7818 Interval estimation and hypothesis testing - Set

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1 Iterval estimatio ad hypothesis testig - Set revised Nov 9, 010 You might wat to read some of the chapter i MGB o Parametric Iterval Estimatio. There are subtle di ereces across questios. uderstad those di ereces. Make sure to look for ad 1. Assume the radom variable X is ormally distributed with ukow mea ad variace 16; that is f X ( : ) = ;16 (). Let X 1 ; X ; :::; X be a radom sample from this populatio. Determie Pr[X < < X + ]. aswer: we kow that Z = X = p is ormally distributed with mea zero ad variace of oe, so has o parameters. I this case, Z = X i this case Pr[X < < X + ] = Pr[(X ) ] < 0 < (X ) + ] (X ) = Pr[ 4= p = Pr[ = Pr[ 4= p = Pr[ 4= p (X ) < 0 < 4= p + 4= p ] 4= p (X ) < 4= p < + 4= p ] > (X ) 4= p > 4= p 4= p < Z < + 4= p ] 4= p.1 So, This probability depeds o the sample size, but ot o. For ay sample size, we could look up the aswer i the stadard-ormal table. For eample, if = 9, Pr[ 4= p < Z < + 4= p ] = Pr[ 4= p < Z < + 9 4= p ] = 9 Pr[ : < Z < :] = NormalDist(:) NormalDist( :) = 0:97. What does this mea? 97:% of the radom itervals (X ) to (X +) will cotai, givig us co dece that lies i this iterval. Note that this iterval varies with X, so is radom. If = 100, Pr[ 4= p < Z < + 4= p ] = Pr[ 4= p 100 < Z < + 4= p 100 ] = Pr[ 7: < Z < 7:] = NormalDist(7:) NormalDist( 7:) = 1: ad e ectively 100% of these radom itervals will cotai. 1 While appears i Z, Z is uit ormal for all values of. Further ote the sig cace of kowig. 1

2 . Assume the radom variable X is ormally distributed with ukow mea ad variace 16; that is f X ( : ) = ;16 (). Let X 1 ; X ; :::; X be a radom sample from this populatio. Determie Pr[X < < X 1]. aswer: we kow that Z = X = p is ormally distributed with mea zero ad variace of oe, ad has o parameters. I this case, Z = X 4= p. So, i this case Pr[X < < X 1] = Pr[(X ) < 0 < (X ) 1] (X ) = Pr[ 4= p 4= p (X ) < 0 < 4= p = Pr[ 4= p (X ) < 4= p < 1 4= p ] 1 = Pr[ 4= p < Z < + 4= p ] 1 4= p ] This probability depeds o the sample size, but ot o. For ay sample size, we could look up the aswer i the stadard-ormal table. For 1 eample, if = 9, Pr[ 4= p < Z < + 9 4= p ] = Pr[:7 < Z < +:] = 9 NormalDist(:) NormalDist(:7) = 0:14 4 What does this mea? 1:44% of the radom itervals (X ) to (X 1) will cotai. I this problem, we have derived a co dece iterval o a parameter,.. Assume the radom variable X is ormally distributed with ukow mea ad variace 16; that is f X ( : ) = ;16 (). Let X 1 ; X ; :::; X be a radom sample from this populatio. Specify three di eret :4 co- dece itervals for i terms of the sample average, X. aswer: we kow that Z = X = p is ormally distributed with mea zero ad variace of oe, ad has o parameters. I this case, Z = X 4= p. NormalDist(Z) NormalDist( Z) = :4, Solutio is: f[z = 0:4 4]g. So, oe rage o Z to get 40% of the distributio is 0:44 to 0:44, symmetrical aroud zero. NormalDist(Z) NormalDist(0) = :4, Solutio is: f[z = 1: 81 6]g, So, aother rage o Z to get 40% of the distributio is 0 to 1:816 NormalDist(Z) NormalDist(:1) = :4, Solutio is: f[z = 1: ]g,, So, aother rage o Z to get 40% of the distributio is 0:1 to 1: Note that these three rages o Z are of icreasig legth, ad oly the rst oe is symmetric aroud zero. Now we eed to covert these rages

