, the random variable. and a sample size over the y-values 0:1:10.

Transcription

1 Lecture 3 (4//9) 000 HW PROBLEM 3(5pts) The estimatr i (c) f PROBLEM, p 000, where { } ~ iid bimial(,, is 000 e f the mst ppular statistics It is the estimatr f the ppulati prprti I PROBLEM we used simulatis t arrive at its prbability structure I this prblem we will address its structure usig thery We begi with the fllwig FACT: Fr { } ~ iid bimial(,, the radm variable Y ~ bimial (, (a)(4pts) Fr a true prprti p 000 use the Matlab bipdf t btai a plt f ad a sample size ( y) f Y 000, ver the y-values 0::0 Sluti: [See 3(a)] Figure 3(a) Bimial(=000,p=00) pdf (b)(4pts) Use the ifrmati i yur plt i (a) [r yur umerical values fr ] t arrive at the umerical value f Pr[ Y 3] The cmpare this t the value fr Pr[ p 0003] that yu btaied i (e) Sluti: Prc=-sum(fY(:3))= 0333 My value i (e) is 0355 These values are early idetical f Y ( y) (c)(4pts) The radm variable ~ bimial (, with sample space S {0, } is als called a Berulli ( radm variable Recall the fllwig: Defiiti: E [ g( )] g( x) f ( x) dx Use this defiiti t shw that p fr ~ Berulli( [NOTE: Sice S {0, }, the frmal itegrati becmes: g( x)pr[ x] ] Sluti: E( ) x0 x Pr[ x] 0( ( p (d)(3pts) Use the defiiti i (d) t shw that p( Sluti: E[( ) ] ( x Pr[ x] ( ( ( ( p( x0 S 000 (e)(0pts) Recall that p T cmpute the umerical values fr p ad p usig the theretical expressis 000 give i (d) ad (e), utilize the fllwig tw facts fr ay tw radm variables ad Y: FACT : E( a by ) ae( ) be( FACT : If ad Y are idepedet, the ar( a b a ar( ) b ar( (i) Cmpute these values fr p 0 00 The (ii) cmpare these results t yur simulati-based results i (d) Sluti: p E( p ) E 000 E( ) p p x p( 00(998) 4 p ar( ar ( ) ( ) 00996( ar 000 p p Hece, σ p 0004 These values are idetical t thse I btaied i (d) )

2 Exam : PROBLEM (f)(7pts) T uderstad the distributi f ˆ, e culd ru a large umber f simulatis f it Hwever, i (e) f HOMEWORK we have the fllwig facts: FACT : E( a by ) ae( ) be( FACT : If ad Y are idepedet, the ar( a b a ar( ) b ar( Use (a)(7pts) Fr each plat, fid the prbability that ay give tractr will eed rewr Sluti: (i) Pr[ C 40] Pr[ C 60] = rmcdf(40,50,3) + -rmcdf(60,50,3) = 858e-04 these facts t shw that (i) ˆ, ad (ii) ˆ / 0 [Shw all steps] Sluti: (i) ˆ E ( ) 0 E 0 0 (ii) ˆ ar ( ) /0 0 ar 0 0 (g)(3pts) Suppse that data frm the give site yields cmpute the estimated ucertaity iterval ˆ ˆ ˆ ˆ 370 ad Sluti: ˆ ˆ 4 / ˆ ˆ ˆ [8, 46] ˆ 4 Use the latter umber ad (f-ii) t PROBLEM 4: (a)(7pts) Fr each plat, fid the prbability that ay give tractr will eed rewr Sluti: (i) Pr[ C 40] Pr[ C 60] = rmcdf(40,50,3) + -rmcdf(60,50,3) = 858e-04 HOMEWORK 3: PROBLEM (0pts) Ofte times it is t s easy t determie whether r t behavir is s ut f the rm as t warrat sudig a alarm Such behavir ca be i relati t helicpter vibrati, heart rate, r eve the behavir f a lved e Suppse that the etity s behavir will be deemed t be rmal if a give metric,, is i the rage [30,70] Furthermre, based histrical data, it is deemed that ~ N( 50, 0) (a)(4pts) Fid the prbability that rmal behavir will be deemed t be abrmal This prbability is called the false alarm prbability, sice it is the prbability that e wuld icrrectly presume abrmal behavir Sluti: [Cpy/paste cde HERE] Pra=rmcdf(30,50,0) + (-rmcdf(70,50,0)) = (b)(4pts) Suppse that give frm f abrmal behavir crrespds t Y ~ N( Y 60, 0) Fid the prbability that it will falsely deemed t be rmal behavir Sluti: [Cpy/paste cde HERE] Prb=rmcdf(70,60,0)-rmcdf(30,60,0) = (c)(4pts) Suppse that give frm f abrmal behavir crrespds t Y ~ N( Y 75, 0) Fid the prbability that it will falsely deemed t be rmal behavir Sluti: [Cpy/paste cde HERE] Prc=rmcdf(70,75,0)-rmcdf(30,75,0) = 03085

