Mathacle PSet Stats, Confidence Intervals and Estimation Level Number Name: Date: Unbiased Estimators So we don t have favorite.

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1 PSet Stat, Cofidece Iterval ad Etimatio Ubiaed Etimator So we do t have favorite. IV. CONFIDENCE INTERVAL AND ESTIMATION 4.1. Sigificat Level ad Critical Value z ad The igificat level, ofte deoted by, i a probability meaure that idicate how cofidet the ample tatitic uch a ample mea x or ample proportio p i igificatly differet from the populatio mea or populatio proportio p, repectively. The igificat level i uually aociated with the critical z-core z i N(0,1) ditributio or critical t i t ditributio, depedig o which ditributio i the applicatio i ued. Three cae aociated with are ofte practiced: the left-tailed ad the right-tailed, twotailed. The two-tailed cae i ued i determiig the cofidece iterval i ormal ad t ditributio. df i the degree of freedom i t ditributio. t I Ti-84: Left-tailed: z ivnorm, t ivt, df ; Right-tailed: z ivnorm 1, t ivt 1, df Two-tailed: z ivnorm, t ivt, df 1

2 PSet Stat, Cofidece Iterval ad Etimatio Example For the give, df, ad the type of tail, fid z or t a idicated. Type df z or 0.1 Z - Ditributio, right - tailed - t Graph 0.1 Z - Ditributio, left - tailed Z - Ditributio, two - tailed Z - Ditributio, two - tailed Z - Ditributio, two - tailed t - Ditributio, two - tailed t - Ditributio, two - tailed t - Ditributio, two - tailed t - Ditributio, two - tailed t - Ditributio, two - tailed t - Ditributio, two - tailed t - Ditributio, two - tailed t - Ditributio, two - tailed t - Ditributio, two - tailed 30

3 PSet Stat, Cofidece Iterval ad Etimatio Solutio: Type df z or t 0.1 Z - Ditributio, right - tailed - z 1.8 Graph 0.1 Z - Ditributio, left - tailed - z Z - Ditributio, two - tailed - z Z - Ditributio, two - tailed - z Z - Ditributio, two - tailed - z t - Ditributio, two - tailed 5 t t - Ditributio, two - tailed 0 t t - Ditributio, two - tailed 30 t t - Ditributio, two - tailed 60 t t - Ditributio, two - tailed 5 t t - Ditributio, two - tailed 0 t t - Ditributio, two - tailed 30 t t - Ditributio, two - tailed 5 t t - Ditributio, two - tailed 30 t.750 3

4 PSet Stat, Cofidece Iterval ad Etimatio 4.. Cofidece Iterval (CI) The cofidece level (CL) i defied a 1. I the two-tailed cae whe the i give, the cofidece iterval (CI) i z, z for Z- ditributio, ad the CI i t, t for t-ditributio. The area o each ide of the two-ided cae i. Example To compare Z-ditributio ad t- ditributio, fid ditributio df CI 0.1 Z Z Z t t t t t t 30 Solutio: ditributio df CI 0.1 Z - ( 1.645,1.645) 0.05 Z - ( 1.960,1.960) 0.01 Z - (.576,.576) 0.1 t 5 (.015,.015) 0.1 t 30 ( 1.697,1.697) 0.05 t 5 (.571,.571) 0.05 t 30 (.04,.04) 0.01 t 5 ( 4.03, 4.03) 0.01 t 30 (.750,.750) 4

5 PSet Stat, Cofidece Iterval ad Etimatio For the ditributio N(, ), the z-core ca be ued to traform I ummary: N (, ) N(0,1). Ditributio Cofidece Iterval (CI) (0,1) CI z, z or 0 z N k (t) CI t, t or N (, ) z z 0 t, where k df CI, or z whe ad are kow 4.3. Defiitio of Ubiaed Etimator A etimator T i called a ubiaed etimator for parameter, if the expected value of T i : E [T] The differece of E[T] i called the bia of T. The ituitive meaig of a ubiaed etimator i oe that doe ot ytematically overetimate or uderetimate the The Proportio Etimator for Biomial Ditributio Let N deote the total umber of uit or item i the populatio. Suppoe X i the um of idepedet dicrete radom biomial variable X, X, 1, X that take oe or zero, ad with ucce probability p for each trial: X X X... 1 X Where N. The mea ad variace of X are E X ] p [ ad Var X ) p1 p (. X X1 X X If the ample ditributio i defied a the recaled or the average biomial variable X, the the mea of i X E[ X ] p ˆ [ ˆ p E p] E p 5

