VIII. Iterval Etimatio A. A Few Importat Defiitio (Icludig Some Remider) 1. Poit Etimate - a igle umerical value ued a a etimate of a parameter.. Poit Etimator - the ample tatitic that provide the poit etimate of a parameter. Some poit etimator (ad the parameter they etimate) iclude Parameter μ p Poit Etimator p 3. Preciio - the eacte of a etimator. 4. Accuracy - the correcte of a etimator. Remember that - Preciio ad Accuracy are iverely related - Poit etimator are i) perfectly precie ad ii) almot certaily iaccurate 5. Samplig Error - the abolute differece betwee a parameter ad it poit etimator. How do we addre the eitece of amplig error? 6. Cofidece Iterval rage of value, ued a a etimate of a parameter, that will cotai the true value of the parameter a give proportio of time over may idepedet, idetical repeated trial alo referred to a a iterval etimate 7. Cofidece Level the proportio of time a cofidece iterval ca be epected to cotai the true value of the parameter over may idepedet, idetical repeated trial the correcte of a etimator OR the probability that the iterval etimatio procedure will geerate a iterval that doe cotai the true value of the parameter alo referred to a the cofidece coefficiet Note that (1 Cofidece Level) i ofte referred to a or the igificace level (o Cofidece Level = 1 - ) 1
B. Large-Sample Iterval Etimatio of the Populatio Mea μ Large Sample Cofidece Iterval for μ recall that we ca fid a ymmetric iterval that atifie the formula: ( ) P μ z μ+ z = Tolerace Level =1- Note that, oce we take a ample, we will have value for, z, ad - but ot for μ (which i what we are tryig to etimate) how doe thi help u develop a etimate of μ? f(z) 0.5-/ 0.5-/ / / μ z z μ ( = μ ) there i a 1 - probability that the value of the ample mea will provide a amplig error - μ of o more thaz If = 0.10, the we would epect (if we took 10 eparate idepedet ample) that 0 μ + z + z z f(z) 0.5-/ 0.5-/ / / μ z z μ = μ μ + z 0 1 + z z 3 4 5 6 7 9 10 8
Problem: We are tryig to etimate μ -but the (tolerace) iterval are cotructed aroud μ! How ca we ue μ to etimate μ? What would happe if we built iterval of half-width z aroud each ample mea? Could we eve do o? f(z) 0.5-/ 0.5-/ / / μ z z μ = μ μ + z 0 1 + z z 3 4 5 6 7 9 10 8 The reultig iterval of half-width z ample mea i give by z which by ubtitutio ca be rewritte a aroud the if the paret populatio tadard deviatio i kow ad either i) the paret populatio i ormal or ii) the ample i ufficietly large ( 30). Note that - thi i ofte referred to a a iterval etimate or a cofidece iterval. -(1 )100% i ofte referred to a the cofidece level or cofidece coefficiet. - z i ofte referred to a the margi of error. ± ± z 3
Aother Problem: We eed to kow the paret populatio tadard deviatio - thi (uually) eceitate a ceu! Thi implie a kowledge about the populatio mea μ, which i what we are tryig to etimate!. It ca be how that the ample tadard deviatio i a etremely reliable etimator of the paret populatio tadard deviatio if we take a large ample ( 30). The reultig iterval of half-width ample mea i give by ± z z which by ubtitutio ca be rewritte a aroud the if the paret populatio tadard deviatio i ot kow ad the ample i ufficietly large ( 30). Note that - thi i ofte referred to a a iterval etimate or a cofidece iterval. - (1 )100% i ofte referred to a the cofidece level or cofidece coefficiet. - z i ofte referred to a the margi of error. ± z Eample: Suppoe we take a ample of the age of forty-five cutomer from a populatio whoe tadard deviatio i 6.5 ad collect the followig obervatio: Obervatio # 1 3 4 5 6 7 8 9 10 11 1 13 14 15 X 55 4 4 8 4 40 38 35 39 8 Obervatio # 16 17 18 19 0 1 3 4 5 6 7 8 9 30 X 3 3 41 34 38 3 9 6 38 5 39 Obervatio # 31 3 34 35 38 39 40 41 4 43 44 45 X 7 34 18 34 1 6 31 35 31 4
