Topic 5: Confidence Intervals (Chapter 9)


 Phillip Cannon
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1 Topic 5: Cofidece Iterval (Chapter 9) 1. Itroductio The two geeral area of tatitical iferece are: 1) etimatio of parameter(), ch. 9 ) hypothei tetig of parameter(), ch. 10 Let X be ome radom variable with ukow mea µ. Suppoe we take a ample of ize. Oe could fid the ample mea, x, ad ue it to etimate µ. Such a igle umber ued to etimate a parameter i called a poit etimate. Quetio: Are there other reaoable poit etimate (other tha x ) for µ? Claim: The bet poit etimate of µ i x. The poit etimate doe ot give u a ee of how cloe x might be to µ, i.e. it doe t tell u about the iheret ucertaity i the proce. Therefore, iterval etimate are ued, which provide a rage of value that cotai the parameter, ay µ, with ome pecified degree of cofidece. Thi rage i called a cofidece iterval.. TwoSided Cofidece Iterval Coider ow ome radom variable X with mea µ ad tadard deviatio. Suppoe we wat to etimate µ (the etimatio of follow later). Uig the cetral limit theorem X ~ N( µ, ) Therefore, (X µ) Z = (1) i a tadard ormal. Recall that by defiitio P( zα < Z < zα) = 1 α () e.g. becaue z.05 = 1.96 P( 1.96 < Z < 1.96) = 1.05 =.95 1
2 Subtitutig (1) ito (), ad uig ome algebra, oe ca how that () implie PX ( z < µ < X+ z α α ) = 1 α (3) e.g. lettig α =.05 PX ( 1.96 < µ < X ) =.95 The iterpretatio of (3) i tricky. Notice that we calculate a radom iterval ( X 1.96, X ). If we do thi repeatedly, we expect that 95% of the iterval cotai the ukow parameter µ. It doe ot mea that µ aume a value withi the iterval with probability A illutratio of thi i give i Figure 9.1. I geeral, a 100 (1 α ) % cofidece iterval for µ i ( X z, X + z α α ). (4) The cofidece iterval could be writte a X ± E where E = zα. (5) E i alo called the boud o error (BOE). Suppoe oe wat to reduce E. Thi could obviouly be doe by icreaig α or icreaig. I fact, may be determied from E. Notice from (5), = zα. E Therefore, = [ zα ]. (6) E Coider a example. Let X be the choleterol level of a U.S. male who moke. Suppoe µ i ukow, but that we kow = 46 (we actually aume thi value becaue we preumably kow = 46 for populatio of all U.S. adult male). Suppoe we take a ample of =, ad fid x = 17. The the bet poit etimate of µ i 17. Alo, a 95% cofidece iterval for µ i (uig X = 17 )
3 1.96(46) 17 ± = 17 ± 6 = (191,43). A 99% cofidece iterval for µ i.58(46) 17 ± = (183, 51) Why i it wider? Suppoe we are deigig the tudy, ad we wat to kow µ withi 10 uit. (Note that E=10 ad the iterval width i 0). Subtitutig, E, ad α =.01 ito (6), oe ha.58(46) = ( ) = Therefore our recommedatio i to obtai a ample for = 141 male. 3. OeSided Cofidece Iterval Sometime (though ot ofte), oe deire jut a upper or a lower boud o µ. Uig algebraic maipulatio a before, oe ha: A 100(1α )% upper (oeided) cofidece boud o µ i X + zα, (7) ad a correpodig lower cofidece boud i X zα. (8) e.g. for U.S. male, to fid a 95% upper boud, recall that z.05 = Therefore the upper boud from the ample i 1.65(46) 17 ± = = Studet t ditributio Coider a radom variable X which i kow to be ormally ditributed, but where either µ or are kow. To make iferece about µ, we could ue the Z traformatio, provided i kow. However, i cae of ukow, which i the commo cae, we replace by to fid 3
4 ( x µ ) Thi tatitic ha the t ditributio with 1 degree of freedom (df). Several fact of iteret are 1) The tditributio i alo ymmetric about 0, but it ha thicker tail, i.e. it a bit flatter. Thi i becaue there i added variability itroduced by uig the radom variable rather tha a cotat i the deomiator (ee Figure 9.). ) The tditributio approache Z a become large. Thi i becaue a icreae, we have more iformatio ad approache. Why? Recall we had a exteive Z table, but it impractical to give a table for each t 1 tatitic. Therefore, table typically give oly upper percet poit. For example, from Table A4, oe ha 10,.05.8 By ymmetry, t 10,.975 =.8, ad thu oe ca fid upper ad lower percetile. Oe ca cotruct cofidece iterval baed o the tditributio i the ame way a before. The expreio comparable to (3) i PX ( t < µ < X+ t ) = α α 1 α. I geeral, a 100(1α )% cofidece iterval for µ, with approximately ormal x, i ( X tα, X + tα ). (9) For example, let X deote a ifat plama alumium level, which i aumed to be approximately ormal. Suppoe we take a radom ample of = 10 ifat, ad fid x = 37. ad = To fid a 95% cofidece iterval, we fid 9,.05.6 Hece, the iterval etimate i.6(7.13) 37. ± 10 = 37. ± 5.1 = ( 3.1, 4.3) 4
5 Notice that i thi problem, the iterval width i a radom variable. Hece, if two experimeter took differet ample, their iterval width would differ, ulike the cae of kow. I geeral, the iterval baed o t would be wider tha that baed o Z. Why? Though the width differ, we expect 100(1α )% of the iterval to cotai µ. See Figure Cae Study Thi ha a ice cae tudy o the efficacy of uig a particular drug for treatig ADD childre. 5
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