{"id":34958,"date":"2019-10-31T22:01:31","date_gmt":"2019-10-31T19:01:31","guid":{"rendered":"https:\/\/prohoster.info\/blog\/kogda-stoit-proveryat-gipotezu-o-ne-menshej-effektivnosti\/"},"modified":"2019-10-31T22:01:31","modified_gmt":"2019-10-31T19:01:31","slug":"kogda-stoit-proveryat-gipotezu-o-ne-menshej-effektivnosti","status":"publish","type":"post","link":"https:\/\/prohoster.info\/sq\/blog\/news\/kogda-stoit-proveryat-gipotezu-o-ne-menshej-effektivnosti","title":{"rendered":"Kur duhet kontrolluar hipoteza se nuk \u00ebsht\u00eb m\u00eb e efektshme?","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

\"Kur
\nNj\u00eb artikull nga ekipi i Stitch Fix sugjeron p\u00ebrdorimin e qasjes s\u00eb k\u00ebrkimeve klinike t\u00eb mosefektivitetit (non-inferiority trials) n\u00eb testet A\/B t\u00eb marketingut dhe produktit. Kjo qasje \u00ebsht\u00eb v\u00ebrtet e aplikueshme kur testojm\u00eb nj\u00eb zgjidhje t\u00eb re me p\u00ebrfitime q\u00eb nuk jan\u00eb t\u00eb matshme nga testet.<\/p>\n

Shembulli m\u00eb i thjesht\u00eb \u00ebsht\u00eb ulja e kostove. P\u00ebr shembull, automatizojm\u00eb procesin e caktimit t\u00eb m\u00ebsimit t\u00eb par\u00eb, por nuk duam t\u00eb ulet ndjesh\u00ebm konversioni i p\u00ebrgjithsh\u00ebm. Ose testojm\u00eb ndryshime t\u00eb orientuara ndaj nj\u00eb segmenti t\u00eb caktuar t\u00eb p\u00ebrdoruesve, duke mbajtur n\u00ebn kontroll q\u00eb konversioni n\u00eb segmentet e tjera t\u00eb mos bjer\u00eb shum\u00eb (kur testojm\u00eb disa hipoteza, mos harroni rregullimet).<\/p>\n

Zgjedhja e kufirit t\u00eb sakt\u00eb t\u00eb mosefektivitetit shkakton v\u00ebshtir\u00ebsi shtes\u00eb n\u00eb faz\u00ebn e dizajnit t\u00eb testit. Pyetja se si t\u00eb zgjidhet \u0394 n\u00eb artikull nuk \u00ebsht\u00eb trajtuar shum\u00eb mir\u00eb. Duket se kjo zgjedhje nuk \u00ebsht\u00eb plot\u00ebsisht transparente, as n\u00eb provat klinike. P\u00ebrmbledhje<\/a><\/noindex> Publikimet mjek\u00ebsore mbi non-inferiority tregojn\u00eb se vet\u00ebm n\u00eb gjysm\u00ebn e publikimeve zgjedhja e kufijve \u00ebsht\u00eb e arsyetuar, dhe shpesh k\u00ebto arsyetime jan\u00eb t\u00eb pasqaruara ose t\u00eb pa detajuara.<\/p>\n

N\u00eb \u00e7do rast, kjo qasje duket interesante, pasi p\u00ebrmes zvog\u00eblimi t\u00eb madh\u00ebsis\u00eb s\u00eb nevojshme t\u00eb mostr\u00ebs mund t\u00eb rritet shpejt\u00ebsia e testimit, dhe k\u00ebshtu edhe shpejt\u00ebsia e marrjes s\u00eb vendimeve. \u2014 Darya Mukhina, analiste produkti p\u00ebr aplikacionin mobil Skyeng.<\/i>
\n<\/a><\/noindex>
\nEkipi i Stitch Fix e do t\u00eb testoj\u00eb gj\u00ebra t\u00eb ndryshme. E gjith\u00eb komuniteti teknologjik n\u00eb thelb e do t\u00eb kryej\u00eb teste. Cila version i faqes t\u00ebrheq m\u00eb shum\u00eb p\u00ebrdorues \u2013 A apo B? A sjell versioni A i modelit rekomandues m\u00eb shum\u00eb para sesa versioni B? Pothuajse gjithmon\u00eb p\u00ebr t\u00eb verifikuar hipotezat p\u00ebrdorim qasjen m\u00eb t\u00eb thjesht\u00eb nga kursi i baz\u00ebs s\u00eb statistik\u00ebs: <\/p>\n

\"Kur
\n
\nMegjith\u00ebse rrall\u00eb e p\u00ebrdorim k\u00ebt\u00eb term, ky lloj testimi quhet \"verifikimi i hipotez\u00ebs p\u00ebr superioritetin\". Me k\u00ebt\u00eb qasje, ne supojm\u00eb se midis dy alternativave nuk ka ndonj\u00eb ndryshim. Ne mbajm\u00eb k\u00ebt\u00eb ide dhe e braktisim vet\u00ebm n\u00eb rast se t\u00eb dh\u00ebnat e marra jan\u00eb mjaft bind\u00ebse p\u00ebr k\u00ebt\u00eb \u2013 dometh\u00ebn\u00eb, demonstrojn\u00eb se nj\u00eb nga alternativat (A ose B) \u00ebsht\u00eb m\u00eb e mir\u00eb se tjetra.<\/p>\n