How would we cotruct a 95% cofidece iterval for thee data? Sice the radom variable X (age) i quatitative (iterval or ratio), we recogize the parameter we wih to etimate i the populatio mea μ. We alo recogize that we have a large ample ( = 45 30) ad we kow the paret populatio tadard deviatio i = 6.5. So we decide to etimate the iterval by uig ± z We fid the appropriate level of z / from the Stadard Normal Ditributio table for the 95% level of cofidece z / = 1.96 ad calculate the ample mea i i=1 55 +4+ + 4+ L+ = = =. 45 ad plug the reult ito the formula 6.5 ± z =.± 1.96 45 =.± 1.899 = 31.301, 35.099 Eample: If we ak thirty-three radomly elected idividual to rate a product o a 100 poit (1-100) cale, ad our repoe are Obervatio # 1 3 4 5 6 7 8 9 10 11 X 7 78 79 66 83 74 77 79 69 8 91 Obervatio # 1 13 14 15 16 17 18 19 0 1 X 76 65 73 79 8 86 76 74 86 76 Obervatio # 3 4 5 6 7 8 9 30 31 3 X 7 74 78 80 70 74 89 94 83 5
How would we cotruct a 90% cofidece iterval for thee data? Sice the radom variable X (product ratig) i quatitative (iterval or ratio), we recogize the parameter we wih to etimate i the populatio mea μ. We alo recogize that we have a large ample ( = 30) but we do t kow the paret populatio tadard deviatio. So we decide to etimate the iterval by uig ± z We fid the appropriate level of z / from the Stadard Normal Ditributio table for the 90% level of cofidece z / = 1.64 or 1.65 (or 1.645) ad calculate the ample mea i i=1 7+78+79 +66 + L +83 = = = 78.6 calculate the ample tadard deviatio ( i -) i=1 = -1 ( 7-78.6 ) + ( 78-78.6 ) + ( 79-78.6 ) + ( 66-78.6 ) + L+ ( 83-78.6) = -1 =6.97 ad plug the reult ito the formula 6.97 ± z = 78.6 ± 1.645 = 78.6 ± 1.997 = 76.603, 80.597 6
Eample: If we ak thirty-three radomly elected idividual to rate a product o a 100 poit (1-100) cale, ad calculate from our reult a ample mea of 78.6 ad a ample variace of 48.63, what would be the appropriate 86% cofidece iterval? We have a large ample ( = 30), we do ot kow the paret populatio tadard deviatio, ad z / = 1.48, o Note that we were 48.63 provided the ample ± z = 78.6 ± 1.48 variace ad ot the ample tadard = 78.6 ± 1.797 deviatio i thi problem! = 76.803, 80.397 Notice that a the cofidece level decreae, the cofidece iterval arrow - why? C. Small-Sample Iterval Etimatio of the Populatio Mea μ If the paret populatio tadard deviatio i ot kow ad the paret populatio i ormally ditributed, but the ample i ot ufficietly large ( 30) we mut ue a pecial probability ditributio. The t-ditributio (or the Studet-t Ditributio) - family of probability ditributio that are ued to cotruct iterval etimate of the populatio mea whe the populatio tadard deviatio i ukow ad the paret populatio i ormally or ear-ormally ditributed. - the t-ditributio i imilar to the ormal ditributio ecept that it i horter ad wider (to allow for the fact that we are uig a etimate of the paret populatio tadard deviatio ). - the height ad width of the t-ditributio i determied by a parameter called the degree of freedom. for a cofidece iterval of the populatio mea, the degree of freedom are - 1. a the degree of freedom icreae, the t-ditributio become more ormal (taller ad arrower). value for the t-ditributio for variou combiatio of / ad degree of freedom icreae are give i Appedi A, Table of ASW. 7