6 PSet Stat, Cofidece Iterval ad Etimatio ˆ. Therefore, the ample proportio i a ubiaed etimator for the populatio proportio p. It variace i That i, the etimator ha the property of Ep p X Var[ X ] p(1 p) p(1 p) ˆ [ ˆ p Var p] Var The proportio problem hould really be modeled a hypergeometric problem. That i, the k uccee i draw without replacemet. The tadard deviatio for hypergeometric model i N p(1 p) N 1 N Where N 1 i the modificatio factor. Whe the amplig fractio i mall, ay N 10% N, the tadard deviatio ca be approximated a 1 N 1 1 N p(1 p) p(1 p) Thi i the the 10% rule, ad i thi cae the biomial approximatio i atifactory. Whe p ( 1 p) 1, or equivaletly 10 ad ( 1 ) 10a oted later, the biomial ditributio ca be approximated by the Z ditributio. The z-core i thi cae ca be calculated a z p p p(1 p) (1 ) ˆp ca be ued to approximate atified. The igificat level i (1 ), whe p i ukow ad the 10% rule i 6

7 PSet Stat, Cofidece Iterval ad Etimatio p p(1 p) p z ˆ (1 p) z The cofidece iterval i the ˆ(1 ˆ) ˆ(1 ˆ) ˆ p p, ˆ p CI p z p z p or i the iterval form: (1 ) (1 ) z p z or i the form of error term: z (1 ) I ummary, the tep to obtai CI for the ample proportio (tatitic) with ample ize are 1.) The ample i a imple radom ample..) The ample ize i le tha 10% of populatio. 3.) 10, ( 1 ) 10to approximate the biomial ditributio to ormal ditributio. Example THS admiitrator wated to kow how may 10 th grader ad 11 th grader did either iterhip or commuity ervice i the pat ummer. A radom ample of 75 tudet idicated that 60 tudet did oe of the two. Fid the 95% cofidece iterval for the chool proportio. Aume that the chool populatio of thee two clae i 800 ad all two grade are equally likely to do ummer iterhip or commuity ervice. Solutio: p ˆ 60/ , 0. 05, z ( z ivnorm(0.05) ) z 75 10% , , 1 p ˆ 7

8 PSet Stat, Cofidece Iterval ad Etimatio or ˆ p z (1 ) CI 0.709, 0.8(1 0.8) That i, the chool i 95% cofidet that the true proportio of thoe two grade who did the ummer iterhip or commuity ervice i betwee ad Example A previou tudy ha uggeted that about 19.3% of tee (aged 1-19) are obee. How large of a ample will be eeded i order to etimate the true proportio of obee tee with 95% cofidece ad a margi of error of o more tha 1%? Solutio: p ˆ 0.193, 0. 05, z ( z ivnorm(0.05) ) z z z (1 ) ( ) The Mea Etimator for the Normal Ditributio N(, ) Suppoe X, X, 1, X are cotiuou idepedet radom variable from a ormal ditributio with the kow mea ad the kow variace. The X 1 X... X X i the ubiaed mea etimator: X X X E[ X i ] 1... E[ X ] E The variace i X [ ] 1 X... X EX x Var[ X ] Var i 8

9 PSet Stat, Cofidece Iterval ad Etimatio Similar to the proportio problem, the correct hypergeometric model hould add the correctio factor to the variace: N x N 1 Whe the 10% rule i atified, the tadard deviatio i 1 N N x N N The Cetral Limit Theorem ay that whe the ample ize i large, thi ubiaed x etimator X ~ N(, ) ~ N(0,1). x For the cofidece level 1, z. The cofidece iterval i, or CI x z x z, x z x z or x z The coditio to obtai CI for the ample mea (tatitic) x with ample ize are 1.) The ample i a imple radom ample..) The ample ize i le tha 10% of populatio. 3.) i kow ad 30for the coditio of uig ormal ditributio. Example The Preidet of a large uiverity wihe to etimate the average age of the tudet preetly erolled. For the pat tudie, the tadard deviatio i kow to be 9