3 3 (d)(9pts) (i) Use the Matlab cmmad gampdf t btai a plt f the pdf fr (t Q) ver the rage [0,00] The (ii) use gamiv t fid the iterval [ x, x ], such that Pr[ x,] Pr[ x,] 0 05 This iterval is called the p=095 tw-sided iterval fr NOTE: Regardless f yur aswers i (c) use parameter values a=8 ad b=5 Sluti : (i): See 3(d) (ii): q = gamiv(05,8,5) = 408 S: x 4 q = gamiv(975,8,5) =74578 S: x 746 Figure 3(d) Plt f f (x) HOMEWORK 4 PROBLEM (0pts) Let Y U dete the act f recrdig the actual weight f a pacage The radm variable relates t the measured weight f the pacage, while the radm variable U relates t measuremet errr assciated with the particular scale beig used t measure weight Suppse that ~ N( 5, 0) ad U ~ N( U 0, U 005) (a)(3pts) Cmpute [See Fact i Appedix ] Y Sluti: Y U (b)(3pts) Recall that if ad U are idepedet f each ther the ar( ar( ) ar( U) Assumig that they are, ideed, idepedet, cmpute Y [See Fact 3 i Appedix ] Sluti: Hece, 08 Y U Y PROBLEM 3(0pts) This prblem cdeses much f the thery we will eed fr the ext few wees [See Appedix ] It is all based a sigle ccept; amely the liearity f E(*): E( a by c) ae( ) be( Y ) c (a)(4pts) Recall the defiiti f variace: E[( ) ] Use the liearity f E(*) t shw that E ( ) Sluti: E[( ) ] E( ) E( ) E( ) E( ) Hece: E ( ) (b)(4pts) The cvariace betwee ad Y is defied as: Y E[( )( Y Y )] Use liearity f E(*) t shw that Y E( Y ) Y Sluti: E[( )( Y )] E( Y Y ) E( Y ) E( E( ) E( Y ) Y Y Y Y Y Y Y (c)(4pts) Let W a by c (i) Use liearity f E(*) t shw that W a by c The (ii) use this ad liearity f E(*) t shw that W a b Y ab Y Sluti: (i): Frm liearity f E(*): E( W ) E( a by c) E( a ) E( by ) c) ae( ) be( Y ) c (ii): E[( W ) ] E{[( a by c) ( a b c)] } E{[ a( ) b( Y )] } This gives: W W Y Y E[ a ( ) b ( Y ) ab( )( Y )] Usig liearity f E(*) gives: W Y Y a E[( ) ] b E[( Y ) ] abe[( )( Y )] a b ab W Y Y Y Y