F(z or t) 0 tadard ormal ditributio t ditributio for df = 5 t ditributio for df = 10 t ditributio for df = 0 z or t The reultig iterval of half-width ample mea i give by ± t t aroud the which by ubtitutio ca be rewritte a ± t if the paret populatio tadard deviatio i ot kow ad the ample i ot ufficietly large ( < 30), but the paret populatio i relatively ormally ditributed. Note that - thi i ofte referred to a a iterval etimate or a cofidece iterval. -(1 )100% i ofte referred to a the cofidece level or cofidece coefficiet. - i ofte referred to a the margi of error. t degree of Area i Upper Tail freedom 0.10 0.05 0.05 0.01 0.005 0.001 3.078 6.314 1.706 31. 63.656 318.89 1 1.6.90 4.303 6.965 9.95.38 3 1.638.353 3.18 4.541 5.1 10.14 4 1.5.13.776 3.747 4.604 7.173 5 1.476.015.571 3.5 4.03 5.894 6 1.440 1.943.447 3.143 3.707 5.08 7 1.415 1.895.5.998 3.499 4.785 8 1.397 1.860.306.896 3.355 4.501 9 1.383 1.8.6. 3.50 4.97 10 1. 1..8.764 3.169 4.144 11 1.3 1.796.01.718 3.106 4.05 1 1.356 1.78.179.6 3.055 3.930 13 1.350 1.771.160.650 3.01 3.85 14 1.345 1.761.145.64.977 3.787 15 1.341 1.753.131.60.947 3.7 16 1.7 1.746.10.583.91 3.686 17 1.3 1.740.110.567.898 3.646 18 1.0 1.734.101.55.878 3.610 19 1.38 1.79.093.539.861 3.579 0 1.35 1.75.086.58.5 3.55 1 1. 1.71.080.518.831 3.57 1.31 1.717.074.508.9 3.505 3 1.319 1.714.069.500.807 3.485 4 1.318 1.711.064.49.797 3.467 5 1.316 1.708.060.485.787 3.450 6 1.315 1.706.056.479.779 3.435 7 1.314 1.703.05.473.771 3.41 8 1.313 1.701.048.467.763 3.408 9 1.311 1.699.045.46.756 3.396 30 1.310 1.697.04.457.750 3.385 40 1.303 1.6.01.43.704 3.307 60 1.96 1.671.000.390.660 3.3 10 1.89 1.658 1.980.358.617 3.160 1.8 1.645 1.960..576 3.090 8
If we ak 17 radomly elected idividual to rate a product o a 100 poit (1-100) cale, ad we wih to calculate the 90% cofidece iterval, what would be the appropriate value of t? We have a ample of = 17 obervatio, o the degree of freedom are - 1 = 17-1 = 16. Thu t / = 1.746. degree Area i Upper Tail of freedom 0.10 0.05 0.05 0.01 0.005 1 3.078 6.314 1.706 31. 63.656...... 15 1.341 1.753.131.60.947 16 1.7 1.746.10.583.91 17 1.3 1.740.110.567.898 18 1.0 1.734.101.55.878 Eample: Suppoe we ak evetee radomly elected idividual to rate a product o a 100 poit (1-100) cale, ad collect the followig ample reult: Obervatio # 1 3 4 5 6 7 8 9 X 74 79 85 69 77 77 79 69 Obervatio # 10 11 1 13 14 15 16 17 X 85 91 76 69 70 79 85 Alo uppoe that product ratig have hitorically bee relatively ormally ditributed. How would we cotruct a 90% cofidece iterval for thee data? Sice the radom variable X (product ratig) i quatitative (iterval or ratio), we recogize the parameter we wih to etimate i the populatio mea μ. We alo recogize that we have a mall ample ( = 17 < 30) from a ormal populatio, ad we do t kow the paret populatio tadard deviatio. So we decide to etimate the iterval by uig ± t 9