Verifikimi i hipotez\u00ebs mbi superioritetin \u00ebsht\u00eb i p\u00ebrshtatsh\u00ebm p\u00ebr zgjidhjen e shum\u00eb problemeve. Ne l\u00ebshojm\u00eb versionin B t\u00eb modelit rekomandues vet\u00ebm n\u00ebse ai \u00ebsht\u00eb duksh\u00ebm m\u00eb i mir\u00eb se versioni A q\u00eb po p\u00ebrdoret tashm\u00eb. Por n\u00eb disa raste, ky qasje nuk funksionon aq mir\u00eb. Le t\u00eb shqyrtojm\u00eb disa shembuj.<\/p>\n

1) Ne p\u00ebrdorim nj\u00eb sh\u00ebrbim t\u00eb jasht\u00ebm<\/b>, i cili ndihmon n\u00eb identifikimin e kartave bankare t\u00eb falsifikuara. Ne kemi gjetur nj\u00eb sh\u00ebrbim tjet\u00ebr, i cili kushton ndjesh\u00ebm m\u00eb pak. N\u00ebse sh\u00ebrbimi m\u00eb i lir\u00eb funksionon me t\u00eb nj\u00ebjt\u00eb efektshm\u00ebri si ai q\u00eb po p\u00ebrdorim tani, ne do ta zgjedhim at\u00eb. Ai nuk duhet domosdoshm\u00ebrisht t\u00eb jet\u00eb m\u00eb i mir\u00eb se sh\u00ebrbimi q\u00eb po p\u00ebrdorim. <\/p>\n

2) Ne duam t\u00eb heqim dor\u00eb nga burimi i t\u00eb dh\u00ebnave<\/b> A dhe ta z\u00ebvend\u00ebsojm\u00eb at\u00eb me burimin e t\u00eb dh\u00ebnave B. Ne mund ta shtyjm\u00eb heqjen dor\u00eb nga A n\u00ebse B jep rezultate shum\u00eb t\u00eb dob\u00ebta, por vazhdimi i p\u00ebrdorimit t\u00eb A nuk \u00ebsht\u00eb i mundur. <\/p>\n

3) Ne do t\u00eb doja t\u00eb kalonim nga qasja A n\u00eb qasjen B jo sepse presim rezultate m\u00eb t\u00eb mira nga B, por sepse kjo na jep m\u00eb shum\u00eb fleksibilitet operativ. Nuk kemi arsye t\u00eb besojm\u00eb se B do t\u00eb jet\u00eb m\u00eb e keqe, por ne nuk do t\u00eb b\u00ebjm\u00eb kalimin n\u00ebse nuk \u00ebsht\u00eb k\u00ebshtu.<\/b>4) Ne kemi b\u00ebr\u00eb disa p\u00ebrmir\u00ebsime t\u00eb cil\u00ebsis\u00eb<\/p>\n

n\u00eb dizajnin e faqes s\u00eb internetit (versioni B) dhe besojm\u00eb se ky version e kalon versionin A. Ne nuk presim ndryshime n\u00eb konvertimin ose n\u00eb ndonj\u00eb tregues ky\u00e7 performanc\u00eb, me t\u00eb cil\u00ebt zakonisht e vler\u00ebsojm\u00eb faqen e internetit. Por besojm\u00eb se ka p\u00ebrfitime n\u00eb parametrat q\u00eb jan\u00eb ose t\u00eb pa matsh\u00ebm, ose teknologjit\u00eb tona nuk jan\u00eb t\u00eb mjaftueshme p\u00ebr t'i matur ato.<\/b> N\u00eb t\u00eb gjitha k\u00ebto raste, hulumtimi mbi superioritetin nuk \u00ebsht\u00eb zgjidhja m\u00eb e p\u00ebrshtatshme. Por shumica e specialist\u00ebve n\u00eb k\u00ebto situata e p\u00ebrdorin at\u00eb si standard. Ne kryejm\u00eb eksperimentin me kujdes p\u00ebr t\u00eb p\u00ebrcaktuar sakt\u00ebsisht madh\u00ebsin\u00eb e efektit. N\u00ebse do t\u00eb ishte e v\u00ebrtet\u00eb q\u00eb versionet A dhe B funksionojn\u00eb n\u00eb m\u00ebnyr\u00eb shum\u00eb t\u00eb ngjashme, ekziston mund\u00ebsia q\u00eb t\u00eb mos arrijm\u00eb t\u00eb heqim dor\u00eb nga hipoteza zero. A arrijm\u00eb n\u00eb p\u00ebrfundimin se A dhe B funksionojn\u00eb n\u00eb m\u00ebnyr\u00eb t\u00eb ngjashme? Jo! Mosarritja p\u00ebr t\u00eb hequr dor\u00eb nga hipoteza zero dhe pranim i hipotez\u00ebs zero nuk jan\u00eb nj\u00ebsoj.<\/p>\n

