Chapter 8 Interval Estimation

Transcription

1 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

2 Solutios: 1. a. σ = σ / = 5/ 40 =.79 x At 95%, zσ / = 196. ( 5 / 40) = a. 3 ± ( 6 / 50 ) 3 ± 1.4 or 30.6 to ± 1.96 ( 6 / 50 ) 3 ± 1.66 or to c. 3 ±.576 ( 6 / 50 ) 3 ±.19 or 9.81 to a. 80 ± 1.96 ( 15 / 60 ) 80 ± 3.8 or 76. to ± 1.96 ( 15 / 10 ) 80 ±.68 or 77.3 to 8.68 c. Larger sample provides a smaller margi of error. 4. Sample mea x = = 156 Margi of Error = = ( σ / ) = 4 = 1.96 σ / 4 = 1.96(15) / 4 = 7.35 = (7.35) = a σ / = 1.96(5/ 49) = ± 1.40 or 3.40 to x ± z ( σ / ) ± 1.96(3.5/ 300 ) 8.5 ±.4 or 8.1 to

3 Iterval Estimatio z ( 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 = (4000 / ) = 500 Solvig for, shows = 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) = 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 ± (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 =.90 c..05 d..01 e. 1 (.05) =.95 f. 1 (.05) =

4 1. a c..457 d. Use.05 colum, ad e. Use.05 colum, ad a. Σx 80 x = i = = 10 8 x ( x x) i i ( x x) i Σ( xi x) 84 s = = = c. t.05( s / ) =.365(3.464 / 8) =.9 d. x ± t ( s / ) ±.9 or 7.1 to x ± t ( s / ) / df = 53 α a..5 ± (4.4 / 54).5 ± 1 or 1.5 to ±.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

5 Iterval Estimatio 15. x ± t ( s / ) / α 90% cofidece df = 64 t.05 = ± (5. / 65) 19.5 ± 1.08 or 18.4 to % cofidece df = 64 t.05 = ± (5. / 65) 19.5 ± 1.9 or 18.1 to a. t ( s / ) df = 99 t = (8.5/ 100) = 1.69 x ± t ( s / ) ± 1.69 or to c. At 95% cofidece, the populatio mea flyig time for Cotietal pilots is betwee ad 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 = ±.010 (.163/ 50) 6.34 ±.61 or 5.73 to Usig Miitab or Excel, x = 3.8 ad s =.57 a. x = 3.8 miutes t ( s / ) df = 9 t = (.57 / 30) =.84 c. x ± t ( s / ) ±.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

6 19. a. t ( / ).05 s df = 599 Use df row, t.05 = (175 / 600) = 14 x ± t ( s / ) ± 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 = ±.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 = = = t.05 =.093 df = 19 95% cofidece iterval: x ± t.05 ( s / ) 130 ±.093 (65.39 / 0) 130 ± or to 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 = ±.064 (.9 / 5) 95% Cofidece Iterval: 3.35 ±.95 or.40% to 4.30% 8-6

7 Iterval Estimatio 3. z.05σ (1.96) (40) = = = Use = 6 E a. Plaig value of σ = Rage/4 = 36/4 = 9 z. 05σ ( 196. ) ( 9) = = = Use = 35 E 3 ( 196. ) ( 9) c. = = Use = (1.96) (6.84) = = Use = 80 (1.5) (1.645) (6.84) = = Use = 3 6. a. z.05σ (1.96) (.15) = = = 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) = = Use 97 (.03) For reportig purposes, the ewspaper might decide to roud up to a sample size of , , Plaig value σ = = a. c. z.05σ (1.96) (3750) = = = E (500) Use = 17 (1.96) (3750) = = (00) Use = 1351 (1.96) (3750) = = (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) = = = Use = 343 E () 8-7

8 c. (1.96) (.5) = = () Use = 487 (.576) (.5) = = () 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 ( ) = 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. = = Use = 38 ( 196. ) ( 6. 5) = = Use = Plaig value σ = = z.05σ (1.96) (13.75) = = = Use = 81 E (3) 31. a. p = 100/400 =.5 p(1 p).5(.75) = = c. p ± z. 05 p( 1 p).5 ± 1.96 (.017).5 ±.044 or.076 to a..70 ± (.30) ±.067 or.6733 to ± (.30) ±.0318 or.668 to z.05 p (1 p ) (1.96) (.35)(.65) = = = Use = 350 E (.05) 8-8

