I-athikili evela ethimbeni Lokulungisa i-Stitch iphakamisa ukusebenzisa indlela yokuhlola engeyona ephansi ekumaketheni nasekuhloleni umkhiqizo we-A/B. Le ndlela isebenza ngempela uma sihlola isixazululo esisha esinezinzuzo ezingakalwa ngokuhlolwa.
Isibonelo esilula ukuncishiswa kwezindleko. Isibonelo, senza ngokuzenzakalelayo inqubo yokunika isifundo sokuqala, kodwa asifuni ukwehlisa ngokuphawulekayo ukuguqulwa kokugcina ukuya ekugcineni. Noma sihlola izinguquko eziqondiswe engxenyeni eyodwa yabasebenzisi, kuyilapho senza isiqiniseko sokuthi ukuguqulwa kwamanye amasegimenti akwehli kakhulu (uma uhlola imibono eminingana, ungakhohlwa mayelana nezichibiyelo).
Ukukhetha imajini elungile engekho ngaphansi kwengeza izinselele ezengeziwe phakathi nesigaba sokuklama sokuhlola. Umbuzo wokuthi ungakhetha kanjani i-Δ awuhlanganisiwe kahle esihlokweni. Kubonakala sengathi lokhu kukhetha akukhona obala ngokuphelele nasezivivinyweni zomtholampilo. ukushicilelwa kwezokwelapha kokungewona okuphansi kubika ukuthi ingxenye kuphela yokushicilelwe ithethelela ukukhethwa komngcele, futhi ngokuvamile lezi zizathu azicacile noma aziningiliziwe.
Kunoma yikuphi, le ndlela ibonakala ithakazelisayo ngoba ... ngokunciphisa usayizi wesampula odingekayo, ingakhuphula isivinini sokuhlola, futhi, ngakho-ke, isivinini sokwenza izinqumo. - UDaria Mukhina, umhlaziyi womkhiqizo wohlelo lokusebenza lweselula lwe-Skyeng.
Ithimba le-Stitch Fix lithanda ukuhlola izinto ezahlukene. Wonke umphakathi wezobuchwepheshe uthanda ukwenza izivivinyo ngokomgomo. Iyiphi inguqulo yesayithi eheha abasebenzisi abaningi - A noma B? Ingabe inguqulo A yemodeli yokuncoma yenza imali eningi kunenguqulo B? Ukuhlola imibono, cishe sisebenzisa indlela elula kakhulu evela esifundweni sezibalo eziyisisekelo:
Nakuba singavami ukusebenzisa leli gama, lolu hlobo lokuhlola lubizwa ngokuthi "ukuhlolwa kwe-hypothesis ephezulu." Ngale ndlela, sicabanga ukuthi awukho umehluko phakathi kwalezi zinketho ezimbili. Sinamathela kulo mbono futhi siwushiye kuphela uma idatha iphoqa ngokwanele ukwenza kanjalo—okungukuthi, ibonisa ukuthi enye yezinketho (A noma B) ingcono kunenye.
Ukuhlola i- superiority hypothesis kulungele izinkinga ezihlukahlukene. Sikhipha kuphela inguqulo B yemodeli yokuncoma uma ngokucacile ingcono kunenguqulo A esivele isetshenziswa. Kodwa kwezinye izimo, le ndlela yokusebenza ayisebenzi kahle kangako. Ake sibheke izibonelo ezimbalwa.
1) Sisebenzisa isevisi yomuntu wesithathu, esiza ekuboneni amakhadi asebhange omgunyathi. Sithole enye isevisi ebiza kancane kakhulu. Uma isevisi eshibhile isebenza kanye naleyo esiyisebenzisayo njengamanje, sizoyikhetha. Akudingekile ukuthi ibe ngcono kunesevisi oyisebenzisayo.
2) Sifuna ukushiya umthombo wedatha A futhi esikhundleni saso sifake umthombo wedatha ongu-B. Singase sibambezeleka ukushiya u-A uma u-B ekhiqiza imiphumela emibi kakhulu, kodwa akunakwenzeka ukuqhubeka sisebenzisa u-A.