4 4 PROBLEM 4 (d)(5pts) I the results widw that yu pasted i (c), yu are give what are called 95% cfidece buds i relati t each parameter These buds are estimates f the rage abut each estimate Fr each parameter, recver the assciated estimate f The cmpare it t the value f that yu btaied frm simulatis i (c) Sluti: Frm the Results bx: m ad b Frm simulatis: ad m 005 b 7 PROBLEM 5(5pts) This prblem addresses what is w as: The Cetral Limit Therem: Let the radm variables be iid with mea ad variace The atural estimatr f is ( ) ( ) (/ ) Defie where ( ) The Z D Z ~ N(0, ) { Z } { } (a)(3pts) The tati abve differs frm that used i equati (7-) p43 f the b Oe differece is the use f the tati The authrs use Hwever, i view f the abve, these are clearly e ad the same We use t emphasize the fact that it is a estimatr f the uw parameter Use liearity f E(*) t shw that E( ( ) ) Sluti: ( ) E( ) E(/ ) E (/ ) ( ) (/ ) () (b)(3pts) The ther differece is ur use f Z ( ) ( ), as ppsed t the authrs use f / Use the geeral relati (5-7) p85 f the b (r the apprpriate fact i the table i Appedix ) t shw that / Sluti: ( ar ( ) ) ar (/ ) (/ ) ar ( ) (/ ) ( ) (/ ) ( ) / S: ( ) / () (c)(4pts) As a applicati f this, let =the act f detectig a geetic marer i a strad f DNA with Pr( ) p 00 The fr strads f DNA the average umber f detected marers ( ) is a scaled bimial radm variable with parameters ad p 0 0 Frm the abve results, it fllws that has mea ( ) p ad stadard deviati p( / Fid the value f such that ( ) 3 ( ) 0 ( ) Sluti: 3 p 3 p( / 9(/p 89 0 ( ) ( ) () (d)(3pts) T shw that 000 is large eugh t claim that the CLT hlds (ie ( ) ~ N( p, p( / ), use Matlab s bipdf cmmad t () btai a stem plt fr the pdf fr [Cmpute this ly ver the values x 0:: /50, ad mae sure that yu scale these values whe yu plt it] Sluti [See (g)] () Figure 5(d)) Plt f the pdf fr (h)(pts) Yur plt i (d) shuld have the verall shape f a bell curve Hece, e might reasably argue that the sample size 000is large eugh t claim ( ) ~ N( p, p( / ) fr p 0 0 Eve s, it shuld be clear frm yur () Figure 5(d) that has a discrete distributi Hece, while the CLT will be a gd apprximati fr the prbability f a evet that icludes a sizeable umber f lumps, it will be a pr e fr small itervals, such as itervals that ctai ly a sigle lump Fr example, the lump f prbability at the value 00 is 057 Assumig that

5 5 ( ) ( ) ~ N( p, p( / ), cmpute the prbability f ], ad the resultig % errr Sluti: muphat=p; stdphat=sqrt(p*(-/); Prd=rmcdf(008,muphat,stdphat)-rmcdf(009,muphat,stdphat)=007 % errr = +60% Appedix Useful Facts Sme Ptetially Useful Defiitis ad Facts Related t Radm ariables ad Y: Defiiti : If ad Y are idepedet, the f ( x, y) f ( x) f ( ) (, Y ) Y y Defiiti If ad Y are ucrrelated, the E( E( ) E( Defiiti 3 The crrelati cefficiet betwee ad Y is Y Y /( Y ) Defiiti 4 Fr ay fucti g( x, y), g(, Y )] E ( x, y) dxdy [ g( x, y) f (, Y ) SY S Fact E( a by c)) ae( ) be( c (ie E(*) is a liear perati) Fact Cv( a by c, W) a W b YW Fact 3 If ad Y are ucrrelated, the ar a by c) a ( b Y [I geeral: ar( a by c) a b ab ] Fact 4 E ) ( Y Y EAM PROBLEM 3(5pts) The perfrmace f a wid turbie is tied directly t the velcity f the wid impigig the blades at ay give time Utilizati f wid speed requires that it be measured at discrete times Let (,, ) dete the act f measurig wid speed at successive time itervals Δ secds apart frm e ather (a)(4pts) Let (/ ) Use the liearity f E(*) t shw that E( ) (/ ) Recall that this liearity icludes the fllwig tw prperties: (i) E( c ) ce( ), ad (ii) E( E( ) E( I each step f yur sluti, idetify which f e these prperties is beig applied Sluti: ( i) ( ii) E( ) E(/ ) (/ ) E (/ ) E( ) (/ ) W (b)(3pts) Assume that Δ is sufficietly large, s that the elemets f ca be assumed t be mutually ucrrelated O p85 f the b, (i) give the equati that will allw yu t the (ii) shw that ar ( ) (/ ) Sluti: (i): O p85: fr Y c, we have ar ( car ( ) (ii): ar ( W ) ar (/ ) ar (/ ) (/ ) ar ( ) (/ ) ( ar ( ) (/ )