10 PSet Stat, Cofidece Iterval ad Etimatio year. A ample of 50 tudet i elected radomly, ad ample mea i foud to be 3. year. Fid the 95% cofidece iterval of the chool populatio mea. Solutio:, 50, x 3., 0. 05, z 1. 96, CI? or z CI.6, 3.8 That i, the Preidet ca ay with 95% cofidece that the average age of tudet i betwee.6 ad 3.8 year old. 50 Example From the lat example, the Preidet would like to be 99% cofidet that the etimate of average age hould be accurate withi 1 year whe the tadard deviatio of the age i 3 year. How large a ample i eceary? Solutio: 3, 0. 01, z. 58,? z z The ample ize eed to be at leat The Mea Etimator for t-ditributio For a radom variable X ~ T( 1), where T ( 1) i the t-ditributio with the degree of freedom k 1 ad with ample ize of, the mea E [ X ] 0, the variace k 6 Var[ X ] for k, the kewe i 0 for k 3, ad the kurtoi i for k 4. k k 4 10

11 PSet Stat, Cofidece Iterval ad Etimatio Whe variace i ukow for X, X, 1, 1 X xi x 1 i1, the ubiaed variace etimator i: That i, E[ ] S, ad X1 X... X x, the etimate of tadard error of X N x N 1, i Whe the 10% rule i atified, the tadard error of x i x Note that the ample tadard deviatio etimator 1 1 x i x i1 i ot a ubiaed etimator of the populatio tadard error S. Whe the ample ize i le tha 30, ad/or i ukow, imilar to ue ormal ditributio, the mea etimator X ca be tudetized by the t-ditributio x ~ T ( 1), ad the cofidece level i determied by x 1 k t. The cofidece iterval i 1 1 CI x t, x t or 1 1 x t x t 11

12 PSet Stat, Cofidece Iterval ad Etimatio or x t 1 Example A ample of 8 THS teacher travel a average (mea) of 14.3 mile to chool. The tadard deviatio of their travel time wa mile. Fid the 95% cofidece iterval of true mea or populatio mea. Solutio: 7 8,, x 14. 3, 0. 05, k 8 1 7, t. 05, CI? t or CI 13.5, Compare the Differece of Two Mea with Uequal Kow Variace ad Large ample Size It i much more commo to compare the differece of two mea or proportio tha to etimate the mea or proportio themelve. For two ormal ditributio Nx1, 1 ad, 1 ad N x, whe 1, 30 ad variace are kow ad uequal, the differece of the two mea x x1to etimate the 1 differece of populatio mea 1 ha the ditributio N x1x,. The 1 cofidece iterval i 1 1 CI ( x1 x) z, ( x1 x) z 1 1 1

13 PSet Stat, Cofidece Iterval ad Etimatio or ( x1 x) z 1 1 Example A reearch team i itereted i the differece betwee erum uric acid level i patiet with ad without Dow' ydrome. A ample of 1 idividual with Dow' ydrome yielded a mea of x1 4.5 mg/100 ml. A ample of 15 ormal idividual of the ame age ad ex were foud to have a mea value of x 3.4 mg/100 ml. If it i reaoable to aume that the two populatio of value are ormally ditributed with variace equal to 1 ad 1.5, fid the 95 percet cofidece iterval for. 1 Solutio: 1, x 4.5, , x 3.4, x1 x ( x1 x) z (0.48) , or CI (0.6,1.94) 1 That i, the differece betwee the two populatio mea i 1.1 ad we are 95% cofidet that the true differece betwee the mea lie betwee 0.6 ad Compare the Differece of Two Mea with Small Sample Size ad/or with Ukow Equal Variace For ample ize 1, 30 ad/or with ukow equal variace, the pooled tadard error i ued for etimatig differece of populatio mea 1 from the differece of the two mea x1 x: p ( 1) ( 1)