3 i tems of Z ito rages i terms of X. Start with the rst. :4 = Pr[ :44 < Z < :44] = Pr[ :44 < X 4= p < :44] = Pr[ :44(4= p ) < X < :44(4= p )] = Pr[ X :44(4= p ) < < X + :44(4= p )] = Pr[X :44(4= p ) < < X + :44(4= p )] = Pr[X :0976= p < < X + :0976= p ] So, 40% of the radom itervals X :0976= p to X + :0976= p will iclude. Now cosider the secod 40% rage o Z :4 = Pr[0 < Z < 1:816] = Pr[0 < X 4= p < 1:816] = Pr[0 < X < 1:816(4= p )] = Pr[ X < < X + 1:816(4= p )] = Pr[X 1:816(4= p ) < < X] = Pr[X :164= p < < X] So, 40% of the radom itervals X cosider the third 40% rage o Z :164= p to X will iclude. Now :4 = Pr[ :1 < Z < 1:] = Pr[ :1 < X 4= p < 1:] = Pr[ :1(4= p ) < X < 1:(4= p )] = Pr[ X :1(4= p ) < < X + 1:(4= p )] = Pr[X 1:(4= p ) < < X + :1(4= p )] = Pr[X 6: 1 = p < < X + :4= p ] So, 40% of the radom itervals X. 6: 1 = p to X +:4= p will iclude Note that i this problem we have ideti ed three di eret :4 co dece iterval o the parameter, u. 4. Let G represet gubers, where G is a radom variable with f G (g) = :1 if 0 g 4 : if 4 < g 6 0 otherwise

4 Charlotte tells you that she has radomly sampled oe G from this distributio, ad its realized value is 7. How likely is it that she is either mistake or lyig? Jesse tells you that he has radomly sampled oe G from this distributio, ad its realized value is :6. How likely is it oe will get a radom draw from this distributio that is greater tha or equal to :6 (what is the probability of drawig a g :6 if the true distributio is f G (g)? Eplai. aswer: The 7 could ot have come from this populatio. The :6 could have. Picture f G (g) f(g) % of the desity is i the rage 4 < g 6. Break this up ito equal rages with 10% i each: 4 < g 4:4, 4:4 g 4:8, 4:8 g :, : g :6, ad :6 g 6:0. So there is a 10% chace of radomly drawig a g :6 from f G (g). Said the other way, there is a 90% chace that the draw g is betwee 0 ad :6. If oe set her level of sigi cace at :10 oe would reject, just, the ull hypothesis that :6 is a radom draw from f G (g). If oe set her sigi cace level at :0, oe would ot reject the ull hypothesis.. Cosider agai, the old issue of how may time oe gets married. Assume agai that the umber of times oe gets married, M, has a Poisso distributio. Further assume you have radomly sampled oe member of the populatio, Ralph, ad that he has bee married oe time. Fid a % co dece iterval for. You choose. g 4