6 6 NOTE: Frm this pit, we will assume that the elemets f are idetically distributed Hece, frm (a) it shuld be clear that E( ) Frm (b) it shuld als be clear that, if the elemets f ca be assumed t be mutually idepedet, the ar( ) / (c)(3pts) Suppse that at ay time idex, we have ~ Weibull( a 37, b 9) [A Ggle search will reveal that the Weibull pdf is the pdf f chice i relati t mdelig wid] These parameter values crrespd t 30 mph ad 4 mph Suppse further, that the time betwee successive measuremets is 60sec This samplig iterval is sufficietly large that we ca assume that the elemets f ca be assumed t be mutually ucrrelated The fr a 0 miute bservati time we have a ttal f 0 measuremets Use the abve NOTE t cmpute the umerical values f the mea ad stadard deviati f the sample mea Sluti: E( ) 30mph & Std ( ) / 4 / 0 65 mph (d)(6pts) The pdf fr ay is give shw at right Clearly, it des t have the symmetry f the rmal pdf The Cetral Limit Therem (CLT) says that if the sample size,, is sufficietly large, the matter what the pdf might be fr, the pdf fr will have a pdf that is apprximately rmal This issue here is that e might claim that 0is NOT AT ALL large eugh fr this assumpti t hld (i) Use 0 5 simulatis f t arrive at the simulati-based pdf verlaid agaist a rmal pdf fr The (ii) use yur results t determie whether yu thi the CLT is applicable Sluti:[See Determiati: There is a slight differece, but verall, I thi that the CLT hlds reasably well Figure 3(d) Simulati-based ad rmal mdel pdf fr PROBLEM 4 (a)(3pts) It was reprted that ˆ ˆ Y 53 Assumig that 5 / ad Y 7 / 30 8, cmpute the 3 value fr ˆ ˆ Y Sluti: ar( ˆ ˆ ) ar( ˆ ) ar( ˆ ) /00 /00 (9 8 ) / Hece: Y Y Y

7 7 Hmewr 5 STAT 305D Sprig 09 Due 4/(R) SOLUTION PROBLEM (5pts) (a)(8pts) Let ~ Ber( p ) This is e f the mst cmm radm variables i statistics It crrespds t ay biary evets (eg yes/, gd/bad/ failure/success, etc) Recall that ~ (/ ) bi(,, with ad p / p( / I rder t ive the CLT it is reasable t require that 3 0(ie esure that there is miimal prbability that is less tha zer; which is impssible) (i) Arrive at the expressi fr as a fucti f p fr this cditi t hld (ii) Plt it fr p=0: 00: 099 (iii) Use the data cursr t estimate the p-value fr =30 Sluti: [See (a)] ( ) (i): p p p 9( Hece: (iii): p ( 30) 03 p Figure (a) Plt f vs p fr 3 0 NOTHING NEW HERE (b)(pts) Yu shuld have fud that fr p 05, the sample size 30 is sufficiet fr 3 0t hld Nw, we w that ~ (/ 30) bi( 30, p 05) Assumig the CLT hlds, we have ~ N (05,0079) (i) Obtai a stem plt f the pdf fr ~ (/ 30) bi( 30, p 05) (ii) Use the data cursr t fid the lcati,, ad crrespdig value f that has the largest prbability (iii) Overlay the pdf fr ~ N (05,0079) x 0 (iv) Frm ~ N (05,0079) cmpute Pr[ x0/ 30 x0 / 30] (v) Cmpute the percet errr f the CLT-based ad true prbability Sluti: [See (b)] (ii): Pr[ 03] 066 ; (iv): p0 = 0399 ; err =95% Figure (b) Plts fr the scaled bi ad rmal pdf s NOTHING NEW HERE (c)(5pts) Cmpute percet errr f the CLT-based ad true Pr[ x0/ 30 x0 / 30] Sluti: [See (c)] Pr[ x0/ 30 x0/ 30] Pr[6 bi(30,5) 8] 47 ad Pr[ x0/ 30 rm x0/ 30] 399 : err= -3% NOTHING NEW HERE