14 PSet Stat, Cofidece Iterval ad Etimatio ad The cofidece iterval i x x ( ) 1 1 p T( ). 1 or CI ( x x ) t, ( x x ) t p 1 p ( ) 1 x1 x t p 1 Example A experimet wa doe to compare the mea umber of tapeworm i the tomach of heep that had bee treated for worm veru thoe ot treated. There were 7 heep i the treatmet group ad 7 i the cotrol group. The mea ad tadard deviatio are Treatmet Cotrol x What i the cofidece iterval for the differece of two mea at igificat level 0.1? Solutio: The ample ize i mall ad it i reaoable to aume the variace are equal. So the pooled etimate will be ued. x 8.57, x 40.0, T C 198.6, T 15.33, 1 7, 7, 0.1, C t t t

15 PSet Stat, Cofidece Iterval ad Etimatio The pooled tadard error i p ( 1 1) 1 ( 1) (7 1)(198.6) (7 1)(15.33) ( x x ) t 1 T C p ( ) 1.78(14.387) Compare the Differece of Two Proportio For two biomial ditributio b( 1, p1, x 1) ad b(, p, x ), whe 1 ad are le tha 10% of populatio ad 1 1, 1 (1 ˆ ˆ 1), p, (1 p) 10, the differece of the two proportio ˆ 1 pto etimate the differece of populatio mea 1(1 1) (1 ) p1 p ha the ditributio N ˆ 1 p, 1. The cofidece iterval i 1(1 1) (1 ) 1(1 1) ˆ (1 p) CI ˆ 1 p z, ˆ 1 p z 1 1 or 1 1 z 1 (1 ) (1 ) 1 Example Idepedet radom ample of 100 luxury car ad 50 o-luxury car i a certai city are examied to ee if they have bumper ticker. Of the 50 o-luxury car, 15 have bumper ticker ad of the 100 luxury car, 30 have bumper ticker. What i a 90% cofidece iterval for the differece i the proportio of o-luxury car with 15

16 PSet Stat, Cofidece Iterval ad Etimatio bumper ticker ad the proportio of luxury car with bumper ticker for the populatio of car repreeted by thee ample? Solutio: , 1 0.5, 100, 0.3, Aume that 1 ad are le tha 10% of total umber of luxury car ad o-luxury car, repectively. z z , 1 1, 1 (1 ˆ ˆ 1), p, (1 p) 10. 1(1 1) ˆ (1 p) ˆ 1 p z 1 0.5(1 0.5) 0.3(1 0.3) Summary ad Example Cae #1: ue the ample proportio to etimate the populatio proportio Give: the igificat level, the ample ize, the favorable outcome x Ditributio: b(, p, x ), aume the ample proportio follow the biomial ditributio. x Sample Statitic: Populatio Parameter: proportio p Cofidece Iterval (CI): z (1 ) Coditio: 10% of populatio, 10, ( 1 ) 10 Cae #: ue the ample mea to etimate the populatio mea Give: the igificat level, the ample ize, the ample mea x N x,, aume the ample mea follow the ormal ditributio. Ditributio: Sample Statitic: x Populatio Parameter: mea 16

17 PSet Stat, Cofidece Iterval ad Etimatio Cofidece Iterval (CI): x Coditio: 30, i kow z Cae #3: ue the ample mea to etimate the populatio mea Give: the igificat level, the ample ize, the ample mea x x Ditributio: ~ T ( 1), aume x follow the t ditributio with df 1. Sample Statitic: x Populatio Parameter: mea 1 Cofidece Iterval (CI): x t Coditio: 30, ad/or i ukow Ditributio b(, p, x ) N x, x x ~ T ( 1) N x1x, 1 1 x x ( ) 1 1 p T( ) 1 Sample Statitic ˆp Cofidece Iterval z x z x 1 x t x x x 1 x 1 (1 ) ( x1 x) z ( x x ) t p x, x are the ample mea Commet 1.) 10% of pop.) 10 3.) ( 1 ) 10 1.) i kow.) 30 1.) k 1deg. of freedom.) 5 30 or 3.) i ukow 1.) 1, are kow ad uequal..) 1, 30 1.) k 1 i deg. of freedom.) 5 1, 30 ad/or 1, are equal. 17