5 aswer: So what do we kow? The Poisso desity is f M (m) = e m m! for iteger values of m = 0; 1; ; ::. With a sample of oe idividual married oce, the maimum likelihood estimate of is 1. Ad the estimated desity fuctio for M is f M (m) = e 1 1 m m! = 1 m! e 1 = Pr[M = m]. Cosider, what this meas, the probability of ever marryig is PoissoDe (0; 1) = e 1 = 0: Oce is PoissoDe (1; 1) = 0:67 88 ad twice is PoissoDe (; 1) = 0:18 94 The estimated CDF is PoissoDist (; ) = P k e k=0 k!. So, for eample, the estimated probability of beig married oe time or less is PoissoDist (1; 1) = 0:7 76. Now that we have a feel for thigs, let s see how the probability of beig married oe time or less varies with. If, for eample, our estimate of was 4, the estimated probability of beig married oe time or less is PoissoDist (1; 4) = 9: = :09, a little over 9%. Now let s ask aother questio. What is the probability that oe will be married oe or more times as a fuctio of. It is 1 PoissoDist (1; ). For eample the probability of beig married oe or more time if is : is 1 PoissoDist (1; :) = 9: = :09, 9%. Flippig these questios, what would the estimated have to be for the estimated probability of someoe beig married oe or less times be :0? I kept playig aroud with values of i PoissoDe (1; ) util I stumbled o PoissoDe (1; 4:) = 0: That is, if is 4: there is oly a % chace that oe will be married oe or less times, pretty low. Now let s gure out what would have to be for :0 to be the probability of beig married oe or more times. We wat to gure what value of will make 1 PoissoDe (1; ) = 0:0. Tryig di eret umbers I got 1 PoissoDist (1; :6) = 0:01 16 So, what has bee determiesd? A 90% co dece iterval fro is :6 to 4:. Note that this iterval is a rv; it varys from sample to sample, holdig costat the umber of observatios at oe. Iterpretig, 90% of these estimated itervals will cotai. What else ca we say about itervals. A 9% co dece iterval for is :6, aother is 4:. A % co dece iterval is 4: aother is :6. To make sure you uderstad, redo this problem assumig Ralph was married twice. 6. Redo the previous problem assumig the sample has two rather tha oe observatio. 7. As P [X y] decreases it becomes less ad less likely that y is a radom draw from the populatio described by f X (; ; ). Eplai to me why this is a reasoable way to decide whether y is a radom draw from populatio described by f X (; ; ).

6 8. For fu, calculate the 9% ad 99% co dece itervals for X for a few di eret speci c speci catios of f X (; ; ). For eample, do it assumig X has a Poisso distributio with a mea of (ad a mea of 6), or assumig X has a t distributio with some speci c parameter. Cosider both oe-sided ad two-sided itervals. (Note that the Poisso is a discrete distributio, so the co dece iterval will be a ite set of itegers that spa some rage.) (Further ote that i this problem we are derivig co dece itervals o X, ot co dece itervals o a parameter.) 9. Grade poit average i the Departmet s udergraduate math-eco course is cotiuously distributed betwee 0 ad 4, where 0 is a F ad 4 is a A. The distributio of grades i Weird Shirley s sectio is always symmetric ad bowl shaped ad we kow its fuctioal form. Put simply, Shirley gives lots of high grades, lots of low grades ad few grades i betwee. Imagie that oe radomly draws the grade for oe udergraduate i oe udergraduate ecoomics course. Discuss i geeral terms how you would decide/test whether this grade was a grade i Weird Shirley s math-eco class. Discuss i geeral terms why ad how what you would do di ers from a case where Shirley s grades are ormally distributed. aswer: The ull hypothesis would be the grade is from Shirley s class. The more the grade is i the middle of the rage, the greater reaso for thikig the grade is ot from Shirley s class. Oe might reject the ull if the probability that the grade was give by Shirley is less tha %. I would d the the GPA rage for Shirley s classes such that % of the grades were i that rage ad the midpoit of that rage was the average grade i Shirley s classes,. If the kid s grade was i this rage, I would reject the ull that he was i Shirley s class. Here the critical regio is i the middle, we are accustomed to seeig the critical rage at oe or both eds of the distributio. I more detail: the co dece iterval is two itervals: a iterval to the right of the mea ad etedig to 4 ad a iterval to the left of the mea etedig to 0. We wat to choose these two itervals such that they are symmetric (the distributio is symmetric) ad cover 9% of the desity. Ad subject to these two costraits, the legth of these two itervals together is miimized. 10. Assume X 1 ; X ; :::; X is a radom sample from a ormal distributio with ukow but kow. Fid a 9% co dece iterval for i terms of X ad iterpret this iterval. p aswer: (See MGB page 74 ad page 81) Cosider the statistic S = X, where p is the stadard deviatio of X. If X is ormally distributed the X p has a stadard ormal distributio, ad Pr " # 1:96 < X p < 1:96 = :9 6

