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# Prob & Statistics Engineering STT 351

MSU

GPA 3.72

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This 3 page Class Notes was uploaded by Aryanna Jerde on Saturday September 19, 2015. The Class Notes belongs to STT 351 at Michigan State University taught by Raoul Lepage in Fall. Since its upload, it has received 78 views. For similar materials see /class/207817/stt-351-michigan-state-university in Statistics and Probability at Michigan State University.

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Date Created: 09/19/15

Supplemental review of CI and testing 120108 We ve studied a variety of confidence interval CI procedures Their objective when used at 95 confidence is to establish a margin of error for a statistical estimator 0 i t or z Var In this notation 3 denotes an estimator of some parameter 0 and l Var denotes the estimator of the standard deviation of estimator When using bootstrap we typically never see t or z or V Var since the bootstrap method deliv ers an estimator of the entire package t or z Var CI Here is a brief summary of most CI studied in this course ex indicates that the CI is exact with perfect calculations otherwise the CI is approximate as n gt 00 and other assumptions including N n gt 00 when sampling finite populations i t or Z Var caveat gtltI ex eqpr Wr sam of n from normal pop gtlt N are eqpr Wr sample of 11 Z S 11 eqpr Withoutr sample of n 11 Z 1 113 11 gtlt l 3 quotQgt lt eqpr Wr sample of 11 1 1 eqpr Withoutr sample of n quotQgt N it 2 review120108 nb 2 S y i Z Indep X y sam eqpr Wr Ax1Ax p p quotx same as just above I3y1 I3y quoty 0 i 95 th or other percentile of WAquot 0 as approp V zK1Wi 5 T zilwi 131 1 13 V 7 711x f i Z Sfy 1 r2 eqpr Wr prs X y prop39l eqpr Wr strat sam A7 i Z prop39l eqpr Wr strat sam Z x i 139 entry of xtr c1 sresid2 with 6i ind N0 02 g PseudoInverseXy 9x resid y y Tests from CI Here is a brief description of how Cl may be used to test hypotheses of the kind stud ied in tis course The idea is a simple one A Cl is trying to locate cover a parame ter 0 If it is a 95 Cl then PCI covers 0 095 So PCI misses 0 1 095 005 If could devise a test of for example the null hypothesis that 0 17 versus the two sided alternative hypothesis 0 2 17 which rejects H0 0 17 if Cl fails to cover 17 If truly 0 17 such a test commits type I error precisely when Cl fails to cover 17 This has probability 005 as above So a 005 for such a use of CI to test review120108 nb 3 If instead we wish to perform a one sided test H0 0 17 versus Ha 0gt 17 we could harness the CI in the following way reject H0 if CI falls entirely to the right of 17 For this test a 1 0952 0025 since the 005 probability of having the CI fail to cover 17 is about equally divided between missing to the left or missing to the right Testing H0 0 17 versus Ha 0lt 17 we would reject H0 if CI falls entirely to the left of 17 Once again a 0025 O 1 data and tests We might make a test for H0 p 01 versus Ha p 2 01 V 13 1 i7 I Then a 005 but that is not the preferred test If p 01 then the population sd is V 01 X 09 and need not be estimated by V f 1 3 The preferred test V 01x 09 f n by fails to cover 01 rejecting H0 if CI 3 i z fails to cover 01 rejects H0 if f i z Two points a The preferred test looks like it uses a CI but does not b The preferred test more closely achieves a 005 The one sided counterpart H0 p 01 versus Ha p gt 01 V01gtlt09 l reject H0 if f i z falls entirely right of 01 is preferred over the CI test as well It more accurately achieves a 0025 than does the one sided CI test Comments z t may take values other than 196 or the t values related to 95 confidence As

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