\u0412\u043e \u0432\u0441\u0435\u0445 \u044d\u0442\u0438\u0445 \u0441\u043b\u0443\u0447\u0430\u044f\u0445 \u0438\u0441\u0441\u043b\u0435\u0434\u043e\u0432\u0430\u043d\u0438\u0435 \u043f\u0440\u0435\u0432\u043e\u0441\u0445\u043e\u0434\u0441\u0442\u0432\u0430 \u2014 \u043d\u0435 \u0441\u0430\u043c\u043e\u0435 \u043f\u043e\u0434\u0445\u043e\u0434\u044f\u0449\u0435\u0435 \u0440\u0435\u0448\u0435\u043d\u0438\u0435. \u041d\u043e \u0431\u043e\u043b\u044c\u0448\u0438\u043d\u0441\u0442\u0432\u043e \u0441\u043f\u0435\u0446\u0438\u0430\u043b\u0438\u0441\u0442\u043e\u0432 \u0432 \u0442\u0430\u043a\u0438\u0445 \u0441\u0438\u0442\u0443\u0430\u0446\u0438\u044f\u0445 \u0438\u0441\u043f\u043e\u043b\u044c\u0437\u0443\u044e\u0442 \u0435\u0433\u043e \u043f\u043e \u0443\u043c\u043e\u043b\u0447\u0430\u043d\u0438\u044e. \u041c\u044b \u0442\u0449\u0430\u0442\u0435\u043b\u044c\u043d\u043e \u043f\u0440\u043e\u0432\u043e\u0434\u0438\u043c \u044d\u043a\u0441\u043f\u0435\u0440\u0438\u043c\u0435\u043d\u0442, \u0447\u0442\u043e\u0431\u044b \u043f\u0440\u0430\u0432\u0438\u043b\u044c\u043d\u043e \u043e\u043f\u0440\u0435\u0434\u0435\u043b\u0438\u0442\u044c \u0432\u0435\u043b\u0438\u0447\u0438\u043d\u0443 \u044d\u0444\u0444\u0435\u043a\u0442\u0430. \u0415\u0441\u043b\u0438 \u0431\u044b \u0431\u044b\u043b\u043e \u0432\u0435\u0440\u043d\u043e, \u0447\u0442\u043e \u0432\u0435\u0440\u0441\u0438\u0438 A \u0438 B \u0440\u0430\u0431\u043e\u0442\u0430\u044e\u0442 \u043e\u0447\u0435\u043d\u044c \u0441\u0445\u043e\u0436\u0438\u043c \u043e\u0431\u0440\u0430\u0437\u043e\u043c, \u0435\u0441\u0442\u044c \u0432\u0435\u0440\u043e\u044f\u0442\u043d\u043e\u0441\u0442\u044c, \u0447\u0442\u043e \u043d\u0430\u043c \u043d\u0435 \u0443\u0434\u0430\u0441\u0442\u0441\u044f \u043e\u0442\u043a\u043b\u043e\u043d\u0438\u0442\u044c \u043d\u0443\u043b\u0435\u0432\u0443\u044e \u0433\u0438\u043f\u043e\u0442\u0435\u0437\u0443. \u0414\u0435\u043b\u0430\u0435\u043c \u043b\u0438 \u043c\u044b \u0432\u044b\u0432\u043e\u0434, \u0447\u0442\u043e \u0410 \u0438 \u0412 \u0432 \u0446\u0435\u043b\u043e\u043c \u0440\u0430\u0431\u043e\u0442\u0430\u044e\u0442 \u043e\u0434\u0438\u043d\u0430\u043a\u043e\u0432\u043e? \u041d\u0435\u0442! \u041d\u0435\u0432\u043e\u0437\u043c\u043e\u0436\u043d\u043e\u0441\u0442\u044c \u043e\u0442\u043a\u043b\u043e\u043d\u0438\u0442\u044c \u043d\u0443\u043b\u0435\u0432\u0443\u044e \u0433\u0438\u043f\u043e\u0442\u0435\u0437\u0443 \u0438 \u043f\u0440\u0438\u043d\u044f\u0442\u0438\u0435 \u043d\u0443\u043b\u0435\u0432\u043e\u0439 \u0433\u0438\u043f\u043e\u0442\u0435\u0437\u044b \u2014 \u043d\u0435 \u043e\u0434\u043d\u043e \u0438 \u0442\u043e \u0436\u0435.<\/p>\n

Studimet mbi volumet e mostrave (t\u00eb cilat natyrisht i keni kryer), zakonisht kryhen me kufij m\u00eb t\u00eb rrept\u00eb p\u00ebr gabimin e par\u00eb (probabiliteti i refuzimit t\u00eb hipotez\u00ebs nul, i njohur shpesh si alfa) sesa p\u00ebr gabimin e dyt\u00eb (probabiliteti i paaft\u00ebsis\u00eb p\u00ebr t\u00eb refuzuar hipotez\u00ebn nul, n\u00ebse hipoteza nul \u00ebsht\u00eb e gabuar, e njohur shpesh si beta). Nj\u00eb vler\u00eb tipike p\u00ebr alfan \u00ebsht\u00eb 0,05, nd\u00ebrsa nj\u00eb vler\u00eb tipike p\u00ebr betan \u00ebsht\u00eb 0,20, e cila korrespondon me fuqin\u00eb statistike 0,80. Kjo do t\u00eb thot\u00eb se mund t\u00eb mos zbulojm\u00eb ndikimin e v\u00ebrtet\u00eb t\u00eb madh\u00ebsis\u00eb q\u00eb kemi specifikuar n\u00eb llogaritjet tona t\u00eb fuqis\u00eb, me probabilitet 20% dhe kjo \u00ebsht\u00eb nj\u00eb boshll\u00ebk relativisht serioz n\u00eb informacion. Si shembull, le t\u00eb shqyrtojm\u00eb k\u00ebto hipoteza:<\/p>\n