8 8 PROBLEM (5pts) I may maufacturig ad prducti peratis the bias is usually the cause f pr quality Csider the simple act f cuttig a piece f wd Ideally, the cut shuld be exactly perpedicular (ie ) Suppse that is w Cuttig bias ccurs whe Sice is, i fact, uw, it is essetial t peridically estimate it, t esure that thigs are t gettig ut f had 0 scraps f wd, (i) use the Matlab cmmad rmrd t simulate measuremets f (a)(5pts) Fr 0 { } ~ iid N( 0, 05 ) The (ii) cmpute a 95% -sided Cfidece Iterval (CI) estimate fr Specifically, cmplete the fllwig steps: Step : Idetify a estimatr,, fr the uw parameter : Aswer: [Yes, this was itetial] NOTHING NEW HERE Step : Frm the Hady-Dady Table i App idetify a apprpriate test statistic, W, that icludes Aswer: W Z ~ N(0,) / 0 What is ew here is (i) the jarg (ie a test statistic) ad a table t chse frm Step 3: Fr a 95% CI, set 005 Pr[ w W ] Pr[ W w ], ad the recver values fr the pe iterval [Nte: Yu d t eed t use the symbl W It is better t use a symbl that describes yur particular test statistic] Sluti: 05 Pr[ Z z ] Pr[ Z ] z=rmiv(05,0,) = -96 ; z=rmiv(975,0,) =96 0 z ONCE YOU IDENTIFY Z THERE IS NOTHING NEW HERE ad ( w, w ) Step 4: Cvert the evet [ w W w ] it the equivalet evet [ c ( ) c( )] This is the 95% CI fr Sluti: [ z Z z] z z z z z z z z Isertig z, 96 ad / % CI gives: NOTHING NEW HERE We have bee discussig equivalet evets sice wee # This is algebra QUESTION: Is this really ew? ANSWER: Csider PROBLEM (g) f HW3: Suppse that data frm the give site yields ˆ 370 ad 4 Use the latter umber ad (f-ii) t cmpute the estimated ucertaity iterval ˆ ˆ ˆ Sluti: ˆ ˆ 4 / ˆ ˆ ˆ [8, 46] ˆ NOTICE that i this prblem we had ˆ ˆ ˆ whereas i the abve we have z 96 z Step 5: Cmpute the CI estimate by substitutig the value fr it the CI i Step 4 This is the CI estimate fr Sluti: Frm the (a) I btaied: Hece: 95%CI (b)(4pts) Per p75 i the b, the term precisi is defied as e-half f the width f the CI= [ c ( ) c( )] : 5[ c ( ) c ( )] Fr a precisi 0, use yur expressi i Step 4 t fid the apprpriate sample size 0

