Iterval Estimatio Learig Objectives 1. Kow how to costruct ad iterpret a iterval estimate of a populatio mea ad / or a populatio proportio.. Uderstad ad be able to compute the margi of error. 3. Lear about the t distributio ad its use i costructig a iterval estimate wheσ is ukow for a populatio mea. 4. Be able to determie the size of a simple radom sample ecessary to estimate a populatio mea ad/or a populatio proportio with a specified margi of error. 5. Kow the defiitio of the followig terms: cofidece iterval cofidece coefficiet cofidece level margi of error degrees of freedom
Solutios: 1. a. σ = σ / = 5/ 40 =.79 x At 95%, zσ / = 196. ( 5 / 40) = 155.. a. 3 ± 1.645 ( 6 / 50 ) 3 ± 1.4 or 30.6 to 33.4 3 ± 1.96 ( 6 / 50 ) 3 ± 1.66 or 30.34 to 33.66 c. 3 ±.576 ( 6 / 50 ) 3 ±.19 or 9.81 to 34.19 3. a. 80 ± 1.96 ( 15 / 60 ) 80 ± 3.8 or 76. to 83.8 80 ± 1.96 ( 15 / 10 ) 80 ±.68 or 77.3 to 8.68 c. Larger sample provides a smaller margi of error. 4. Sample mea 160 + 15 x = = 156 Margi of Error = 160 156 = 4 1.96( σ / ) = 4 = 1.96 σ / 4 = 1.96(15) / 4 = 7.35 = (7.35) = 54 5. a. 1.96 σ / = 1.96(5/ 49) = 1.40 4.80 ± 1.40 or 3.40 to 6.0 6. x ± z ( σ / ).05 8.5 ± 1.96(3.5/ 300 ) 8.5 ±.4 or 8.1 to 8.9 8 -
Iterval Estimatio z 7..05 ( σ / ) = 1.96(4000 / 60) = 101 A larger sample size would be eeded to reduce the margi of error. Sectio 8.3 ca be used to show that the sample size would eed to be icreased to = 46. 1.96(4000 / ) = 500 Solvig for, shows = 46 8. a. Sice is small, a assumptio that the populatio is at least approximately ormal is required. z.05 ( σ / ) = 1.96(5/ 10) = 3.1 z c..005 ( σ / ) =.576(5/ 10) = 4.1 9. x ± z.05 ( σ / ) 3.37 ± 1.96 (.8/ 10) 3.37 ±.05 or 3.3 to 3.4 σ 10. a. x ± z α / 119,155 ± 1.645 (30,000 / 80) 119,155 ± 5517 or $113,638 to $14,67 119,155 ± 1.96 (30,000 / 80) 119,155 ± 6574 or $11,581 to $15,79 c. 119,155 ±.576 (30,000 / 80) 119,155 ± 8640 or $110,515 to $17,795 d. The cofidece iterval gets wider as we icrease our cofidece level. We eed a wider iterval to be more cofidet that it will cotai the populatio mea. 11. a..05 1 -.10 =.90 c..05 d..01 e. 1 (.05) =.95 f. 1 (.05) =.90 8-3
1. a..179-1.676 c..457 d. Use.05 colum, -1.708 ad 1.708 e. Use.05 colum, -.014 ad.014 13. a. Σx 80 x = i = = 10 8 x ( x x) i i ( x x) 10 0 0 8-4 1 4 15 5 5 13 3 9 11 1 1 6-4 16 5-5 5 84 i Σ( xi x) 84 s = = = 3.464 1 7 c. t.05( s / ) =.365(3.464 / 8) =.9 d. x ± t ( s / ).05 10 ±.9 or 7.1 to 1.9 14. x ± t ( s / ) / df = 53 α a..5 ± 1.674 (4.4 / 54).5 ± 1 or 1.5 to 3.5.5 ±.006 (4.4 / 54).5 ± 1. or 1.3 to 3.7 c..5 ±.67 (4.4 / 54).5 ± 1.6 or 0.9 to 4.1 d. As the cofidece level icreases, there is a larger margi of error ad a wider cofidece iterval. 8-4
Iterval Estimatio 15. x ± t ( s / ) / α 90% cofidece df = 64 t.05 = 1.669 19.5 ± 1.669 (5. / 65) 19.5 ± 1.08 or 18.4 to 0.58 95% cofidece df = 64 t.05 = 1.998 19.5 ± 1.998 (5. / 65) 19.5 ± 1.9 or 18.1 to 0.79 16. a. t ( s / ) df = 99 t.05.05 = 1.984 1.984 (8.5/ 100) = 1.69 x ± t ( s / ).05 49 ± 1.69 or 47.31 to 50.69 c. At 95% cofidece, the populatio mea flyig time for Cotietal pilots is betwee 47.31 ad 50.69 hours per moth. This is clearly more flyig time tha the 36 hours for Uited pilots. With the greater flyig time, Cotietal will use fewer pilots ad have lower labor costs. Uited will require relatively more pilots ad ca be expected to have higher labor costs. 