9 Iterval Estimatio 34. Use plaig value p* =.50 (1.96) (.50)(.50) = = Use = 1068 (.03) 35. a. p = 81/611 =.4599 (46%) z.05 p(1 p).4599(1.4599) = = c. p ± ±.033 or.467 to a. p = 46/00 =.3 p(1 p).3(1.3) = = p ± z.05 p(1 p).3 ± 1.96(.098).3 ±.0584 or.1716 to a. p = 473/1100 =.43 z.05 p(1 p).43(1.43) = 1.96 = c. p ± ±.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 = Cofidece iterval:.640 ±.0738 or.568 to

10 c. 39. a (.64)(.358) = = 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) = = = Use 971 E (.03) 40. z.05 p(1 p).86(1.86) = 1.96 = p ± ±.067 or.8333 to Margi of error = z.05 p(1 p).09(1.09) = 1.96 = ±.0150 or.075 to a. p *(1 p*).50(1.50) = = 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) = = Use (.50)(1.50) = = Use (.50)(1.50) = = (.50)(1.50) = = p(1 p) (.53)(.47) 43. a. Margi of Error = zα / = 1.96 = % Cofidece Iterval:.53 ±.053 or.5047 to

11 Iterval Estimatio Margi of Error = 1.96 (.31)(.69) 1500 = % Cofidece Iterval:.31 ±.034 or.866 to.3334 c. Margi of Error = 1.96 (.05)(.95) 1500 = % 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 = Cofidece iterval: x ± margi of error ± 4.00 or $9.77 to $ a. x ± t.05 ( s / ) df = 63 t.05 = ± ( / 64 ) 5.45 ± or $33.84 to $71.06 Yes. the lower limit for the populatio mea at Niagara Falls is $33.84 which is greater tha $ a. t.05 ( s / ) df = 99 t.05 = (4980 / 100) = 998 x ± ± 998 or $4,479 to $6,455 c. 367($5,467) = $93,514,84 d. Harry Potter beat Lost World by $ = $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 = A poit estimate of the mea P/E ratio for NYSE stocks o Jauary 19, 004 is

12 s Margi of error = t.05 =.045 = % 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 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 (13.381, ) a. x = 14 miutes to 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 = miutes 8-1

13 Iterval Estimatio c. x ± t.05 ( s / ) df = 199 t ± 1.96 ( / 00 ) 49.8 ±. or to 5.0 (. 33) (. 6) 50. = = Use = 37 1 ( 196. ) ( 8) 51. = = Use = 6 (. 576) ( 8) = = Use = 107 ( 196. ) ( 675) 5. = = Use = a. p ± ± 1.96 p( 1 p) (.47)(.53) ±.0461 or.439 to ±.576 (.47)(.53) ±.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 = c..540 ±.0508 or.491 to a. p =.74 Margi of error = z.05 p(1 p) (.74)(.6) = 1.96 = % Cofidece Iterval:.74 ±.0 or.7 to

14 p =.48 p(1 p) (.48)(.5) Margi of error = z.005 =.576 = % 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 = a. p = 455/550 =.873 p(1 p).873(1.873) Margi of error = 1.96 = 1.96 = % Cofidece iterval:.873 ±.0316 or.7957 to a. (1.96) (.3)(.7) = = Use = 017 (.0) p = 50/017 =.578 c. p ± 196. p( 1 p).578 ± 1.96 (.578)(.74) ±.0191 or.387 to a. (.33) (.70)(.30) = = Use = 167 (.03) (.33) (.50)(.50) = = Use = 1509 (.03) 59. a. p = 110/00 = ± 1.96 (.55)(.45) ±.0689 or.4811 to.6189 (1.96) (.55)(.45) = = Use = 381 (.05) 8-14

15 Iterval Estimatio 60. a. p = 618/1993 =.3101 p ± 196. p( 1 p) ± 1.96 (.3101)(.6899) ±.003 or.898 to.3304 c. = z p (1 p ) E (1.96) (.3101)(.6899) z = = Use = 819 (.01) No; the sample appears uecessarily large. The.0 margi of error reported i part (b) should provide adequate precisio. 8-15