3) Singathanda ukusuka endleleni yokumodelaIndlela ka-A ukuya ku-B hhayi ngoba silindele imiphumela engcono evela ku-B, kodwa ngoba isinika ukuguquguquka okukhulu kokusebenza. Asinaso isizathu sokukholelwa ukuthi u-B uzoba kubi kakhulu, kodwa ngeke senze uguquko uma kunjalo.
4) Senze izinguquko ezimbalwa zekhwalithi esakhiweni sewebhusayithi (inguqulo B) futhi sikholelwa ukuthi le nguqulo iphakeme kunenguqulo A. Asilindele izinguquko ekuguquleni noma yiziphi izinkomba zokusebenza ezibalulekile esivame ukuhlola ngazo iwebhusayithi. Kodwa sikholwa ukuthi kunezinzuzo kumapharamitha okungenzeka azinakulinganiswa noma ubuchwepheshe bethu abanele ukukala.
Kuzo zonke lezi zimo, ucwaningo oluphezulu alusona isisombululo esifanele kakhulu. Kodwa ochwepheshe abaningi ezimweni ezinjalo basebenzisa ngokuzenzakalelayo. Siqhuba ngokucophelela isilingo ukuze sinqume kahle usayizi womphumela. Ukube bekuyiqiniso ukuthi izinguqulo A no-B zisebenza ngezindlela ezifanayo kakhulu, kunethuba lokuthi besingehluleka ukwenqaba i-null hypothesis. Ingabe siphetha ngokuthi u-A no-B benza ngokuyisisekelo okufanayo? Cha! Ukwehluleka ukwenqaba i-null hypothesis nokwamukela i-null hypothesis akuyona into efanayo.
Izibalo zosayizi wesampula (okuyiqiniso, ozenzile) ngokuvamile zenziwa ngemingcele eqinile yephutha loHlobo I (amathuba okwehluleka ukwenqaba i-null hypothesis, evame ukubizwa ngokuthi i-alpha) kunephutha lohlobo lwe-II (amathuba okwehluleka ukwenqaba). i-null hypothesis, uma kunikezwe isimo sokuthi i-null hypothesis ingamanga, ngokuvamile ebizwa ngokuthi i-beta). Inani elijwayelekile le-alpha ngu-0,05, kuyilapho inani elivamile le-beta lingu-0,20, elihambisana namandla ezibalo angu-0,80. Lokhu kusho ukuthi kunamathuba angu-20% okuthi sizophuthelwa umphumela wangempela wenani esilicacisile ekubalweni kwethu kwamandla, futhi lokho kuyigebe elibi kakhulu olwazini. Njengesibonelo, ake sicabangele ama-hypotheses alandelayo:
H0: ubhaka wami AKUKHO ekamelweni lami (3)
H1: ubhaka wami usegumbini lami (4)
Uma ngisesha igumbi lami futhi ngathola ubhaka wami, kuhle, ngingawenqaba umbono ongenalutho. Kodwa uma ngiqalaza ekamelweni futhi ngingawutholi ubhaka wami (Umfanekiso 1), isiphi isiphetho okufanele ngifinyelele kuso? Ngiqinisekile ukuthi ayikho? Ngabe ngibukeke ngiqinile ngokwanele? Kuthiwani uma ngiseshe kuphela u-80% wegumbi? Ukuphetha ngokuthi i-backpack ayikho ekamelweni kungaba yisinqumo sokuxhamazela. Akumangalisi ukuthi singakwazi "ukwamukela i-null hypothesis."
Indawo esiyiseshile
Asiwutholanga ubhaka - kufanele samukele i-null hypothesis?
Umfanekiso 1: Ukusesha u-80% wegumbi kucishe kufane nokusesha ngamandla angu-80%. Uma ungawutholi ubhaka ngemva kokubheka u-80% wegumbi, ungaphetha ngokuthi awukho?
Ngakho-ke usosayensi wedatha kufanele enzeni kulesi simo? Ungakwazi ukwandisa kakhulu amandla ocwaningo, kodwa-ke uzodinga usayizi wesampula omkhulu kakhulu futhi umphumela uzobe ungagculisi.