17. Usig Miitab or Excel, x = 6.34 ad s =.163 x ± t ( s / ) df = 49 t.05.05 =.010 6.34 ±.010 (.163/ 50) 6.34 ±.61 or 5.73 to 6.95 18. Usig Miitab or Excel, x = 3.8 ad s =.57 a. x = 3.8 miutes t ( s / ) df = 9 t.05.05 =.045.045 (.57 / 30) =.84 c. x ± t ( s / ).05 3.8 ±.84 or.96 to 4.64 d. There is a modest positive skewess i this data set. This ca be expected to exist i the populatio. While the above results are acceptable, cosiderig a larger sample ext time would be a good strategy. 8-5
19. a. t ( / ).05 s df = 599 Use df row, t.05 = 1.96 1.96 (175 / 600) = 14 x ± t ( s / ).05 649 ± 14 or 635 to 663 c. At 95% cofidece, the populatio mea is betwee $635 ad $663. This is slightly above the prior year s $63 level, so holiday spedig is icreasig. The poit estimate of the slight icrease is $649 - $63 = $17 or.7% per household. 0. x = Σ x / = miutes i Σ( xi x) s = = 1.1 miutes 1 x ± t.05 ( s / ) df = 19.00 ±.093 (1.1 / 0).00 ±.5 or 1.48 to.5 miutes 1. x i Σ 600 x = = = 130 liters of alcoholic beverages 0 Σ( xi x) 8144 s = = = 65.39 1 0 1 t.05 =.093 df = 19 95% cofidece iterval: x ± t.05 ( s / ) 130 ±.093 (65.39 / 0) 130 ± 30.60 or 99.40 to 160.60 liters per year. a. x = 3.35 So 3.35% is a poit estimate of the mea retur for the populatio. x ± t.05 ( s / ) t.05 =.064 df = 4 From the sample, s =.9 3.35 ±.064 (.9 / 5) 95% Cofidece Iterval: 3.35 ±.95 or.40% to 4.30% 8-6
Iterval Estimatio 3. z.05σ (1.96) (40) = = = 61.47 Use = 6 E 10 4. a. Plaig value of σ = Rage/4 = 36/4 = 9 z. 05σ ( 196. ) ( 9) = = = 34. 57 Use = 35 E 3 ( 196. ) ( 9) c. = = 77. 79 Use = 78 5. (1.96) (6.84) = = 79.88 Use = 80 (1.5) (1.645) (6.84) = = 31.65 Use = 3 6. a. z.05σ (1.96) (.15) = = = 17.64 Use 18. E (.07) If the ormality assumptio for the populatio appears questioable, this should be adjusted upward. c. (1.96) (.15) = = 34.6 Use 35 (.05) (1.96) (.15) = = 96.04 Use 97 (.03) For reportig purposes, the ewspaper might decide to roud up to a sample size of 100. 45, 000 30, 000 7. Plaig value σ = = 3750 4 a. c. z.05σ (1.96) (3750) = = = 16.09 E (500) Use = 17 (1.96) (3750) = = 1350.56 (00) Use = 1351 (1.96) (3750) = = 540.5 (100) Use = 5403 d. Samplig 5403 college graduates to obtai the $100 margi of error would be viewed as too expesive ad too much effort by most researchers. 8. a. zα / σ (1.645) (.5) = = = 34.48 Use = 343 E () 8-7
c. (1.96) (.5) = = 486.0 () Use = 487 (.576) (.5) = = 839.84 () Use = 840 d. The sample size gets larger as the cofidece is icreased. Do ot recommed 99% cofidece. The sample size must be icreased by (487 to 343) = 144 to go from 90% to 95%. This may be reasoable. However, icreasig the sample size by (840-487) = 353 to go from 95% to 99% would be viewed as too expesive ad time cosumig for the 4% gai i cofidece. ( 196. ) ( 6. 5) 9. a. = = 37. 5 Use = 38 ( 196. ) ( 6. 5) = = 150. 06 Use = 151 1 60 5 30. Plaig value σ = = 13.75 4 z.05σ (1.96) (13.75) = = = 80.70 Use = 81 E (3) 31. a. p = 100/400 =.5 p(1 p).5(.75) = =.017 400 c. p ± z. 05 p( 1 p).5 ± 1.96 (.017).5 ±.044 or.076 to.94 3. a..70 ± 1.645.70(.30) 800.70 ±.067 or.6733 to.767.70 ± 1.96.70(.30) 800.70 ±.0318 or.668 to.7318 33. z.05 p (1 p ) (1.96) (.35)(.65) = = = 349.59 Use = 350 E (.05) 8-8
