VIII. Interval Estimation A. A Few Important Definitions (Including Some Reminders)

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1 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

2 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 z z

3 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 z z 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

4 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 # X Obervatio # X Obervatio # X

5 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= L+ = = =. 45 ad plug the reult ito the formula 6.5 ± z =.± =.± = , 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 # X Obervatio # X Obervatio # X

6 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= L +83 = = = 78.6 calculate the ample tadard deviatio ( i -) i=1 = -1 ( ) + ( ) + ( ) + ( ) + L+ ( ) = -1 =6.97 ad plug the reult ito the formula 6.97 ± z = 78.6 ± = 78.6 ± = ,

7 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 provided the ample ± z = 78.6 ± 1.48 variace ad ot the ample tadard = 78.6 ± deviatio i thi problem! = , 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

8 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

9 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 / = degree Area i Upper Tail of freedom 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 # X Obervatio # X 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

10 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 / = ad calculate the ample mea i i= L+ = = = calculate the ample tadard deviatio ( i -) i=1 = -1 ( ) + ( ) + ( ) + ( ) + L+ ( -78.6) = 17-1 =6.97 ad plug the reult ito the formula 6.97 ± t = 78.6 ± = 78.6 ±.953 = ,.553 Notice that a thi cofidece level i wider tha the cofidece iterval of (76.603, ) 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

11 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) = = 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

12 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 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?

13 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 = = 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 p ± z = 0.45± = 0.45 ± = 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

14 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( ) = = 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 Ue Krook urvey reult to build a appropriate cofidece iterval at the 90% cofidece level for Mr. Keeth. 14

15 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= L+8 = = =

16 ad calculate the ample tadard deviatio ( i -) i=1 = -1 ( ) + ( ) + ( ) + ( ) + L+ ( ) = 48-1 = 0.64 ad plug the reult ito the formula 0.64 ± z = 74.77± = 74.77± 4.90 = , 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 p = =

17 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 / = o the 90% cofidece iterval i p 1- p p ± z = 0.75 ± = 0.75 ± = 0.647, 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