Inqaku elivela kwiqela le-Stitch Fix licebisa ukusebenzisa indlela yovavanyo olungekho ngaphantsi kwintengiso kunye novavanyo lwe-A / B yemveliso. Le ndlela isebenza ngokwenene xa sivavanya isisombululo esitsha esineenzuzo ezingenakulinganiswa ngovavanyo.
Umzekelo olula kukulahleka kwamathambo. Umzekelo, masizenzele inkqubo yokwabela isifundo sokuqala, kodwa asifuni ukuyiyeka kakhulu inguquko ukuya esiphelweni. Okanye sivavanya utshintsho olujolise kwicandelo elinye labasebenzisi, ngelixa siqinisekisa ukuba ukuguqulwa kwamanye amacandelo akubi kakhulu (xa uvavanya iingcamango ezininzi, ungalibali malunga nezilungiso).
Ukukhetha i-right non-inferiority bound yongeza imingeni eyongezelelweyo kwinqanaba loyilo lovavanyo. Umbuzo wendlela yokukhetha i-Δ ayifakwanga kakuhle kwinqaku. Kubonakala ngathi olu khetho alubonakali ngokupheleleyo kulingo lwezonyango. ushicilelo lwezonyango kwiingxelo ezingekho phantsi kweengxelo zokuba kuphela isiqingatha sopapasho sithethelela ukhetho lomda kwaye kaninzi ezi zizathu azicacanga okanye azicaciswanga.
Kwimeko nayiphi na into, le ndlela ibonakala inomdla, njengoko Ngokunciphisa ubungakanani besampulu efunekayo, inokunyusa isantya sovavanyo, kwaye, ngenxa yoko, isantya sokwenza izigqibo. - UDaria Mukhina, umhlalutyi wemveliso weSkyeng app mobile.
Iqela le-Stitch Fix lithanda ukuvavanya izinto ezahlukeneyo. Uluntu lonke lwetekhnoloji luthanda ukuqhuba iimvavanyo. Yeyiphi inguqulelo yesiza etsala abasebenzisi abaninzi - A okanye B? Ngaba uguqulelo A lwemodeli yokuncoma lwenza imali eninzi kunenguqulelo B? Phantse ngalo lonke ixesha, ukuvavanya iingqikelelo, sisebenzisa eyona ndlela ilula ukusuka kwikhosi yezibalo ezisisiseko:
Nangona singafane silisebenzise eli gama, olu hlobo lovavanyo lubizwa ngokuba “kukuvavanya ingqikelelo yokulunga”. Ngale ndlela, sicinga ukuba akukho mahluko phakathi kwezi zimbini iinketho. Sinamathela kule ngcamango kwaye siyishiya kuphela ukuba iziphumo ziqinisekisa ngokwaneleyo ukuba ziqinisekise-oko kukuthi, kubonisa ukuba enye inketho (A okanye B) ingcono kunomnye.
Uvavanyo lwe-hypothesis ephezulu lufanelekile ukusombulula iingxaki ezahlukeneyo. Sikhulula inguqulo ye-B yemodeli ye-recommender kuphela ukuba ingcono ngokucacileyo kune-A esele isetyenziswa. Kodwa kwezinye iimeko, le ndlela ayisebenzi kakuhle. Makhe sijonge imizekelo embalwa.
1) Sisebenzisa inkonzo yomntu wesithathu, enceda ekuchongeni amakhadi ebhanki enkohliso. Sifumene enye inkonzo exabisa kancinci kakhulu. Ukuba inkonzo enexabiso eliphantsi isebenza kunye naleyo siyisebenzisayo ngoku, siya kuyikhetha. Akunyanzelekanga ukuba ibengcono kunenkonzo oyisebenzisayo.
2) Sifuna ukulahla umthombo wedatha A kwaye endaweni yayo ngomthombo wedatha B. Singalibazisa ukulahla u-A ukuba u-B uvelisa iziphumo ezimbi kakhulu, kodwa akunakwenzeka ukuqhubeka sisebenzisa u-A.
3) Singathanda ukuhamba kwindlela yokwenza imodeliIndlela ka-A ukuya ku-B, hayi kuba silindele iziphumo ezingcono ukusuka ku-B, kodwa kuba isinika ukuguquguquka okungaphezulu kokusebenza. Asinasizathu sokukholelwa ukuba i-B iya kuba yimbi ngakumbi, kodwa asiyi kutshintsha ukuba kunjalo.
4) Senze utshintsho oluthile lomgangatho uyilo lwewebhusayithi (Ushicilelo B) kwaye sikholelwa ukuba le nguqulelo ingaphezulu kunenguqulelo A. Asilindelanga utshintsho kwiinguqulelo okanye naziphi na ii-KPIs esiqhele ukulinganisa ngazo iwebhusayithi. Kodwa sikholelwa ukuba kukho izibonelelo kwiiparameters ezingenakulinganiswa, okanye itekhnoloji yethu akwanelanga ukulinganisa.