We fid the appropriate level of t / from the Studet t Ditributio table for the 90% level of cofidece ad 1 = 16 degree of freedom t / = 1.746 ad calculate the ample mea i i=1 74+79 +85 +69 + L+ = = = 78.6 17 calculate the ample tadard deviatio ( i -) i=1 = -1 ( 74-78.6 ) + ( 79-78.6 ) + ( 85-78.6 ) + ( 69-78.6 ) + L+ ( -78.6) = 17-1 =6.97 ad plug the reult ito the formula 6.97 ± t = 78.6 ± 1.746 17 = 78.6 ±.953 = 75.647,.553 Notice that a thi cofidece level i wider tha the cofidece iterval of (76.603, 80.597) derived uig the tadard ormal ditributio - why? ye ye o i kow? ue to etimate i large ( 30)? ye ye o o i kow? i the populatio approimately ormal? o ue ue ± z ± z ue to etimate icreae the ample ize to at leat 30 or ue a oparametric approach ue ue ± z ± t 10
Ca we determie the ample ize eceary to eure a deired cofidece level ad margi of error? If we aume a give populatio tadard deviatio ad tate the deired margi of error (deoted a E), ample ize eceary to eure a deired cofidece level ad margi of error i z = E Thi i othig but a algebraic maipulatio of the defiitio of the margi of error z E= Eample: Suppoe we wih to determie the miimum ample ize eceary to eure a 95% cofidece level ad a margi of error of.5. If we believe that the populatio tadard deviatio i 1, how large mut the ample ize be? We have have that = 1.0, E =.5, ad z / = 1.96, o or 7 (why?). ( 1.96) ( 1.0) = = 71.063.5 D.Iterval Etimatio of the Populatio Proportio p Recall that if the ample ize i ufficietly large (p 5 ad (1-p) 5), that p i approimately ormally ditributed with a mea of p ad tadard deviatio of p1-p p = o the reultig (1 )100% cofidece iterval i p ± z p which through ubtitutio ca be rewritte a p± z p1-p 11
Aother Problem: How do we ue the populatio proportio p to etimate the populatio proportio p? We imply ubtitute p for p, i.e. p1-p p± z ad p 5 ad (1- p ) 5 Aother alterative i to ue p = 0.50 i p 1- p 0.50 1-0.50 0.5 p = = = (Why?) Eample: Suppoe we poll oe-hudred radomly elected regitered voter idividual ad ak if they ited to vote for Cadidate Joe, ad our ample reult are a follow: Obervatio # Vote For Joe? Obervatio # Vote For Joe? Obervatio # Vote For Joe? Obervatio # Vote For Joe? 1 6 50 76 7 51 77 3 8 5 78 4 9 53 79 5 30 54 80 6 31 55 7 3 56 8 8 57 83 9 34 58 10 35 59 85 11 60 86 1 61 87 13 38 6 14 39 63 89 15 40 64 90 16 41 65 91 17 4 66 9 18 43 67 93 19 44 68 94 0 45 69 95 1 46 70 96 47 71 97 3 48 7 98 4 49 73 99 5 50 74 100 1
Sice the radom variable of iteret (whether a repodet idicated /he would vote for Joe) i omial, we recogize the parameter we wih to etimate i the populatio proportio p. We alo recogize that the ample proportio i 45 p = = 0.45 100 o p = 100(0.45) = 45 5 ad (1 p) = 100(1-0.45) = 55 5 Cochra rule are atified ad we ca etimate the cofidece iterval uig p1-p p± z For a 9% level of cofidece we have that z / = 1.75 o the 9% cofidece iterval i p 1- p 0.45 1-0.45 p ± z = 0.45± 1.75 100 = 0.45 ± 0.087 = 0.3, 0.5 What do you thik of the likelihood that Joe will wi? Ca we determie the ample ize eceary to eure a deired cofidece level ad margi of error? If we aume a give populatio tadard deviatio ad tate the deired margi of error (deoted a E), ample ize eceary to eure a deired cofidece level ad margi of error i z p ( 1-p) = E Agai, thi i othig but a algebraic maipulatio of the defiitio of the margi of error E=z p1-p 13