\"Kur
\n
\nH0: \u00e7anta ime nuk \u00ebsht\u00eb n\u00eb dhom\u00ebn time (3)
\nH1: \u00e7anta ime \u00ebsht\u00eb n\u00eb dhom\u00ebn time (4)<\/i><\/p>\n

N\u00ebse e kam kontrolluar dhom\u00ebn time dhe kam gjetur \u00e7ant\u00ebn time \u2014 shk\u00eblqyer, mund t\u00eb refuzoj hipotez\u00ebn nul. Por n\u00ebse kam kontrolluar dhom\u00ebn dhe nuk kam arritur t\u00eb gjej \u00e7ant\u00ebn time (figura 1), \u00e7far\u00eb p\u00ebrfundimi duhet t\u00eb nxjerr? A jam i sigurt se nuk \u00ebsht\u00eb aty? A kam k\u00ebrkuar mjaftuesh\u00ebm me vigjilenc\u00eb? \u00c7far\u00eb n\u00ebse kam kontrolluar vet\u00ebm 80% t\u00eb dhom\u00ebs? T\u00eb arrijm\u00eb n\u00eb p\u00ebrfundimin se \u00e7anta me siguri nuk \u00ebsht\u00eb n\u00eb dhom\u00eb do t\u00eb ishte nj\u00eb vendim i pamenduar. Nuk \u00ebsht\u00eb \u00e7udi q\u00eb nuk mund t\u00eb \u201cpranojm\u00eb hipotez\u00ebn nul.\u201d
\n\"Kur
\nZona q\u00eb kemi kontrolluar
\nNuk gjet\u00ebm \u00e7ant\u00ebn \u2013 a duhet t\u00eb pranojm\u00eb hipotez\u00ebn nul?<\/i><\/p>\n

Figura 1. T\u00eb kontrollosh 80% t\u00eb dhom\u00ebs \u00ebsht\u00eb p\u00ebraf\u00ebrsisht si t\u00eb kryesh nj\u00eb studim me fuqi 80%. N\u00ebse nuk gjet\u00ebt \u00e7ant\u00ebn, pas kontrollit t\u00eb 80% t\u00eb dhom\u00ebs, a mund t\u00eb arrijm\u00eb n\u00eb p\u00ebrfundim se nuk \u00ebsht\u00eb aty?<\/i> <\/p>\n

Pra, \u00e7far\u00eb duhet t\u00eb b\u00ebj\u00eb specialisti i t\u00eb dh\u00ebnave n\u00eb k\u00ebt\u00eb situat\u00eb? Mund t\u00eb rris\u00eb ndjesh\u00ebm fuqin\u00eb e studimit, por at\u00ebher\u00eb do t'ju duhet nj\u00eb mostr\u00eb shum\u00eb m\u00eb e madhe, dhe rezultati akoma do t\u00eb jet\u00eb i pak\u00ebnaqsh\u00ebm.<\/p>\n

Fatkeq\u00ebsisht, k\u00ebto probleme jan\u00eb studiuar prej koh\u00ebsh n\u00eb bot\u00ebn e k\u00ebrkimeve klinike. Ila\u00e7i B \u00ebsht\u00eb m\u00eb i lir\u00eb se ila\u00e7i A; pritet q\u00eb ila\u00e7i B t\u00eb shkaktoj\u00eb m\u00eb pak efekte an\u00ebsore se ila\u00e7i A; ila\u00e7i B \u00ebsht\u00eb m\u00eb i leht\u00eb p\u00ebr tu transportuar sepse nuk ka nevoj\u00eb t\u00eb ruhet n\u00eb frigorifer, nd\u00ebrsa ila\u00e7i A \u2013 ka. Le t\u00eb verifikojm\u00eb hipotez\u00ebn p\u00ebr efikasitetin e nj\u00ebjt\u00eb. Kjo \u00ebsht\u00eb e nevojshme p\u00ebr t\u00eb treguar se versioni B \u00ebsht\u00eb po aq i mir\u00eb sa versioni A \u2014 t\u00eb pakt\u00ebn brenda nj\u00eb kufiri t\u00eb paracaktuar \u201ct\u00eb efikasitetit t\u00eb nj\u00ebjt\u00eb\u201d, \u0394. Pak m\u00eb von\u00eb do t\u00eb flasim m\u00eb n\u00eb detaje se si t\u00eb p\u00ebrcaktojm\u00eb k\u00ebt\u00eb kufi. Por p\u00ebr tani, le t\u00eb supojm\u00eb se kjo \u00ebsht\u00eb ndryshimi minimal q\u00eb \u00ebsht\u00eb praktikisht i r\u00ebnd\u00ebsish\u00ebm (n\u00eb kontekstin e provave klinike zakonisht quhet r\u00ebnd\u00ebsi klinike).<\/p>\n

Hipotezat p\u00ebr efikasitetin e nj\u00ebjt\u00eb p\u00ebrmbysin gjith\u00e7ka:<\/p>\n