18 PSet Stat, Cofidece Iterval ad Etimatio p ( 1) ( 1) ad 1, are the ample deviatio 3.) 1, are ukow ad equal b(, p, x ) 1 1 z 1(1 ˆ ˆ 1) p(1 p) 1 1.) 1, 10% of pop.), 10, 1 1 (1 ) 10, 1 1 (1 ) 10 Example I wat to cotruct a 99% cofidece iterval for the proportio of America who thik that the govermet ha placed too may regulatio o buiee, ad I wat a margi of error of o more tha 3%. Aume the populatio proportio i 0.5. How large of a ample will thi require? Solutio: 0.01, p 0.5, z z , z p(1 p) 3%? 0.5(0.5).575(0.5).576 3% Example A tudy of 530 people aged 60 or older i US foud 14 with rheumatoid arthriti. Cotruct 90% cofidece iterval for the actual proportio of all people aged 60 ad older who have rheumatoid arthriti. Solutio: 0.1, Or , z z , 530, CI? 0.033( ) CI 0.01,

19 PSet Stat, Cofidece Iterval ad Etimatio Example [AP practice quetio, College Board] A large compay i coiderig opeig a frachie i St. Loui ad wat to etimate the mea houehold icome for the area uig a imple radom ample of the houehold. Baed o iformatio from a pilot tudy, the compay aume that the tadard deviatio of houehold icome i $7, 00. What i the leat umber of houehold that hould be urveyed to obtai a etimate that i withi $00 of the true mea houhold icome with 95 percet cofidece? Solutio: The variace i kow. 7, 00, 0.05, 1.96, z z z Example [AP practice quetio, College Board] Courtey ha cotructed a cricket out of paper ad rubber bad. Accordig to the itructio for makig the cricket, whe it jump it will lad o it feet half of the time ad o it back the other half of the time. I the 50 jump, Courtey cricket laded o it feet 35 time. I the ext 10 jump, it laded o it feet oly twice. Baed o thi experiece, Courtey ca coclude that (A) the cricket wa due to lad o it feet le tha half the time durig the fial 10 jump, ice it had haded too ofte o it feet durig the firt 50 jump. (B) a cofidece iterval for etimatig the cricket true probability of ladig o it feet i wider after the fial 10 jump tha it wa before the fial 10 jump. (C) a cofidece iterval for etimatig the cricket true probability of ladig o it feet after the fial 10 jump i exactly the ame a it wa before the fial 10 jump. (D) a cofidece iterval for etimatig the cricket true probability of ladig o it feet i more arrow after the fial 10 jump tha it wa before the fial 10 jump. 19

20 PSet Stat, Cofidece Iterval ad Etimatio (E) a cofidece iterval for etimatig the cricket true probability of ladig o it feet baed o the iitial 50 jump doe ot iclude 0., o there mut be a defect i the cricket cotructio to accout for the poor howig i the fial 10 jump. Solutio: The awer i D. The proportio i aumed to be p 0.5. The error term (width) of the cofidece iterval i calculated by p(1 p) z. So, whe the ample ize i icreaig, the error term will be decreaig. That i, the CI i arrowig whe ample ize i icreaig. Example [015 AP Stat FRQ, #] To icreae buie, the ower of a retaurat i ruig a promotio i which a cutomer bill ca be radomly elected to receive a dicout. Whe a cutomer bill i prited, a program i the cah regiter radomly determie whether the cutomer will receive a dicout o the bill. The program wa writte to geerate a dicout with a probability of 0., that i, givig 0 percet of the bill a dicout i the log ru. However, the ower i cocered that the program ha a mitake that reult i the program ot geeratig the iteded log-ru proportio of 0.. The ower elected a radom ample of bill ad foud that oly 15 percet of them received dicout. A cofidece iterval for p, the proportio of bill that will receive a dicout i the log ru, i All coditio for iferece were met. a.). Coider the cofidece iterval i. Doe the cofidece iterval provide covicig tatitical evidece that the program i ot workig a iteded? Jutify your awer. ii. Doe the cofidece iterval provide covicig tatitical evidece that the program geerate the dicout with a probability of 0.? Jutify your awer. A ecod radom ample of bill wa take that wa four time the ize of the origial ample. I the ecod ample, 15 percet of the bill received the dicout. b.) Determie the value of the margi of error baed o the ecod ample of bill that would be ued to compute a iterval for p with the ame cofidece level a that of the origial iterval. 0