Eample: Suppoe we wih to determie the miimum ample ize eceary to eure a 95% cofidece level ad a margi of error of.05 for Cadidate Joe. If we believe that Joe i curretly favored by 50% of the populatio, how large mut the ample ize be? We have have that p = 0.50, E = 0.05, ad z / = 1.96, o ( 1.96) 0.50( 1-0.50) = = 15.64 0.05 or 15 (why?). Agai, if we do t have a good etimate (gue) for the populatio proportio p, we hould alway ue p = 0.50 (why?). Eample: Raymod Keeth, Brad maager for Imai Diot perfume, wat to ae the market potetial of a ew fragrace Eau de Commodé. Hitorically a coumer that rate a Diot perfume 80 or better o a 100 poit cale after amplig the product i uually willig to coider purchaig the cet. Mr. Keeth commiio Tommie Krook Reearch to collect ratig of Eau de Commodé from forty-eight frequet perfume purchaer. The reult of Krook urvey follow. Repodet # Ratig Repodet # Ratig Repodet # Ratig Repodet # Ratig 1 13 5 48 65 14 85 6 66 38 44 3 99 15 8 7 89 39 9 4 91 16 69 8 94 40 5 8 17 9 65 41 6 18 83 30 4 7 8 19 83 31 90 43 66 8 0 100 3 44 67 9 90 1 45 80 10 8 34 46 11 80 3 80 35 40 47 53 1 4 94 8 48 Ue Krook urvey reult to build a appropriate cofidece iterval at the 90% cofidece level for Mr. Keeth. 14
Should you build a cofidece iterval of the populatio mea or populatio proportio of Eau de Commodé? How would we cotruct a 90% cofidece iterval for the populatio mea ratig μ Eau de Commodé of Eau de Commodé with thee data? We recogize that we have a large ample ( = 48 30) but we do t kow the paret populatio tadard deviatio. So we decide to etimate the iterval by uig ± z We fid the appropriate level of z / from the Stadard Normal Ditributio table for the 90% level of cofidece z / = 1.64 or 1.65 (or 1.645) ad calculate the ample mea i i=1 +9+99+91+ L+8 = = = 74.77 48 15
ad calculate the ample tadard deviatio ( i -) i=1 = -1 ( -74.77 ) + ( 9-74.77 ) + ( 99-74.77 ) + ( 91-74.77 ) + L+ ( 8-74.77) = 48-1 = 0.64 ad plug the reult ito the formula 0.64 ± z = 74.77± 1.645 48 = 74.77± 4.90 = 69.868, 79.67 What doe thi mea for Mr. Keeth? How would we cotruct a 90% cofidece iterval for the populatio proportio of coumer p Eau de Commodé who will coider purchaig Eau de Commodé with thee data? We recogize that the ample proportio i 0.75 (why?) Repodet # Ratig Repodet # Ratig Repodet # Ratig Repodet # Ratig 1 13 5 48 65 14 85 6 66 38 44 3 99 15 8 7 89 39 9 4 91 16 69 8 94 40 5 8 17 9 65 41 6 18 83 30 4 7 8 19 83 31 90 43 66 8 0 100 3 44 67 9 90 1 45 80 10 8 34 46 11 80 3 80 35 40 47 53 1 4 94 48 8 p = = 0.75 48 16
o p = 48(0.75) = 5 ad (1 p) = 48(1-0.75) = 1 5 Cochra rule are atified ad we ca etimate the cofidece iterval uig p± z p1-p For a 90% level of cofidece we have that z / = 1.645 o the 90% cofidece iterval i p 1- p 0.75 1-0.75 p ± z = 0.75 ± 1.645 48 = 0.75 ± 0.103 = 0.647, 0.853 What doe thi mea for Mr. Keeth? Which i more appropriate the 90% cofidece iterval of the populatio mea or the 90% cofidece iterval of the populatio proportio? I which radom variable hould Mr. Keeth be itereted: The idividual ratig coumer give Eau de Commodé? Whether idividual coumer give Eau de Commodé a ratig of 80 or higher? Awer thi quetio ad you have determied whether the cofidece iterval of the populatio mea or the cofidece iterval of the populatio proportio i more appropriate for thi problem. 17