7 Rearragig (pivotig aroud ) " # :9 = Pr 1:96 < X p < 1:96 = Pr 1:96 p < X < 1:96 p = Pr X 1:96 p < < X + 1:96 p = Pr X + 1:96 p > > X 1:96 p = Pr X 1:96 p < < X + 1:96 p So, what does this mea? the iterval X 1:96 p to X + 1:96 p varies from sample to sample ad 9% of these itervals will cotai. For a give sample, 1:96 p to + 1:96 p is a umerical rage (e.g. to 8) ad it highly likely that will fall i this rage, but ot for sure. (As a aside, ote the distictio betwee X p ad X ad ote that both have a stadard ormal distributio if X is ormally distributed.) 11. Assume that X is ormally distributed with kow ad kow. Determie ad iterpret the 9% co dece iterval for X. aswer: If X is ormally distributed the X distributio. So Pr 1:96 < X < 1:96 = :9 Pivotig this aroud X :9 = Pr 1:96 < X < 1:96 = Pr [ 1:96 < X < 1:96 ] = Pr [ 1:96 < X < + 1:96 ] has a stadard-ormal The rage 1:96 to +1:96 is a speci c umerical rage determied by ad ; it does ot vary from sample to sample. If oe radomly draws a X from the populatio, there is a 9% chage that the draw X,, will fall i this rage. 1. What s the di erece betwee ad :9 = Pr [ 1:96 < X < + 1:96 ] :9 = Pr 1:96 p < X < + 1:96 p 7

8 Why? As part of your aswer, iterpret both itervals. aswer: The rage 1:96 to + 1:96 is a speci c umerical rage determied by ad ; it does ot vary from sample to sample. If oe radomly draws a X from the populatio, there is a 9% chage the draw X,, will fall i this rage. The iterval 1:96 p to +1:96 p is a umerical rage; it does ot vary from sample to sample. How to iterpret? If oe draws a radom sample of size ad takes its mea, there is a 9% chace that the observed mea will be i this rage. The rage o the latter is obviously smaller tha the rage o the former. 1. Assume that X is ormally distributed with ukow ad kow. Iterpret :9 = Pr [X 1:96 < < X + 1:96 ] aswer: Everytime oe draws a ew X oe gets a di eret iterval, ad 9% of these itervals will cotai the populatio mea,. 14. Assume X 1 ; X ; :::; X is a radom sample from a ormal distributio with ukow ad ukow. Fid a 9% co dece iterval for i terms of X ad iterpret this iterval. aswer: De e the statistic M = X b p where b = (X i X) ( 1) is a estimate of. The statistic M has a t distributio with 4 degrees of freedom ( 1) see MGB page??? So, :9 = Pr 4 :776 < X < :776 b p = Pr :776 b p < X < :776 b p = Pr X :776 b p < < X + :776 b p = Pr X + :776 b p > > X = Pr X :776 b p :776 b p < < X + :776 b p 9% of the itervals, X :776 p b to X + :776 p b will cotai. 1. Assume X 1 ; X ; :::; X is a radom sample from a ormal distributio with kow ad ukow. Fid a 9% co dece iterval for ad iterpret it. :766 is the critical t value for :0 whe parameter of the t is 4. 8