Ngenhlanhla, izinkinga ezinjalo sekuyisikhathi eside zafundwa emhlabeni wocwaningo lwezokwelapha. Umuthi B ushibhile kunomuthi A; Umuthi B kulindeleke ukuthi ubangele imiphumela emibi embalwa kune-Drug A; Isidakamizwa esingu-B kulula ukusithwala ngoba asidingi ukuba sifakwe esiqandisini, kodwa sidinga umuthi A. Ake sihlole i-hypothesis yokungabi ngaphansi. Lokhu okokukhombisa ukuthi inguqulo B inhle njengenguqulo A—okungenani ngaphakathi kwemajini ethile echazwe ngaphambilini engeyona ephansi, Δ. Sizokhuluma kabanzi mayelana nokuthi ungawumisa kanjani lo mkhawulo ngokuhamba kwesikhathi. Kodwa okwamanje ake sicabange ukuthi lona umehluko omncane kakhulu onenjongo (ngokwengqikithi yokuhlolwa komtholampilo, lokhu ngokuvamile kubizwa ngokuthi ukubaluleka komtholampilo).
I-non-inferiority hypotheses iphendulela yonke into ekhanda layo:
Manje, esikhundleni sokuthatha ngokuthi awukho umehluko, sizothatha ngokuthi inguqulo B yimbi kunenguqulo A, futhi sizonamathela kulokhu kucabangela kuze kube yilapho sibonisa ukuthi akunjalo. Lesi yisikhathi lapho kunengqondo ukusebenzisa ukuhlolwa kwe-hypothesis yohlangothi olulodwa! Empeleni, lokhu kungenziwa ngokwakha isikhawu sokuzethemba nokunquma ukuthi ingabe isikhawu sikhulu kuno-Δ (Umfanekiso 2).
Khetha u-Δ
Ungakhetha kanjani i-Δ efanele? Inqubo yokukhetha ye-Δ ihlanganisa ukulungisiswa kwezibalo kanye nokuhlola okuqinile. Emhlabeni wocwaningo lwezokwelapha, kuneziqondiso zokulawula ezibeka ukuthi i-delta kufanele imele umehluko omncane obalulekile womtholampilo—ozokwenza umehluko ekusebenzeni. Nasi isicaphuna esivela kuzinkombandlela zase-Europe ukuze uzihlole ngaso: “Uma umehluko ukhethwe ngendlela efanele, isikhawu sokuzethemba esiphakathi -∆ kanye no-0… sisanele ukukhombisa ukungabi ngaphansi. Uma lo mphumela ubonakala ungamukeleki, kusho ukuthi ∆ ayikhethwanga ngendlela efanele.”
I-delta akufanele neze yeqe usayizi womthelela wenguqulo A ehlobene nokulawula kwangempela (i-placebo/akukho ukwelashwa), njengoba lokhu kusiholela ekutheni sithi inguqulo B yimbi kunokulawula kwangempela, kuyilapho ngesikhathi esifanayo ikhombisa “ukungabi ngaphansi .” Ake sicabange ukuthi ngenkathi inguqulo A yethulwa, yathathelwa indawo inguqulo 0 noma isici sasingekho nhlobo (bona Umfanekiso 3).
Ngokusekelwe emiphumeleni yokuhlola i-hypothesis yokuphakama, usayizi womphumela u-E wembulwa (okungukuthi, cishe μ^A−μ^0=E). Manje u-A uyindinganiso yethu entsha, futhi sifuna ukwenza isiqiniseko sokuthi u-B muhle njengo-A. Enye indlela yokubhala μB−μA≤−Δ (null hypothesis) ithi μB≤μA−Δ. Uma sicabanga ukuthi lokho kwenza kuyalingana noma kukhulu kuno-E, bese kuba ngu-μB ≤ μA−E ≤ placebo. Manje siyabona ukuthi isilinganiso sethu sika-μB sidlula ngokuphelele u-μA−E, ngaleyo ndlela senqabe ngokuphelele i-hypothesis eyize futhi kusivumele ukuba siphethe ngokuthi u-B muhle njengo-A, kodwa ngesikhathi esifanayo u-μB angase abe ≤ μ i-placebo, okungeyona i-placebo. sidingani. (Umfanekiso 3).