\"Kur
\n
\nTani, n\u00eb vend q\u00eb t\u00eb supozojm\u00eb se nuk ka ndryshim, ne supozojm\u00eb se versioni B \u00ebsht\u00eb m\u00eb i dob\u00ebt se versioni A, dhe do t\u00eb q\u00ebndrojm\u00eb te kjo supozim derisa t\u00eb tregojm\u00eb se nuk \u00ebsht\u00eb k\u00ebshtu. Ky \u00ebsht\u00eb momenti kur ka kuptim t\u00eb p\u00ebrdorim testimin e hipotez\u00ebs nj\u00ebansh\u00ebm! N\u00eb praktik\u00eb, kjo mund t\u00eb realizohet duke nd\u00ebrtuar nj\u00eb interval besueshm\u00ebrie dhe duke p\u00ebrcaktuar n\u00ebse ky interval \u00ebsht\u00eb n\u00eb t\u00eb v\u00ebrtet\u00eb m\u00eb i madh se \u0394 (shihni figur\u00ebn 2).
\n\"Kur
\n<\/p>\n

Zgjedhja e \u0394<\/h3>\n

\nSi t\u00eb zgjidhni sakt\u00ebsisht \u0394? Procesi i se\u00e7itjes s\u00eb \u0394 p\u00ebrfshin arsyetim statistik dhe vler\u00ebsim t\u00eb m\u00ebsuar. N\u00eb bot\u00ebn e k\u00ebrkimeve klinike ekzistojn\u00eb udh\u00ebzime t\u00eb rregullta, sipas t\u00eb cilave delta duhet t\u00eb p\u00ebrfaq\u00ebsoj\u00eb ndryshimin m\u00eb t\u00eb vog\u00ebl klinikisht t\u00eb r\u00ebnd\u00ebsish\u00ebm \u2014 nj\u00eb ndryshim q\u00eb do t\u00eb ket\u00eb r\u00ebnd\u00ebsi n\u00eb praktik\u00eb. K\u00ebtu \u00ebsht\u00eb nj\u00eb citat nga udh\u00ebzimi evropian, p\u00ebrmes t\u00eb cilit mund t\u00eb verifikoni veten: \"N\u00ebse ndryshimi \u00ebsht\u00eb zgjedhur si\u00e7 duhet, intervali i besueshm\u00ebris\u00eb, i cili \u00ebsht\u00eb plot\u00ebsisht midis \u2013\u2206 dhe 0..., \u00ebsht\u00eb ende i mjaftuesh\u00ebm p\u00ebr t\u00eb demonstruar efikasitetin e nj\u00ebjt\u00eb. N\u00ebse ky rezultat nuk duket i p\u00ebrshtatsh\u00ebm, kjo do t\u00eb thot\u00eb se \u2206 nuk \u00ebsht\u00eb zgjedhur si\u00e7 duhet.\"<\/p>\n

Deltaja me shprehsh\u00ebm nuk duhet t\u00eb tejkaloj\u00eb madh\u00ebsin\u00eb e efektit t\u00eb versionit A n\u00eb krahasim me kontrollin e v\u00ebrtet\u00eb (placebo \/ munges\u00eb trajtimi), sepse kjo na \u00e7on n\u00eb p\u00ebrfundimin q\u00eb versioni B \u00ebsht\u00eb m\u00eb i dob\u00ebt se kontrolli i v\u00ebrtet\u00eb, nd\u00ebrkoh\u00eb q\u00eb tregon \"n\u00eb asnj\u00eb rast m\u00eb pak efikasitet\". Supozoni se kur u prezantua versioni A, n\u00eb vend t\u00eb tij ishte versioni 0 ose funksioni nuk ekzistonte fare (shih figur\u00ebn 3). <\/p>\n

Nga rezultatet e prov\u00ebs p\u00ebr hipotez\u00ebn e superioritetit u identifikua madh\u00ebsia e efektit E (dmth, supozimisht \u03bc^A\u2212\u03bc^0=E). Tani A \u00ebsht\u00eb standardi yn\u00eb i ri, dhe ne duam t\u00eb sigurohemi se B nuk \u00ebsht\u00eb m\u00eb i dob\u00ebt se A. Nj\u00eb m\u00ebnyr\u00eb tjet\u00ebr p\u00ebr t\u00eb shkruar \u03bcB\u2212\u03bcA\u2264\u2212\u0394 (hipoteza nul) \u00ebsht\u00eb \u03bcB\u2264\u03bcA\u2212\u0394. N\u00ebse supozoni se e b\u00ebn t\u00eb barabart\u00eb ose m\u00eb t\u00eb madhe se E, at\u00ebher\u00eb \u03bcB \u2264 \u03bcA\u2212E \u2264 placebo. Tani ne shohim se vler\u00ebsimi yn\u00eb p\u00ebr \u03bcB tejkalon plot\u00ebsisht \u03bcA\u2212E, duke e mohuar k\u00ebshtu plot\u00ebsisht hipotez\u00ebn nul dhe duke lejuar p\u00ebrfundimin q\u00eb B nuk \u00ebsht\u00eb m\u00eb i dob\u00ebt se A, por n\u00eb t\u00eb nj\u00ebjt\u00ebn koh\u00eb \u03bcB mund t\u00eb jet\u00eb \u2264 \u03bc placebo, gj\u00eb q\u00eb nuk \u00ebsht\u00eb ajo q\u00eb na nevojitet. (Figura 3).<\/p>\n