21 PSet Stat, Cofidece Iterval ad Etimatio c) Baed o the margi of error i part (b) that wa obtaied from the ecod ample, what do you coclude about whether the program i workig a iteded? Jutify your awer. Solutio: a.) i. No. The aumed proportio i 0., ad it i withi the CI. So, there i o tatitical evidece to claim that the program i ot workig. ii. No. Ay umber withi CI could be the probability b.) c.) Now the CI i , o 0. i ot withi the CI. So, there i a covicig evidece that the program i ot workig. Example [011 AP Stat FRQ, #6] Every year, each tudet i a atioally repreetative ample i give tet i variou ubject. Recetly, a radom ample of 9,600 1 th -grade tudet from US were admiitered a multiple-choice US hitory exam. Oe of the multiple-choice quetio i below. (The correct awer i C.) Of the 9,600 tudet, 8 percet awered the multiple-choice quetio correctly. a.). Let p be the proportio of all Uited State twelfth-grade tudet who would awer the quetio correctly. Cotruct ad iterpret a 99 percet cofidece iterval for p. Aume that tudet who actually kow the correct awer have a 100 percet chace of awerig the quetio correctly, ad tudet who do ot kow the correct awer to the quetio gue completely at radom from amog the four optio. Let k repreet the proportio of all Uited State twelfth-grade tudet who actually kow the correct awer to the quetio. 1

22 PSet Stat, Cofidece Iterval ad Etimatio b.) A tree diagram of the poible outcome for a radomly elected twelfth-grade tudet i provided below. Write the correct probability i each of the five empty boxe. Some of the probabilitie may be expreio i term of k. c.) Baed o the completed tree diagram, expre the probability, i term of k, that a radomly elected twelfth-grade tudet would correctly awer the hitory quetio. d.) Uig your iterval from part (a) ad your awer to part (c), calculate ad iterpret a 99 percet cofidece iterval for k, the proportio of all Uited State twelfth-grade tudet who actually kow the awer to the hitory quetio. You may aume that the coditio for iferece for the cofidece iterval have bee checked ad verified. Solutio: a.). 0.80, 9600(0.80) , (1 ) 9600(1 0.80) , 0.01, z z , (1 ).

23 PSet Stat, Cofidece Iterval ad Etimatio z (1 ) (0.68, 0.9) 0.80(1 0.80) 9600 The CI idicate that 99% cofidece that the populatio proportio i betwee 0.68 ad 0.9. That i, we are 99 percet cofidet that the iterval from 0.68 to 0.9 cotai the populatio proportio of all Uited State twelfth-grade tudet who would awer thi quetio correctly. b.) c.) P( A _ Corr) P( Kow _ Corr) P( Gue _ Corr) k 0.5(1 k) k d.) Sice k, the 3

24 PSet Stat, Cofidece Iterval ad Etimatio k k We are 99 percet cofidet that the iterval from 0.04 to cotai the proportio of all Uited State twelfth-grade tudet who actually kow the awer to the hitory quetio. Example [013 AP Stat FRQ, #1] A evirometal group coducted a tudy to determie whether crow i a certai regio were igetig food cotaiig uhealthy level of lead. A biologit claified lead level greater tha 6.0 part per millio (ppm) a uhealthy. The lead level of a radom ample of 3 crow i the regio were meaured ad recorded. The data are how i the templot below. a.) What proportio of crow i the ample had lead level that are claified by the biologit a uhealthy? b.) The mea lead level of the 3 crow i the ample wa 4.90 ppm ad the tadard deviatio wa 1.1 ppm. Cotruct ad iterpret a 95 percet cofidece iterval for the mea lead level of crow i the regio. Solutio: a.)

25 PSet Stat, Cofidece Iterval ad Etimatio b.) x 4.90, 3 df 31, t df 3 t x t CI (4.417, 5.383) We ca be 95% cofidet that the populatio mea lead level amog all crow i thi regio i betwee ad part per millio 5

26 PSet Stat, Cofidece Iterval ad Etimatio Quiz -- Cofidece Iterval 6