Umfanekiso 3. Ukuboniswa kwezingozi zokukhetha umkhawulo ongekho ngaphansi. Uma i-cutoff iphezulu kakhulu, kungaphethwa ngokuthi u-B ungaphansi kuka-A, kodwa ngesikhathi esifanayo akahlukaniseki ku-placebo. Ngeke sishintshisane ngomuthi osebenza kahle kakhulu kune-placebo (A) ukuze uthole umuthi osebenza njenge-placebo.
Ukukhetha kwe-α
Asiqhubekele ekukhetheni u-α. Ungasebenzisa inani elijwayelekile α = 0,05, kodwa lokhu akulungile ngokuphelele. Njengesibonelo, uma uthenga okuthile ku-inthanethi futhi usebenzisa amakhodi ambalwa esaphulelo ngesikhathi esisodwa, nakuba kungafanele ahlanganiswe - umthuthukisi usanda kwenza iphutha, wabaleka nalo. Ngokwemithetho, inani lika-α kufanele lilingane nengxenye yenani lika-α elisetshenziswa lapho kuhlolwa i-hypothesis yokuphakama, okungukuthi, 0,05 / 2 = 0,025.
Usayizi wesampula
Ungalinganisa kanjani usayizi wesampula? Uma ukholelwa ukuthi umehluko wangempela wencazelo phakathi kuka-A no-B ngu-0, isibalo sosayizi wesampula siyefana nalapho uhlola i-hypothesis yokuphakama, ngaphandle kokuthi ushintsha usayizi womthelela ngemajini engekho ngaphansi, inqobo nje uma usebenzisa. Ukusebenza kahle okungeyona okungaphansi = 1/2αukuphakama (αnon-inferiority=1/2αukuphakama). Uma unesizathu sokukholelwa ukuthi leyo nketho B ingase ibe yimbi kancane kunenketho A, kodwa ufuna ukufakazela ukuthi imbi kakhulu ngokungabi ngaphezu kuka-Δ, usenhlanhleni! Lokhu empeleni kunciphisa usayizi wesampula yakho ngoba kulula ukukhombisa ukuthi u-B mubi kuno-A uma empeleni ucabanga ukuthi kubi kakhulu kunokulingana.
Isibonelo ngesixazululo
Ake sithi ufuna ukuthuthukela kunguqulo B, inqobo nje uma ingekho ngaphezu kwephuzu elingu-0,1 elibi kunenguqulo A esikalini sokwaneliseka kwamakhasimende esinamaphuzu angu-5... Ake sibhekane nale nkinga sisebenzisa i-hypothesis yokuphakama.
Ukuhlola i-hypothesis yokuphakama, singabala usayizi wesampula kanje:
Okusho ukuthi, uma unokubheka okungu-2103 eqenjini lakho, ungaqiniseka ngo-90% ukuthi uzothola usayizi womphumela ongu-0,10 noma ngaphezulu. Kodwa uma i-0,10 iphakeme kakhulu kuwe, kungase kungafaneleki ukuhlola i-hypothesis yokuphakama kwayo. Ukuze ube sohlangothini oluphephile, unganquma ukwenza ucwaningo ngosayizi womphumela omncane, njengo-0,05. Kulokhu, uzodinga ukubonwa okungu-8407, okungukuthi, isampula izokhula cishe izikhathi ezi-4. Kodwa kuthiwani uma sinamathela kusayizi wethu wesampula wangempela, kodwa sandisa amandla ku-0,99 ukuze siphephe uma sithola umphumela omuhle? Kulesi simo, i-n yeqembu elilodwa izoba ngu-3676, osevele ungcono, kodwa ukwandisa usayizi wesampula ngaphezu kuka-50%. Futhi ngenxa yalokho, namanje ngeke sikwazi ukuphikisa i-null hypothesis, futhi ngeke sithole impendulo yombuzo wethu.
Kuthiwani uma sihlola i-nonferiority hypothesis esikhundleni salokho?
Usayizi wesampula uzobalwa kusetshenziswa ifomula efanayo ngaphandle kwedinominetha.