Kuzo zonke ezi meko, uphando olugqwesileyo ayisosisombululo silungileyo. Kodwa uninzi lweengcali kwiimeko ezinjalo zisebenzisa ngokungagqibekanga. Siqhuba ngononophelo uvavanyo lokumisela ngokuchanekileyo ubukhulu besiphumo. Ukuba bekuyinyani ukuba iinguqulelo A kunye no-B zisebenza ngeendlela ezifanayo kakhulu, amathuba okuba asiyi kuba nako ukukhaba ingqikelelo engeyiyo. Ngaba sigqiba kwelokuba uA no-B ngokubanzi basebenza ngendlela efanayo? Hayi! Ukungaphumeleli ukukhaba i-null hypothesis kunye nokwamkela i-hypothesis engekho akuyona into enye.
Ubalo lobungakanani besampulu (olwenzileyo, kakade) luthande ukuba nemida engqongqo kwimpazamo yoHlobo I (ukuba nokwenzeka kokukhaba ngempazamo ihypothesis engekhoyo, edla ngokubizwa ngokuba yialpha) kunempazamo yoHlobo lwe-II (Inokwenzeka yokusilela ukwala i-null. i-hypothesis, inikwe imeko yokuba i-hypothesis engekhoyo bubuxoki, ehlala ibizwa ngokuba yi-beta). Ixabiso eliqhelekileyo le-alpha ngu-0,05 ngelixa ixabiso eliqhelekileyo le-beta ngu-0,20, elihambelana namandla ezibalo ka-0,80. Oku kuthetha ukuba asinakuyibona isiphumo esiyinyani sexabiso esilibonakalisileyo kwizibalo zethu zamandla okunokwenzeka ngama-20% kwaye oku kusikhewu esibi kakhulu kulwazi. Njengomzekelo, makhe siqwalasele ezi ngqikelelo zilandelayo:
H0: ubhaka wam awukho egumbini lam (3)
H1: ubhaka wam usegumbini lam (4)
Ukuba ndikhangele igumbi lam kwaye ndifumene ubhaka wam, kulungile, ndingayiyeka i-null hypothesis. Kodwa ukuba ndithe ndalaqaza egumbini ndize ndingawufumani ubhaka wam (Umfanekiso 1), ndingafikelela kwesiphi isigqibo? Ndiqinisekile ukuba ayikho? Ngaba ndiyihlolisise kakuhle ngokwaneleyo? Kuthekani ukuba ndikhangele kuphela i-80% yegumbi? Ukugqiba ukuba akukho bhegi kwigumbi kuya kuba sisigqibo esikhawulezileyo. Akumangalisi ukuba singenako "ukwamkela i-hypothesis engekhoyo".
Indawo sayigqogqa
Asiwufumananga ubhaka- ngaba kufuneka samkele i-hypothesis engekhoyo?
Umzobo 1. Ukukhangela i-80% yegumbi malunga nokufana nokwenza uphando nge-80% yamandla. Ukuba awufumananga ubhaka emva kokukhangela malunga ne-80% yegumbi, ngaba unokugqiba ukuba akukho?
Ke kufuneka isazi sedatha senze ntoni kule meko? Ungawanyusa kakhulu amandla ophononongo, kodwa ke uya kufuna ubungakanani besampulu enkulu kakhulu, kwaye isiphumo siya kuhlala singanelisi.
Ngethamsanqa, iingxaki ezinjalo kudala zifundwe kwihlabathi lophando lwezonyango. Ichiza B lixabiso eliphantsi kunechiza A; ichiza B kulindeleke ukuba lenze iziphumo ebezingalindelekanga ezimbalwa kunechiza A; Ichiza B kulula ukulihambisa kuba ayifuni kufakwa efrijini, kodwa ichiza A liyakwenza. Makhe sivavanye i-hypothesis yokungabi ngaphantsi. Oku kukubonisa ukuba uguqulelo B lulungile njengenguqulelo A—ubuncinane ngaphakathi komda othile omiselwe kwangaphambili “ongasebenzi kakuhle”, Δ. Siza kuthetha ngakumbi malunga nendlela yokumisela lo mda kamva. Kodwa okwangoku, masicinge ukuba lo ngowona mahluko omncinci obaluleke kakhulu (kwimeko yolingo lwezonyango, oku kuqhele ukubizwa ngokuba lukubaluleka kwezonyango).
Iingqikelelo malunga nokusebenza okuncinci zijika yonke into ibe phantsi:
Ngoku, endaweni yokucinga ukuba akukho mahluko, sicinga ukuba uguqulelo B lubi kunoguqulelo A, kwaye siya kubambelela kule ngcinga de sibonise ukuba akunjalo. Le yingongoma kanye xa kunengqiqo ukusebenzisa uvavanyo lwe-hypothesis olucalanye! Ngokwenza oku, oku kunokwenziwa ngokwakha ikhefu lokuzithemba kunye nokugqiba ukuba ngaba i-interval ngokwenene inkulu kune-Δ (Figure 2).