9 aswer: Cosider the statistic Q = (X i ) = Xi. This statistic has a Chi-squared distributio with parameter (degrees for freedom) see MGB page 4. Therefore (X i ) :9 = Pr 6 4 :81 < < 1:8 7 I eplaatio, with degrees of freedom, :% of the values of Q are less tha :81 ad :% are greater tha 1:8. Note that all of the terms are positive. Rearragig (X i ) :9 = Pr 6 4 :81 < < 1:8 7 = Pr 1 6 4:81 > 1 > 1:87 (X i ) = Pr 6 4 1: 0 4 > > : (X i ) = Pr Pr " " 1: 0 4 :07 81 The iterval :07 81 (X i ) > > :07 81 (X i ) < < 1: 0 4 (X i ) to 1: 0 4 # (X i ) # (X i ) (X i ) varys across samples (each with observatios). 9% of these itervals will cotai. Note that the iterval depeds o, which was assumed kow. 16. Assume X 1 ; X ; :::; X is a radom sample from a ormal distributio with ukow ad ukow. Fid a 9% co dece iterval for i terms of X ad iterpret it. 9

10 aswer: Cosider the statistic W = (X i X) (X i X) = ( 1)b where b = ( 1) is a estimate of. The statistic W has a Chi-square distributio with parameter (degrees for freedom) 1 see MGB page???. Therefore, for ve observatios " # :9 = Pr :484 < ( 1)b < 11:1 I eplaatio, with observatios, 4 degrees of freedom, :% of the values of Q are less tha :484 ad :% are greater tha 11:1. Note that all of the terms are positive. Rearragig " # :9 = Pr 1 = Pr = Pr h = Pr :484 < ( 1)b < 11:1 :484 > ( 1)b > 1 11:1 : > ( 1)b > 0: : 066 1( 1)b > > 0:090 09( 1)b h i = Pr 0:090 09( 1)b < < : 066 1( 1)b The iterval 0:090 09( 1)b to : 066 1( 1)b varys across samples. 9% of these itervals will iclude. For fu, compare this iterval to :07 81 (X i ) to 1: 0 4 (X i ), which is the comparable iterval whe is kow derived i the previous questio. The secod oe will be a tighter iterval, because more is kow. 17. So you meet a guy at a bar. He tells you that the radom variable X has some desity f X ( : ) but that he does ot tell you its form or the value of. After aother drik he tell that he has a derived a estimator for, b = b (X 1 ; X ; :::; X ) where X 1 ; X ; :::; X is a radom sample. He does ot tell you the form of the estimator but does tell you the samplig distributio of b, f b (: E[ b ]; b ) is Guber distributed (the famous Guber distributio, amed after its fouder Philoeus Guber) with parameters 4 ad 9. You look i the Guber table i the back of some statistics book ad determie a 9% co dece iterval for b is 1 to 9. I words, what does this co dece iterval tell you? i 10

11 Give the iformatio provided, what do you kow about? Eplai aswer: The co dece iterval tell you that everytime you draw a radom sample ad calculate b it has a 9% chace of beig betwee 1 ad 9. (9% of all samples will geerate b that are betwee 1 ad 9). The aswer to the secod questio is othig. We do t kow aythig about the guy s estimator, so do t kow if it has ay desirable properties. For eample, if we kew it was a ubiased estimator, we would kow that = 4, but we do t kow this. 18. Imagie that me come from both Mars ad Pluto, ad o where else. We kow eactly how IQ (the itelligece coe cet) varies i each populatio. That is, the researcher kows f IQm (IQ m ; m ) ad f IQp (IQ p ; p ). To be eplicit, the two fuctioal forms are kow as well as the values of the two parameters ( m ad p ). Further assume that the IQ of me from Mars varys from 0 to 10 (0 IQ m 10) ad the IQ of me from Pluto varys from 97 to 160 (97 IQ m 160). Oe ma is radomly sampled from oe of the two populatios ad preseted to you as a gift. You do t kow whether he is from Mars or Pluto; all you kow is that his IQ is 1. You are idi eret to where he is from; you just wat to kow. (Part 1) Describe a rule or rules you might use to estimate whether he is from Mars or Pluto, where he is most likely to come from. Drawig ad presetig some eample overlappig desity fuctios will clarify thigs for both you ad the reader aswer to part 1: A simple rule, with a soud theoretical foudatio, would be to compare the umerical values of f IQm (1; m ) ad f IQp (1; p ), ad choose the largest. For eample, if f IQm (1; m ) = :07 > :0 = f IQp (1; p ) estimate that he is from Mars. I more detail, f IQm (IQ m ; m ) is the likehood fuctio for a sample of oe if the populatio radomly sampled is Mars ad, f IQp (IQ p ; p ) is the likehood fuctio for a sample of oe if the populatio radomly sampled is Pluto. Which populatio is more likely to geerate a observatio of 1, the oe with the larger f IQi (1; i ), i = 1 or. The rst graph is a eample where where you would estimate Mars. The secod graph is a eample where you would estimte Pluto. There are more complicated estimatio techiques that oe might imagie. For eample, oe might choose plaet i over j if f IQi (1; i ) > f IQj (1; j ) by at least %; otherwise choose by ippig a fair coi. Or, oe might just ip a fair coi (probably ot the most e ciet estimatio techique). (Part ) Now make the problem more iterestig by asssumig two guy are radomly sampled from oe of the two plaets, What is your rule for choosig which plaet they are from if oe has a IQ of 104 ad the other a IQ of 119? 11