Umehluko kufomula esetshenziselwa ukuhlola i-hypothesis yokuphakama imi kanje:
— Z1−α/2 ithathelwa indawo Z1−α, kodwa uma wenza konke ngokwemithetho, ushintsha u-α = 0,05 ngo-α = 0,025, okungukuthi, inombolo efanayo (1,96)
— (μB−μA) ivela edinominethini
— θ (usayizi womphumela) uthathelwa indawo Δ (umkhawulo wokungephansi)
Uma sithatha ngokuthi µB = µA, khona-ke (µB − µA) = 0 kanye nesibalo sesampula sikasayizi wemajini engeyona ephansi yilokho kanye ebesiyokuthola uma sibala ukuphakama ngosayizi womthelela ongu-0,1, kuhle! Singenza isifundo sobukhulu obufanayo ngemibono ehlukene kanye nendlela ehlukile yokufinyelela eziphethweni, futhi sizothola impendulo yombuzo esifuna ukuwuphendula ngempela.
Manje ake sithi empeleni asicabangi ukuthi µB = µA kanye
Sicabanga ukuthi i-µB yimbi kakhulu, mhlawumbe ngamayunithi angu-0,01. Lokhu kwandisa inani lethu, kunciphisa usayizi wesampula eqenjini ngalinye ukuya ku-1737.
Kwenzekani uma inguqulo B empeleni ingcono kunenguqulo A? Siyayichitha inkolelo-ze eyize yokuthi u-B mubi kuno-A ngaphezu kuka-Δ futhi samukela enye inkolelo-mbono yokuthi u-B, uma kubi kakhulu, akabi kakhulu kuno-A ngo-Δ futhi angase abe ngcono. Zama ukufaka lesi siphetho kuphrezentheshini ehlukahlukene futhi ubone ukuthi kwenzekani (ngokujulile, izame). Esimeni esibheke phambili, akekho ofuna ukuxazulula "okungaphezu kuka-Δ kubi futhi mhlawumbe okungcono."
Kulokhu, singenza ucwaningo, olubizwa kafushane kakhulu “ukuhlola umbono wokuthi enye yezinketho iphakeme noma iphansi kunenye.” Isebenzisa amasethi amabili we-hypotheses:
Isethi yokuqala (kufana nokuhlola i-non-inferiority hypothesis):
Isethi yesibili (okufanayo nalapho kuhlolwa i-hypothesis yokuphakama):
Sihlola i-hypothesis yesibili kuphela uma eyokuqala yenqatshwa. Lapho sihlola ngokulandelana, sigcina isilinganiso sephutha sohlobo I (α). Empeleni, lokhu kungafezwa ngokudala isikhawu sokuzethemba esingu-95% somehluko phakathi kwezindlela nokuhlola ukuze kutholwe ukuthi ingabe sonke isikhawu sikhulu kuno-Δ. Uma isikhawu singeqi -Δ, asikwazi ukwenqaba inani elingenalutho bese simisa. Uma sonke isikhawu ngempela sikhulu kuno-−Δ, sizoqhubeka futhi sibone ukuthi ingabe isikhawu siqukethe u-0.
Kukhona olunye uhlobo locwaningo esingaxoxanga ngalo - izifundo zokulingana.
Lezi zinhlobo zezifundo zingathathelwa indawo izifundo ezingezona ezansi futhi ngokuphambene nalokho, kodwa empeleni zinomehluko obalulekile. Uhlolo lokungeyona into ephansi luhlose ukukhombisa ukuthi inketho B okungenani inhle njengo-A. Isilingo sokulinganisa sihlose ukukhombisa ukuthi inketho B okungenani inhle njengo-A. Inketho A inhle njengo-B, okunzima kakhulu. Empeleni, sizama ukuthola ukuthi ingabe sonke isikhawu sokuzethemba somehluko wezindlela siphakathi kuka-−Δ kanye no-Δ. Izifundo ezinjalo zidinga usayizi wesampula omkhulu futhi zenziwa kancane. Ngakho-ke ngokuzayo lapho uqhuba isifundo lapho inhloso yakho enkulu kuwukuqinisekisa ukuthi inguqulo entsha ayibi nakakhulu, ungakulungiseleli "ukwehluleka ukwenqaba i-null hypothesis." Uma ufuna ukuhlola i-hypothesis ebaluleke ngempela, cabangela izinketho ezahlukene.
Source: www.habr.com