\"Kur
\nFigura 3. Demonstrimi i rreziqeve t\u00eb zgjedhjes s\u00eb kufirit t\u00eb jo-m\u00eb pak efikasiteti. N\u00ebse kufiri \u00ebsht\u00eb tep\u00ebr i madh, mund t\u00eb p\u00ebrfundohet se B nuk \u00ebsht\u00eb m\u00eb i dob\u00ebt se A, por nj\u00ebkoh\u00ebsisht \u00ebsht\u00eb i padalluesh\u00ebm nga placebo. Ne nuk do t\u00eb nd\u00ebrrojm\u00eb nj\u00eb bar q\u00eb \u00ebsht\u00eb qartazi m\u00eb efikas se placebo (A), p\u00ebr nj\u00eb bar q\u00eb ka nj\u00eb efikasitet t\u00eb barabart\u00eb me placebo. <\/i><\/p>\n

Zgjedhja e \u03b1 <\/h3>\n

\nT\u00eb kalojm\u00eb n\u00eb zgjedhjen e \u03b1. Mund t\u00eb p\u00ebrdoret vlera standarde \u03b1 = 0,05, por kjo nuk \u00ebsht\u00eb krejt e ndershme. Si, p\u00ebr shembull, kur blini di\u00e7ka n\u00eb internet dhe p\u00ebrdorni menj\u00ebher\u00eb disa kode zbritjeje, ndon\u00ebse ato nuk duhet t\u00eb grumbullohen \u2014 thjesht zhvilluesi ka b\u00ebr\u00eb nj\u00eb gabim dhe kjo ju ka kaluar. Sipas rregullave, vlera \u03b1 duhet t\u00eb jet\u00eb gjysm\u00eb e atij \u03b1 q\u00eb p\u00ebrdoret gjat\u00eb prov\u00ebs p\u00ebr hipotez\u00ebn e superioritetit, dmth 0,05 \/ 2 = 0,025.<\/p>\n

Madh\u00ebsia e mostr\u00ebs <\/h3>\n

\nSi t\u00eb vler\u00ebsoni madh\u00ebsin\u00eb e mostr\u00ebs? N\u00ebse mendoni se ndryshimi i v\u00ebrtet\u00eb i mesatarjeve midis A dhe B \u00ebsht\u00eb 0, at\u00ebher\u00eb llogaritja e madh\u00ebsis\u00eb s\u00eb mostr\u00ebs do t\u00eb jet\u00eb e nj\u00ebjt\u00eb si gjat\u00eb prov\u00ebs p\u00ebr hipotez\u00ebn e superioritetit, p\u00ebrve\u00e7 se z\u00ebvend\u00ebsoni madh\u00ebsin\u00eb e efektit me kufirin e jo-m\u00eb pak efikasiteti, me kusht q\u00eb ta p\u00ebrdorni. \u03b1jo jo-m\u00eb pak efikasitet = 1\/2\u03b1superioritet <\/b>(\u03b1non-inferiority=1\/2\u03b1superiority). N\u00ebse keni arsye t\u00eb besoni se opsioni B mund t\u00eb jet\u00eb disi m\u00eb i dob\u00ebt se opsioni A, por d\u00ebshironi t\u00eb provoni q\u00eb ai nuk \u00ebsht\u00eb m\u00eb i dob\u00ebt se \u0394, at\u00ebher\u00eb keni me fat! Faktikisht, kjo e ul madh\u00ebsin\u00eb e mostr\u00ebs tuaj, sepse \u00ebsht\u00eb m\u00eb e leht\u00eb t\u00eb demonstroni se B \u00ebsht\u00eb m\u00eb i dob\u00ebt se A, n\u00ebse n\u00eb t\u00eb v\u00ebrtet\u00eb besoni se ai \u00ebsht\u00eb disi m\u00eb i dob\u00ebt dhe jo i barabart\u00eb.<\/p>\n

Shembulli me vendim <\/h3>\n

\nSupozoni se d\u00ebshironi t\u00eb kaloni n\u00eb versionin B, me kusht q\u00eb ai t\u00eb jet\u00eb m\u00eb i dob\u00ebt se versioni A jo m\u00eb shum\u00eb se 0,1 pik\u00eb n\u00eb shkall\u00ebn e k\u00ebnaq\u00ebsis\u00eb s\u00eb konsumator\u00ebve nga 5... Le t'i qasemi k\u00ebtij problemi duke p\u00ebrdorur hipotez\u00ebn e superioritetit.<\/p>\n

P\u00ebr t\u00eb testuar hipotez\u00ebn e superioritetit, do t\u00eb llogaritnim madh\u00ebsin\u00eb e mostr\u00ebs si m\u00eb posht\u00eb:<\/p>\n