12 aswer to part : I would compare f IQm (104; m )f IQm (119; m ) with f IQp (104; p )f IQp (119; p ). Each is the likelihood of the sample give the plaet. some thoughts after readig your aswers: f IQm (1; m ), for eample, is a umber, a likelihood, but for a cotiuous distributio it is ot a probability. Some of you tured it ito a probability by assumig some small " ad the itergratig R +" " f IQ m (1 + ; m )d to tur it ito R a probability. You the compared, for the rst part of the questio, +" " f IQ m (1+; m )d 7 R +" " f IQ p (1+; p )d, ad chooe the plaet with the larger oe. This is e, but the itegratio is uecessary. This itegratio approach becomes a bit more complicated whe oe has two observatios. May people wated to aswer the questio with a co dece iterval. This tack is bit problematic: Imagie assumig the ull hypothesis is that the guys are from Mars, ad the geeratig the 9% co dece iterval for IQ m, ad easy task.. Oe the ds oe of two thigs: the observed IQ is or is ot i the iterval, or it is. What does oe coclude i each case, ot much. It the observed IQ is outside the iterval oe might be tempted to coclude the guy is from Pluto, but that would be premature - maybe you could also reject the ull that the guy is from Pluto. There is o atural ull hypothesis for this problem, uless you wat him to be from Mars, or from Pluto. Alteratively, if the guy is i the iterval for Mars he might also be i the iterval for Pluto. May people imposed a ormal, or some some other desity o the two distributios. Doig so correctly is ot aswerig the questio but a restrictive form of the questio, so deservig of oly partial credit. As a aside, either desity fuctio ca be ormal. Also ote that there was othig i the questio that implied that the two desity fuctios had the same fuctioal form. Hako had a iterestig cojecture that got me goig for a bit. He felt we had to take accout of the fact the the male populatios o the two plaets might di er substatively i size, so a perso from the plaet with the larger populatio was more likely to be draw. Whether this is correct depeds o how oe samples. If oe puts both populatios i a big ur ad radomly draws oe idividual, Hako would be correct, but that is ot what the questio implies. It says, "Oe ma is radomly sampled from oe of the two populatios.." that is, the chooser picks a plaet, radomly or otherwise, ad the radomly selects a idividual 19. Deote a radom sample of size from the ormal distributio, f X (; ; ) = X (; ) as X 1 ; X ; :::; X. As a aside a umber of you thought that to do this oe eeded to rst determie the variace of IQ m. While OK, this is ot ecessary: oe has the distributio, just d the the shortest spa that captures 9% of the desity. 1