Iterval Estimatio 34. Use plaig value p* =.50 (1.96) (.50)(.50) = = 1067.11 Use = 1068 (.03) 35. a. p = 81/611 =.4599 (46%) z.05 p(1 p).4599(1.4599) = 1.645 =.033 611 c. p ±.033.4599 ±.033 or.467 to.4931 36. a. p = 46/00 =.3 p(1 p).3(1.3) = =.098 00 p ± z.05 p(1 p).3 ± 1.96(.098).3 ±.0584 or.1716 to.884 37. a. p = 473/1100 =.43 z.05 p(1 p).43(1.43) = 1.96 =.093 1100 c. p ±.093.43 ±.093 or.4007 to.4593 d. With roughly 40% to 46% of employees surveyed idicatig strog dissatisfactio ad with the high cost of fidig successors, employers should take steps to improve employee satisfactio. The survey suggested employers may aticipate high employee turover costs if employee dissatisfactio remais at the curret level. 38. a. p = 9 /16 =.1790 p = 104 /16 =.640 p(1 p) (.64)(.358) Margi of error = 1.96 = 1.96 =.0738 16 Cofidece iterval:.640 ±.0738 or.568 to.7158 8-9
c. 39. a. 1.96 (.64)(.358) = = 353.18 Use = 354 (.05) z.05 p (1 p ) (1.96) (.156)(1.156) = = = 56 E (.03) z.005 p (1 p ) (.576) (.156)(1.156) = = = 970.77 Use 971 E (.03) 40. z.05 p(1 p).86(1.86) = 1.96 =.067 650 p ±.067.86 ±.067 or.8333 to.8867 41. Margi of error = z.05 p(1 p).09(1.09) = 1.96 =.0150 1400.09 ±.0150 or.075 to.1050 4. a. p *(1 p*).50(1.50) = =.06 491 z.05 p *(1 p*) = 1.96(.06) =.044 z.05 p (1 p ) = E September October November Pre-Electio 1.96 (.50)(1.50) = = 600.5 Use 601.04 1.96 (.50)(1.50) = = 1067.11 Use 1068.03 1.96 (.50)(1.50) = = 401.0 1.96 (.50)(1.50) = = 9604.01 p(1 p) (.53)(.47) 43. a. Margi of Error = zα / = 1.96 =.053 1500 95% Cofidece Iterval:.53 ±.053 or.5047 to.5553 8-10
Iterval Estimatio Margi of Error = 1.96 (.31)(.69) 1500 =.034 95% Cofidece Iterval:.31 ±.034 or.866 to.3334 c. Margi of Error = 1.96 (.05)(.95) 1500 =.0110 95% Cofidece Iterval:.05 ±.0110 or.039 to.061 d. The margi of error decreases as p gets smaller. If the margi of error for all of the iterval estimates must be less tha a give value (say.03), a estimate of the largest proportio should be used as a plaig value. Usig p * =.50 as a plaig value guaratees that the margi of error for all the iterval estimates will be small eough. 44. a. Margi of error: z.05 σ 15 = 1.96 = 4.00 54 Cofidece iterval: x ± margi of error 33.77 ± 4.00 or $9.77 to $37.77 45. a. x ± t.05 ( s / ) df = 63 t.05 = 1.998 5.45 ± 1.998 ( 74. 50 / 64 ) 5.45 ± 18.61 or $33.84 to $71.06 Yes. the lower limit for the populatio mea at Niagara Falls is $33.84 which is greater tha $15.60. 46. a. t.05 ( s / ) df = 99 t.05 = 1.984 1.984 (4980 / 100) = 998 x ± 998 5467 ± 998 or $4,479 to $6,455 c. 367($5,467) = $93,514,84 d. Harry Potter beat Lost World by $93.5 7.1 = $1.4 millio. This is a 1.4/7.1(100) = 30% icrease i the first weeked. The words shatter the record are justified. 47. a. From the sample of 30 stocks, we fid x = 1.9 ad s = 14.86 A poit estimate of the mea P/E ratio for NYSE stocks o Jauary 19, 004 is 1.9. 8-11
