatitic are: Decriptive Statitic: GIA, IGI, HRD Variable N Mean Median TrMean StDev SE Mean GIA IGI HRD Variable Minimum Maximum Q Q3 GIA IGI HRD For GIA: x ±.673 ±.456 (.467,.979) 84 of the 5 value fall in thi interval. The proportion i.56. Thi i much maller than the.68 we would expect if the data were normal. x ±.673 ± (.456).673 ±.49 (.8,.635) 5 of the 5 value fall in thi interval. The proportion i.00. Thi i much larger than the.95 we would expect if the data were normal. x ± ± 3(.456).673 ±.7368 (.0645,.409) 5 of the 5 value fall in thi interval. The proportion i.00. Thi i the ame a the.00 we would expect if the data were normal. From thi method, it appear that the data are not normal. For IGI: x ±.3665 ±.63 (.50,.588) 69 of the 78 value fall in thi interval. The proportion i.88. Thi i much larger than the.68 we would expect if the data were normal. x ±.3665 ± (.63).3665 ±.436 (.066,.799) 74 of the 78 value fall in thi interval. The proportion i.95. Thi i the ame a the.95 we would expect if the data were normal. x ± ± 3(.63).3665 ±.6489 (.84,.054) 78 of the 78 value fall in thi interval. The proportion i.00. Thi i the ame a the.00 we would expect if the data were normal. From thi method, it appear that the data are not normal. Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 5

26 For HRD: x ±.89 ±.83 (.698,.9960) 30 of the 79 value fall in thi interval. The proportion i.38. Thi i much maller than the.68 we would expect if the data were normal. x ±.89 ± (.83).89 ±.366 (.4467,.79) 79 of the 79 value fall in thi interval. The proportion i.00. Thi i much larger than the.95 we would expect if the data were normal. x ± 3.89 ± 3(.83).89 ±.5493 (.636,.36) 79 of the 79 value fall in thi interval. The proportion i.00. Thi i the ame a the.00 we would expect if the data were normal. From thi method, it appear that the data are not normal. Next, we look at the ratio of the IQR to. For GIA: IQR = Q U Q L =..3 =.8. IQR.8 = = 3.6 Thi i much larger than the.3 we would expect if the data were.456 normal. Thi method indicate the data are not normal. For IGI: IQR = Q U Q L = =.83. IQR.83 = = 3.84 Thi i much larger than the.3 we would expect if the data were.63 normal. Thi method indicate the data are not normal. For HRD: IQR = Q U Q L = =.59. IQR.59 = = 3. Thi i much larger than the.3 we would expect if the data were.83 normal. Thi method indicate the data are not normal. 6 Chapter 7

27 Finally, uing MINITAB, the normal probability plot for GIA i: Normal Probability Plot for GIA ML Etimate - 95% CI ML Etimate Percent Mean StDev Goodne of Fit AD* Data Since the data do not form a traight line, the data are not normal. Uing MINITAB, the normal probability plot for IGI i: Normal Probability Plot for IGI ML Etimate - 95% CI Percent ML Etimate Mean StDev Goodne of Fit AD* Data Since the data do not form a traight line, the data are not normal. Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 7

28 Uing MINITAB, the normal probability plot for HRD i: Normal Probability Plot for HRD ML Etimate - 95% CI Percent ML Etimate Mean 0.89 StDev Goodne of Fit AD* Data Since the data do not form a traight line, the data are not normal. From the 4 different method, all indication are that the carat ize data are not normal for any of the certification bodie a. The amount of variability of GHQ core tell u how imilar or different the member of the group are on GHQ core. The larger the variability, the larger the difference are among the member on the GHQ core. The maller the variability, the maller the difference are among the member on the GHQ core. b. Let σ = variance of the mental health core of the employed and σ = variance of the mental health core of the unemployed. To determine if the variability in mental health core differ for employed and unemployed worker, we tet: H 0 : σ = σ H a : σ σ c. The tet tatitic i F = Larger ample variance 5.0 = = Smaller ample variance 3.6 =.45 The rejection region require α/ =.05/ =.05 in the upper tail of the F-ditribution with ν = n = 49 = 48 and ν = n = 4 = 4. From Table XI, Appendix B, F The rejection region i F >.6. Since the oberved value of the tet tatitic fall in the rejection region (F =.45 >.6), H 0 i rejected. There i ufficient evidence to indicate that the variability in mental health core differ for employed and unemployed worker for α = Chapter 7