\"Kur
\n
\nK\u00ebshtu, n\u00ebse keni 2103 v\u00ebzhgime n\u00eb grupin tuaj, mund t\u00eb jeni 90% t\u00eb sigurt se do t\u00eb zbuloni nj\u00eb efekt me madh\u00ebsi 0,10 ose m\u00eb shum\u00eb. Por n\u00ebse vlera 0,10 \u00ebsht\u00eb shum\u00eb e madhe p\u00ebr ju, ndoshta nuk ia vlen ta testoni hipotez\u00ebn e superioritetit p\u00ebr t\u00eb. Ndoshta p\u00ebr besueshm\u00ebrin\u00eb, do t\u00eb vendosni t\u00eb kryeni nj\u00eb studim p\u00ebr nj\u00eb efekt m\u00eb t\u00eb vog\u00ebl, p\u00ebr shembull 0,05. N\u00eb k\u00ebt\u00eb rast, do t'ju nevojiten 8407 v\u00ebzhgime, dmth, mostra do t\u00eb rritet gati 4 her\u00eb. Por \u00e7far\u00eb n\u00ebse ne q\u00ebndrojm\u00eb me madh\u00ebsin\u00eb ton\u00eb fillestare t\u00eb mostr\u00ebs, por rrisim fuqin\u00eb n\u00eb 0,99, n\u00eb m\u00ebnyr\u00eb q\u00eb t\u00eb mos kemi dyshime n\u00ebse marrim nj\u00eb rezultat pozitiv? N\u00eb k\u00ebt\u00eb rast, n p\u00ebr nj\u00eb grup do t\u00eb jet\u00eb 3676, q\u00eb \u00ebsht\u00eb tashm\u00eb m\u00eb mir\u00eb, por rrit madh\u00ebsin\u00eb e mostr\u00ebs m\u00eb shum\u00eb se 50%. Dhe si rezultat, ne ende thjesht nuk mund t\u00eb mohojm\u00eb hipotez\u00ebn zero, dhe as t\u00eb marrim nj\u00eb p\u00ebrgjigje p\u00ebr pyetjen ton\u00eb.<\/p>\n

\u00c7far\u00eb n\u00ebse n\u00eb vend t\u00eb k\u00ebsaj testojm\u00eb hipotez\u00ebn e efikasitetit jo m\u00eb t\u00eb vog\u00ebl? <\/p>\n

\"Kur
\n
\nMadh\u00ebsia e mostr\u00ebs do t\u00eb llogaritet sipas t\u00eb nj\u00ebjt\u00ebs formul\u00eb p\u00ebrve\u00e7 em\u00ebruesit.
\nDallimet nga formula e p\u00ebrdorur p\u00ebr testimin e hipotez\u00ebs s\u00eb superioritetit jan\u00eb si m\u00eb posht\u00eb:<\/p>\n

\u2014 Z1\u2212\u03b1\/2 z\u00ebvend\u00ebsohet me Z1\u2212\u03b1, por n\u00ebse b\u00ebni gjith\u00e7ka sipas rregullave, z\u00ebvend\u00ebsoni \u03b1 = 0,05 me \u03b1 = 0,025, dmth, \u00ebsht\u00eb i nj\u00ebjti num\u00ebr (1,96)<\/p>\n

\u2014 n\u00eb em\u00ebrues shfaqet (\u03bcB\u2212\u03bcA)<\/p>\n

\u2014 \u03b8 (madh\u00ebsia e efektit) z\u00ebvend\u00ebsohet me \u0394 (kufi i efikasitetit jo m\u00eb t\u00eb vog\u00ebl)<\/p>\n

N\u00ebse supozoni se \u00b5B = \u00b5A, at\u00ebher\u00eb (\u00b5B \u2212 \u00b5A) = 0 dhe llogaritja e madh\u00ebsis\u00eb s\u00eb mostr\u00ebs p\u00ebr limitin e efikasitetit jo m\u00eb t\u00eb vog\u00ebl \u00ebsht\u00eb pik\u00ebrisht ajo q\u00eb do t\u00eb merrnim n\u00eb llogaritjen e superioritetit p\u00ebr madh\u00ebsin\u00eb e efektit 0,1, shk\u00eblqyesh\u00ebm! Ne mund t\u00eb kryejm\u00eb nj\u00eb studim t\u00eb s\u00eb nj\u00ebjt\u00ebs shkall\u00eb me hipoteza t\u00eb ndryshme dhe nj\u00eb qasje tjet\u00ebr n\u00eb p\u00ebrfundimet, dhe do t\u00eb marrim p\u00ebrgjigje p\u00ebr pyetjen q\u00eb v\u00ebrtet duam t\u00eb p\u00ebrgjigjemi.<\/p>\n

Tani supozoni se ne n\u00eb t\u00eb v\u00ebrtet\u00eb nuk besojm\u00eb se \u00b5B = \u00b5A dhe
\nmendojm\u00eb se \u00b5B \u00ebsht\u00eb pak m\u00eb keq, ndoshta me 0,01 nj\u00ebsi. Kjo rrit em\u00ebrtuesin ton\u00eb, duke e ulur madh\u00ebsin\u00eb e mostr\u00ebs p\u00ebr grupin n\u00eb 1737.<\/p>\n