s 14.86 Margi of error = t.05 =.045 = 5.5 30 95% Cofidece Iterval: 1.9 ± 5.5 or 16.4 to 7.4 The poit estimate is greater tha 0 but the 95% cofidece iterval goes dow to 16.4. So we would be hesitat to coclude that the populatio mea P/E ratio was greater tha 0. Perhaps takig a larger sample would be i order. c. From the sample of 30 stocks, we fid p = 1/30 =.70 A poit estimate of the proportio of NYSE stocks payig divideds is.70. With p = 30(.7) = 1 ad (1 p) = 30(.3) = 9, we would be justified i usig a ormal distributio to costruct a cofidece iterval. A 95% cofidece iterval is p ± z.05 p(1 p).7 ± 1.96 (.7)(.3) 30.7 ±.16 or.54 to.86 While the sample size is large eough to use the ormal distributio approximatio, the sample size is ot large eough to provide much precisio. The margi of error is larger tha most people would like. 48. Variable N Mea StDev SE Mea 95.0% CI Time 150 14.000 3.838 0.313 (13.381, 14.619) a. x = 14 miutes 13.381 to 14.619 c. 7.5 hours = 7.5(60) = 450 miutes per day A average of 450/14 = 3 reservatios per day if o idle time. Assumig perhaps 15% idle time or time o somethig other tha reservatios, this could be reduced to 7 reservatios per day. d. For large airlies, there are may telephoe calls such as these per day. Usig the olie reservatios would reduce the telephoe reservatio staff ad payroll. Addig i a reductio i total beefit costs, a chage to olie reservatios could provide a sizeable cost reductio for the airlie. 49. a. Usig a computer, x = 49.8 miutes Usig a computer, s = 15.99 miutes 8-1
Iterval Estimatio c. x ± t.05 ( s / ) df = 199 t.05 1.96 49.8 ± 1.96 ( 15. 99 / 00 ) 49.8 ±. or 47.58 to 5.0 (. 33) (. 6) 50. = = 36. 7 Use = 37 1 ( 196. ) ( 8) 51. = = 6147. Use = 6 (. 576) ( 8) = = 10617. Use = 107 ( 196. ) ( 675) 5. = = 175. 03 Use = 176 100 53. a. p ± 196..47 ± 1.96 p( 1 p) (.47)(.53) 450.47 ±.0461 or.439 to.5161.47 ±.576 (.47)(.53) 450.47 ±.0606 or.4094 to.5306 c. The margi of error becomes larger. 54. a. p = 00/369 =.540 p(1 p) (.540)(.4580) 1.96 = 1.96 =.0508 369 c..540 ±.0508 or.491 to.598 55. a. p =.74 Margi of error = z.05 p(1 p) (.74)(.6) = 1.96 =.0 1677 95% Cofidece Iterval:.74 ±.0 or.7 to.76 8-13
p =.48 p(1 p) (.48)(.5) Margi of error = z.005 =.576 =.03 1677 95% Cofidece Iterval:.48 ±.03 or.45 to.51 c. The margi of error is larger i part b for two reasos. With p =.48, the estimate of the stadard error is larger. Ad z.005 =.576 is larger tha z.05 = 1.96 56. a. p = 455/550 =.873 p(1 p).873(1.873) Margi of error = 1.96 = 1.96 =.0316 550 95% Cofidece iterval:.873 ±.0316 or.7957 to.8589 57. a. (1.96) (.3)(.7) = = 016.84 Use = 017 (.0) p = 50/017 =.578 c. p ± 196. p( 1 p).578 ± 1.96 (.578)(.74) 017.578 ±.0191 or.387 to.769 58. a. (.33) (.70)(.30) = = 166.74 Use = 167 (.03) (.33) (.50)(.50) = = 1508.03 Use = 1509 (.03) 59. a. p = 110/00 =.55 0.55 ± 1.96 (.55)(.45) 00.55 ±.0689 or.4811 to.6189 (1.96) (.55)(.45) = = 380.3 Use = 381 (.05) 8-14
Iterval Estimatio 60. a. p = 618/1993 =.3101 p ± 196. p( 1 p) 1993.3101 ± 1.96 (.3101)(.6899) 1993.3101 ±.003 or.898 to.3304 c. = z p (1 p ) E (1.96) (.3101)(.6899) z = = 818.64 Use = 819 (.01) No; the sample appears uecessarily large. The.0 margi of error reported i part (b) should provide adequate precisio. 8-15