29 d. We mut aume that the population of mental health core are normally ditributed. We mut alo aume that we elected independent random ample Let σ = variance zinc meaurement from the text-line, from the witne-line, and 3 MINITAB, the decriptive tatitic are: σ = variance zinc meaurement σ = variance zinc meaurement from the interection. Uing Decriptive Statitic: Text-line, Witne-line, Interection Variable N Mean Median TrMean StDev SE Mean Text-lin Witne Interec Variable Minimum Maximum Q Q3 Text-lin Witne Interec a. To determine if the variation in the zinc meaurement for the text-line and the interection differ, we tet: H 0 : H a : σ = σ σ σ 3 3 The tet tatitic i Larger ample variance.053 F = = = =.437 Smaller ample variance The rejection region require α/ =.05/ =.05 in the upper tail of the F-ditribution with ν = n = 3 = and ν = n 3 = 5 = 4. From Table X, Appendix B, F.05 = The rejection region i F > Since the oberved value of the tet tatitic doe not fall in the rejection region (F =.437 >/ 0.65), H 0 i not rejected. There i inufficient evidence to indicate the variation in the zinc meaurement for the text-line and the interection differ at α =.05. b. To determine if the variation in the zinc meaurement for the witne-line and the interection differ, we tet: H 0 : H a : σ = σ σ σ 3 3 The tet tatitic i Larger ample variance.05 F = = = = 5.50 Smaller ample variance The rejection region require α/ =.05/ =.05 in the upper tail of the F-ditribution with ν = n = 6 = 5 and ν = n 3 = 5 = 4. From Table X, Appendix B, F.05 = The rejection region i F > Since the oberved value of the tet tatitic doe not fall in the rejection region (F = 5.50 >/ 9.36), H 0 i not rejected. There i inufficient evidence to indicate the variation in the zinc meaurement for the witne-line and the interection differ at α =.05. Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 9

30 c. There i no indication that the variance of the zinc meaurement for three location differ. d. With only 3, 6, and 5 meaurement, it i very difficult to check the aumption. 7.8 Uing MINITAB, ome preliminary calculation are: Decriptive Statitic: HEATRATE Variable ENGINE N N* Mean SE Mean StDev Minimum Q Median HEATRATE Advanced Aeroderiv Traditional Variable ENGINE Q3 Maximum HEATRATE Advanced Aeroderiv Traditional a. To determine if the heat rate variance for traditional and aeroderivative augmented ga turbine differ, we tet: H 0 : H a : σ = σ σ 3 σ 3 89) The tet tatitic i Larger ample variance 65 F = = = = 4.99 Smaller ample variance 79 The rejection region require α/ =.05/ =.05 in the upper tail of the F ditribution with numerator df = ν = n = 7 = 6 and denominator df = ν 3 = n 3 = 39 = 38. From Table X, Appendix B, F The rejection region i F >.74. Since the oberved value of the tet tatitic fall in the rejection region (F = 4.99 >.74), H 0 i rejected. There i ufficient evidence to indicate the heat rate variance for traditional and aeroderivative augmented ga turbine differ at α =.05. Since the tet in Exercie 7.3 a aume that the population variance are the ame, the validity of the tet i upect ince we jut found the variance are different. b. To determine if the heat rate variance for advanced and aeroderivative augmented ga turbine differ, we tet: H 0 : H a : σ = σ σ σ 3 30 Chapter 7

31 Larger ample variance 65 The tet tatitic i F = = = = 7.4 Smaller ample variance 639 The rejection region require α/ =.05/ =.05 in the upper tail of the F ditribution with numerator df = ν = n = 7 = 6 and denominator df = ν = n = = 0. From Table X, Appendix B, F.05 = 3.3. The rejection region i F > 3.3. Since the oberved value of the tet tatitic fall in the rejection region (F = 7.4 > 3.3), H 0 i rejected. There i ufficient evidence to indicate the heat rate variance for advanced and aeroderivative augmented ga turbine differ at α =.05. Since the tet in Exercie 7.3 b aume that the population variance are the ame, the validity of the tet i upect ince we jut found the variance are different a. The ample are randomly elected in an independent manner from the two population. The ample ize, n and n, are large enough o that x and x each have approximately normal ampling ditribution and o that and provide good approximation to σ and σ. Thi will be true if n 30 and n 30. b.. Both ampled population have relative frequency ditribution that are approximately normal.. The population variance are equal. 3. The ample are randomly and independently elected from the population. c.. The relative frequency ditribution of the population of difference i normal.. The ample of difference are randomly elected from the population of difference. d. The two ample are independent random ample from binomial ditribution. Both ample hould be large enough o that the normal ditribution provide an adequate approximation to the ampling ditribution of ˆp and ˆp. e. The two ample are independent random ample from population which are normally ditributed a. H 0 : σ = σ H a : σ σ The tet tatitic i F = Larger ample variance 0. = = = Smaller ample variance The rejection region require α/ =.05/ =.05 in the upper tail of the F-ditribution with numerator df ν = n = 5 = 4 and denominator df ν = n = 0 = 9. From Table XI, Appendix B, F The rejection region i F >.66. Since the oberved value of the tet tatitic fall in the rejection region (F = 3.84 >.66), H 0 i rejected. There i ufficient evidence to conclude σ σ at α =.05. Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 3