\u00c7far\u00eb do t\u00eb ndodhte n\u00ebse versioni B n\u00eb t\u00eb v\u00ebrtet\u00eb do t\u00eb ishte m\u00eb i mir\u00eb se versioni A? Ne do t\u00eb refuzonim hipotez\u00ebn zero se B \u00ebsht\u00eb m\u00eb keq se A, m\u00eb shum\u00eb se \u0394 dhe do t\u00eb pranojm\u00eb hipotez\u00ebn alternative se B, n\u00ebse \u00ebsht\u00eb m\u00eb keq, nuk \u00ebsht\u00eb m\u00eb keq se \u0394, dhe mund t\u00eb jet\u00eb m\u00eb i mir\u00eb. Provoni ta p\u00ebrfshini k\u00ebt\u00eb konkludim n\u00eb nj\u00eb prezantim nd\u00ebr-funksional dhe shihni se \u00e7far\u00eb del (seriozisht, provo). N\u00eb nj\u00eb situat\u00eb ku duhet t\u00eb orientohemi n\u00eb perspektiv\u00eb, askush nuk d\u00ebshiron t\u00eb pranoj\u00eb \u201cm\u00eb keq se \u0394 dhe, ndoshta, m\u00eb mir\u00eb\u201d.<\/p>\n

N\u00eb k\u00ebt\u00eb rast, ne mund t\u00eb kryejm\u00eb nj\u00eb studim q\u00eb quhet shum\u00eb shkurtimisht \u201cverifikimi i hipotez\u00ebs se nj\u00ebra nga opsionet e tejkalon tjetrin ose i n\u00ebnshtrohet atij\u201d. Ai p\u00ebrdor dy grupe hipotezash:<\/p>\n

Grupi i par\u00eb (e nj\u00ebjt\u00eb si n\u00eb verifikimin e hipotez\u00ebs p\u00ebr efikasitetin e pakt\u00eb): <\/p>\n

\"Kur
\n
\nGrupi i dyt\u00eb (e nj\u00ebjt\u00eb si n\u00eb verifikimin e hipotez\u00ebs p\u00ebr superioritetin):<\/p>\n

\"Kur
\n
\nNe testojm\u00eb hipotez\u00ebn e dyt\u00eb vet\u00ebm n\u00eb rast se hipoteza e par\u00eb refuzohet. N\u00eb testimin e vazhduesh\u00ebm, ne ruajm\u00eb nivelin e p\u00ebrgjithsh\u00ebm t\u00eb gabimeve t\u00eb par\u00eb (\u03b1). N\u00eb praktik\u00eb, kjo mund t\u00eb arrihet duke krijuar nj\u00eb interval besimi prej 95 % p\u00ebr dallimin midis mesataresh dhe duke verifikuar n\u00ebse e gjith\u00eb periudha e tejkalon -\u0394. N\u00ebse intervali nuk tejkalon -\u0394, ne nuk mund t\u00eb refuzojm\u00eb vler\u00ebn zero dhe ndalemi. N\u00ebse e gjith\u00eb periudha me t\u00eb v\u00ebrtet\u00eb e tejkalon \u2212\u0394, ne do t\u00eb vazhdojm\u00eb dhe do t\u00eb shohim n\u00ebse intervali p\u00ebrmban 0. <\/p>\n

Ekziston nj\u00eb tjet\u00ebr lloj studimesh q\u00eb nuk e kemi diskutuar \u2013 studime ekvivalente. <\/p>\n

Studimet e k\u00ebtij lloji mund t\u00eb z\u00ebvend\u00ebsohen me studime p\u00ebr verifikimin e hipotez\u00ebs p\u00ebr efikasitet t\u00eb nj\u00ebjt\u00eb dhe anasjelltas, por ato kan\u00eb nj\u00eb dallim t\u00eb r\u00ebnd\u00ebsish\u00ebm mes tyre. Testi p\u00ebr verifikimin e hipotez\u00ebs p\u00ebr efikasitetin e nj\u00ebjt\u00eb synon t\u00eb tregoj\u00eb se opsioni B \u00ebsht\u00eb t\u00eb pakt\u00ebn po aq i mir\u00eb sa A. Nd\u00ebrsa hulumtimi i ekvivalenc\u00ebs ka si q\u00ebllim t\u00eb tregoj\u00eb se opsioni B \u00ebsht\u00eb t\u00eb pakt\u00ebn po aq i mir\u00eb sa A, dhe opsioni A \u00ebsht\u00eb po ashtu i mir\u00eb sa B, q\u00eb \u00ebsht\u00eb m\u00eb e komplikuar. N\u00eb thelb, ne p\u00ebrpiqemi t\u00eb p\u00ebrcaktojm\u00eb n\u00ebse i gjith\u00eb intervali i besimit p\u00ebr diferenc\u00ebn e mesatareve q\u00ebndron midis \u2212\u0394 dhe \u0394. K\u00ebto studime k\u00ebrkojn\u00eb nj\u00eb madh\u00ebsi m\u00eb t\u00eb madhe sample dhe realizohen m\u00eb rrall\u00eb. Prandaj, her\u00ebn tjet\u00ebr kur t\u00eb realizoni nj\u00eb studim ku q\u00ebllimi juaj kryesor \u00ebsht\u00eb t\u00eb siguroni q\u00eb versioni i ri nuk \u00ebsht\u00eb m\u00eb i dob\u00ebt, mos pranoni \"mosdiskriminimin e hipotez\u00ebs zero\". N\u00ebse doni t\u00eb verifikoni nj\u00eb hipotez\u00eb v\u00ebrtet t\u00eb r\u00ebnd\u00ebsishme, shqyrtoni mund\u00ebsi t\u00eb ndryshme.
\n
Burimi: habr.com<\/a><\/p>","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

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