13 Describe, i words, the iterval X h Pr X i :776 p b < < X + :776 p b = :9. :776 p b < < X + :776 p b where Now describe i words the iterval 1:96 < X i < + 1:96 where Pr [ 1:96 < X i < + 1:96 ] = :9 aswer: The iterval X :776 p b < < X +:776 p b varies from sample to sample ad 9% of these itervals will cotai. For eample, for a speci c sample, this iterval is a speci c umerical rage (e.g. to 8) ad it highly likely that will fall i this rage, but ot for sure. The iterval 1:96 < X i < + 1:96 is ot a rv, but rather a ed iterval that depeds o the values of ad, it does ot vary from sample to sample. X i is the i th draw from a sample. The i th draw i 9% of the samples will be i this ed iterval. So, for this iterval, but ot the rst iterval, oe ca say that "there is a 9% chace that the i th draw from a radom sample will be i this iterval. 0. Assume you are teachig ecoometrics. For a log time, I thought that for a speci c radom variable, X with desity f X (), there was oly oe 0% co dece iterval. Eplai, i a way your studets would uderstad (try ot to be too techical) why there ca be may 0% co dece itervals. Graphs might help. Now that your studets uderstad that there is likely more tha oe 0% co dece iterval, eplai how oe might choose a speci c 0% co dece iterval. Is the best" co dece iterval always cotiuous (o gaps)? Is it always symmetric aroud the mea? Eplai. aswer: Assume some radom variable X has some desity fuctio f X (). Because it is a desity fuctio the area uder it, by de itio, is oe. A 0% co dece iterval is ay rage of values of X such that the area uder that rage is :0. There are ofte a i ite umber of rages that have this property. Cosider for eample a 0% co dece iterval for the stadard ormal. NormalDe (; 0; 1) 1

14 f() I my software the commad NormalDist () is the CDF of the stadard ormal. E.g. NormalDist (0) = 1. So, oe ca epress a 0% co dece iterval as NormalDist () NormalDist (y) = :0 where > y For eample (NormalDist (10) NormalDist (0)) = 0:, so 0 to 10 is a 0% co dece iterval for X (0% of radomly draw X 0 s will be i this rage) (NormalDist (0) NormalDist ( 10)) = 0:, so 10 to 0 is a 0% co - dece iterval (NormalDist (:69) NormalDist ( :69)) = 0:09 81, so :69 to :69 is a 0% co dece iterval. (NormalDist (:) NormalDist ( :9)) = 0:07 4, so :9 to : is a 0% co dece iterval. So, at this poit, I am hopefully coviced that there ca be more tha oe m% co dece itervals. Which oe to choose? Typically, the shortest oe. Why? Because the shortest oe is the oe the provides the most accurate iformatio about where a draw X is most likely to fall. For eample, i our above eample, oe could say that there is a 0% chace that a radomly draw X will fall betwee zero ad 10, but oe ca also say there isa 0% chace that a radomly draw X will fall betwee :69 ad :69. The rst iterval covers a rage of 10 the secod iterval a rage of 1:8. The shorter oe provides much more accurate iformatio about where the X s i this distributio "lie". 14

15 Note the best iterval (the shortest oe) will ot always be a cotiuous iterval. Cosider for eample the shortest 0% co dece iterval for a desity fuctio that is U-shaped. For eample. f X () = 8 < 8:4 if > 0 0 if 0 : 8:4 if < 0 f() By trial ad error I determied that Z :96 ( =8:4)d = 0:0 66, so % of the desity is betwee ad :96. So by symmetry, % of the desity is betwee :96 ad. So, the shortest 0% co dece iterval for this desity is to :96 ad :96 to. So, this is a eample of a iterval that is ot cotiuous. Note, however that it is symmetric aroud the mea. Cosider the followig eample where the shortest 0% co dece iterval is ot symmetric aroud the mea. Assume f X () = ChiSquareDe(; ) 1

16 f() I my software the commad for the CDF of the Chi-Square is ChiSquareDist(; ) ChiSquareDist(4:4; ) = 0:06 6. So, the shortest 0% co dece iterval is 0 to 4:4. Sice the epected value of this Chi-Square is this iterval is ot cetered o the mea ad does ot eve iclude the mea