32 b. No, we hould not ue a mall ample t tet to tet H 0 : (μ μ ) = 0 againt H a : (μ μ ) 0 becaue the aumption of equal variance doe not eem to hold ince we concluded σ σ in part b Some preliminary calculation are: x 0 ˆp = n = 00 =.55; x 30 ˆp = n = 00 =.65; x + x pˆ = = = n + n a. H 0 : (p p ) = 0 H a : (p p ) < 0 ( pˆ pˆ) 0 (.55.65) 0.0 The tet tatitic i z = = =.049 pq ˆˆ +.6(.6) + n n =.04 The rejection region require α =.0 in the lower tail of the z-ditribution. From Table IV, Appendix B, z.0 =.8. The rejection region i z <.8. Since the oberved value of the tet tatitic fall in the rejection region (z =.04 <.8), H 0 i rejected. There i ufficient evidence to conclude (p p < 0) at α =.0. b. For confidence coefficient.95, α =.95 =.05 and α/ =.05/ =.05. From Table IV, Appendix B, z.05 =.96. The 95% confidence interval for (p p ) i approximately: pˆˆ q pˆˆ q ( pˆ ˆ p ) ± zα / + n n.55(.55).65(.65) (.55.65) ± ±.096 (.96,.004) c. From part b, z.05 =.96. Uing the information from our ample, we can ue p =.55 and p =.65. For a width of.0, the margin of error i.005. n = n = ( ) z ( ) α pq + p q + / ( ) (.96).55(.55).65(.65).8476 = = ME = ,99 3 Chapter 7

33 7.90 a. Let p = proportion of Opening Door tudent enrolled full time and p = proportion of traditional tudent enrolled full time. The target parameter for thi comparion i p p. b. Let μ = mean GPA of Opening Door tudent and μ = mean GPA of traditional tudent. The target parameter for thi comparion i μ μ. 7.9 Uing MINITAB, ome preliminary calculation are: Decriptive Statitic: Spillage Variable Caue N N* Mean SE Mean StDev Minimum Q Median Spillage Colliion Fire Grounding HullFail Unknown * 6.00 Variable Caue Q3 Maximum Spillage Colliion Fire Grounding HullFail Unknown * 7.00 a. Let μ = mean pillage for accident caued by colliion and μ = mean pillage for accident caued by fire/exploion. p ( n ) + ( n ) ( ) + ( ) = = = 4, n + n 0+ For confidence coefficient.90, α =.90 =.0 and α/ =.0/ =.05. From Table VI, Appendix B, with df = n + n = 0 + = 0, t.05 =.75. The confidence interval i: 5.7 ± 48.9 ( 4.59, 53.89) ( x x ) ± t.05 p + ( ) ± n n 0 b. Let μ 3 = mean pillage for accident caued by grounding and μ 4 = mean pillage for accident caued by hull failure. p ( n ) + ( n ) ( ) + ( ) = = =, n + n 4+ Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 33

34 To determine if the mean pillage amount for accident caued by grounding i different from the mean pillage amount caued by hull failure, we tet: H 0 : μ 3 μ 4 = 0 H a : μ 3 μ 4 0 The tet tatitic i ( x x ) D ( ) o t = = = = p + + n n 4 The rejection region require α/ =.05/ =.05 in each tail of the t-ditribution with df = n + n = 4 + = 4. From Table VI, Appendix B, t.05 =.064. The rejection region i t <.064 or t >.064. Since the oberved value of the tet tatitic doe not fall in the rejection region (t =.39 </.064), H0 i not rejected. There i inufficient evidence to indicate the mean pillage amount for accident caued by grounding i different from the mean pillage amount caued by hull failure at α =.05. c. The neceary aumption are: We mut aume that the ditribution from which the ample were elected are approximately normal, the ample are independent, and the variance of the two population are equal. Below are the tem-and-leaf plot for each of the ample: Stem-and-leaf of Spillage Caue = Colliion N = 0 Leaf Unit = 0 (6) Stem-and-leaf of Spillage Caue = Fire N = Leaf Unit = (4) Chapter 7

35 Stem-and-leaf of Spillage Caue = Grounding N = 4 Leaf Unit = (5) Stem-and-leaf of Spillage Caue = Hull Failure N = Leaf Unit = 0 (8) Baed on the hape of the tem-and-leaf plot, it doe not appear that the data are normally ditributed. Alo, we know that if the data are normally ditributed, then the Interquartile Range, IQR, divided by the tandard deviation hould be approximately.3. We will compute IQR/ for each of the ample: Colliion: IQR/ = ( ) / 70.4 =.95 Fire: IQR/ = ( ) / 60.7 =.79 Grounding: IQR/ = ( ) / 8.47 =.0 Hull Failure: IQR/ = ( ) / 56.4 =.9 Since all of thee ratio are quite a bit maller than.3, it indicate that none of the ample come from normal ditribution. Thu, it appear that the aumption of normal ditribution i violated. The ample tandard deviation are: Colliion: = 70.4 Fire: = 60.7 Grounding: = 8.47 Hull Failure: = 56.4 Without doing formal tet, it appear that the variance of the group Colliion, Fire, and Hull Failure are probably not ignificantly different. However, it appear that the variance for the grounding group i maller than the other. Inference Baed on Two Sample: Confidence Interval and Tet of Hypothei 35