9 9 Sluti: 05( ) 05 96(05 ) 0 05 z z z z Hece, 96(05 ) THIS IS ALGEBRA NOTHING NEW (c)(6pts) O p74 the authrs emphasize that the CI fr is a radm iterval I view f this, fr each f the fllwig give the rigial equati ad yur crrected equati: (i) (8-5), (ii) (8-), ad (iii) (8-) Aswer: (i): (8-5): x z/ / x z/ / Crrected: z/ / z/ / (ii): (8-): x z/ s / x z/s / Crrected: z/ S / z/s / (iii): (8-): ˆ z ˆ ˆ z ˆ Crrected: ˆ z ˆ z ˆ ˆ / / / / WE HAE EMPHASIZED LOWER S UPPER CASE SINCE WEEK NOTHING NEW HERE PROBLEM 3(5pts) Rughess data fr a sample size 5 gave m ad 7 5 m (a)(0pts) Use the apprpriate statistic i Appedix t arrive at a 90% -sided CI estimate fr Sluti: * * * 09 Pr t T4 t t tiv(95,4) 709 / * * * * Hece, t t / / t t / 30377(75) / WHAT IS NEW HERE IS THE INTRODUCTION TO t AS OPPOSED TO Z (b)(0pts) Repeat (a) i relati t ( ) Sluti: 09 Prx ~ x x chiiv(05,4) 3848 & x chiiv(95,4) 3645 ( ) x x ( ) ( ) x x Hece: ( ) ( ) x x ( ) ( ) x x WHAT IS NEW HERE IS THE INTRODUCTION TO χ AS OPPOSED TO Z

10 0 (c)(5pts) T better appreciate the ccept f a CI, assume the truth mdel ~ N( 3037, 75) Fr sample 5 size, simulate ad plt 00 CI estimates fr each f ad I each plt draw a hriztal lie crrespdig t the true value f the parameter Fially give the fracti f CI estimates that iclude that parameter Sluti: [See 3(c)] The fracti f CIs icludig is 084, ad the fracti f CIs icludig is 090 Figure 3(c) Plts f CI estimates fr (left) ad fr (right) NOTHING REALLY NEW HERE; ONLY THE APPLICATION

11 PROBLEM 4(5pts) Let =The act f measurig the time required t cmplete a test cde used i a ew peratig system Assume that ~ N(, ) where 05msec is w, but where is uw After ruig the cde m ad 03msec (ie 05msec was used) 00 times, it was bserved that 0505 sec (a)(5pts) Use the apprpriate frmula i APPENDI t arrive at a 95% -sided CI estimate fr Sluti: The apprpriate frmula is: (C): / ~ whe is used [ie ( ) ] Step : x=chiiv(05,00) = 749 ; x=chiiv(975,00) = 956 NOTHING NEW ECEPT χ AS OPPOSED TO Z Step : [ x x] [ x / x] [ x / / x / ] [ / x / x ] [ / x / x ] Hece, the CI fr is the radm iterval [ / x, / x], s that the CI estimate is [ / x, / x ] [03 00 /956, / 749 = [ 060, 053 ] (b)(5pts) Yur bss des t uderstad the use f, ad wats yu t use ˆ istead But yu lger have the rigial data Shw that ( ) ( ˆ ) ( ˆ ) Sluti: [( ˆ ) ( ˆ )] ( ˆ ) ( ˆ ) ( ˆ )( ˆ ) where ( ˆ ) ( ˆ )( ˆ ) ( ˆ ) Sice ( ˆ ˆ ) 0 fllws, the result NOTHING NEW HERE THIS IS ALGEBRA (WITH A TRICK GIEN BY THE HINT) 00 (c)(5pts) Frm (b), we have: ( ˆ ) ( ˆ ) , r Arrive at the apprpriate 95% -sided CI estimate fr Sluti: (D): ( ) / ~ ; x=chiiv(05,99) = 7336 ; x=chiiv(975,00) = 840 [ ( ) / x, ( ) / x ] [ /840, / 7336] = [ 065, 054 ] NOTHING NEW ECEPT χ AS OPPOSED TO Z THE APPLICATION IS NEW SUMMARY: THE FORMAL CONCEPT OF A TEST STATISTIC AND THE DEFINITION OF A CONFIDENCE INTERAL ARE NEW HOWEER, THESE ARE NOT NEW CONCEPTS THEY ARE AN APPLICATION OF CONCEPTS THAT WE HAE BEEN ADDRESSING THROUGHOUT THE COURSE