Ukhetho Δ
Unokukhetha njani okulungileyo Δ? Inkqubo yokhetho Δ ibandakanya ulungelelwaniso lweenkcukacha-manani kunye novandlakanyo olubambekayo. Ehlabathini lophando lwezonyango, kukho izikhokelo zesiqhelo ezicebisa ukuba idelta kufuneka ibe ngowona mahluko umncinci ubalulekileyo eklinikhi-eya kuthi isebenze. Nantsi isicatshulwa esisuka kwincwadana yaseYurophu ukuze uzivavanye ngayo: “Ukuba umahluko ukhethwe ngokuchanekileyo, ixesha lokuzithemba eliphakathi ngokupheleleyo-∆ kunye no-0… lisanele ukubonisa ukusebenza kakuhle. Ukuba esi siphumo sibonakala singamkelekanga, oko kuthetha ukuba ∆ akakhethwanga ngokufanelekileyo.”
I-delta kufuneka ngokuqinisekileyo ingagqithisi ubungakanani besiphumo se-Version A ngokunxulumene nolawulo lwenyani (i-placebo/akukho unyango), njengoko oku kusikhokelela ekubeni sigqibe kwelokuba i-Version B imbi kunolawulo lokwenyani, lo gama kwangaxeshanye ibonisa "ukusebenza okuncinci. ". Masithi xa kuqaliswa uguqulelo A, uguqulelo 0 lwalukwindawo yalo, okanye uphawu lwalungekho kwaphela (jonga uMfanekiso 3).
Ngokusekelwe kwiziphumo zokuvavanya i-hypothesis yokuphakama, ubukhulu besiphumo E bubonakaliswe (oko kukuthi, mhlawumbi μ^A-μ^0=E). Ngoku uA ngumgangatho wethu omtsha, kwaye sifuna ukuqiniseka ukuba uB ulungile njengo-A. Enye indlela yokubhala uμB−μA≤−Δ (ingqikelelo engekhoyo) ngu μB≤μA−Δ. Ukuba sithatha ukuba ukwenza kuyalingana okanye kukhulu kuno-E, ngoko ke μB ≤ μA−E ≤ placebo. Ngoku siyabona ukuba uqikelelo lwethu luka-μB lukhulu ngokupheleleyo kuno-μA−E, oluthi ngaloo ndlela lukhanyele ngokupheleleyo ucingelo olungento luze lusivumele ukuba sigqibe kwelokuba u-B akakho ngaphantsi kuno-A, kodwa kwangaxeshanye, u-μB unokuba ≤ μ placebo, leyo asiyiyo into esiyidingayo. (Umfanekiso 3).
Umzobo 3. Umboniso weengozi zokukhetha umda ongekho ngaphantsi kokusebenza kakuhle. Ukuba umda uphezulu kakhulu, kunokugqitywa ukuba i-B ayikho ngaphantsi kwe-A, kodwa kwangaxeshanye ayibonakali kwi-placebo. Asiyi kulitshintsha iyeza elisebenza ngokucacileyo kune-placebo (A) kwichiza elisebenza njenge-placebo.
Ukhetho α
Masidlulele kukhetho luka-α. Ungasebenzisa ixabiso eliqhelekileyo α = 0,05, kodwa oku akulunganga ngokupheleleyo. Njengomzekelo, xa uthenga into kwi-Intanethi kwaye usebenzise iikhowudi ezininzi zesaphulelo ngaxeshanye, nangona kungafuneki ukuba zongezwe - umphuhlisi wenze impazamo, kwaye ubaleke nayo. Ngokwemigaqo, ixabiso le-α kufuneka lilingane nesiqingatha sexabiso le-α, elisetyenziselwa ukuvavanya i-hypothesis yokuphakama, oko kukuthi 0,05 / 2 = 0,025.
Ubungakanani besampuli
Indlela yokuqikelela ubungakanani besampulu? Ukuba ucinga ukuba umahluko oyinyaniso phakathi kuka-A kunye no-B ngu-0, ngoko ke ukubalwa kobungakanani besampulu kuyafana novavanyo lwe-hypothesis ephezulu, ngaphandle kokuba ubuyisela ubungakanani besiphumo ngomda wokungasebenzi kakuhle, ngaphandle kokuba sebenzisa α ayikho ngaphantsi ngokufanelekileyo = 1/2 α ukongama (αnon-inferiority=1/2αukuphakama). Ukuba unesizathu sokukholelwa ukuba olu khetho B lunokuba lubi kancinane kunokhetho A, kodwa ufuna ukubonakalisa ukuba alukho ngaphezu kuka-Δ kubi, ke unethamsanqa! Ngokwenyani, oku kunciphisa ubungakanani besampulu yakho kuba kulula ukubonisa ukuba u-B umbi kuno-A ukuba ucinga ukuba kubi kancinci, akulingani.
Umzekelo wesisombululo
Masithi ufuna ukunyusela kuguqulelo B, ngaphandle kokuba akukho ngaphezulu kwe-0,1 amanqaku embi kunoguqulelo A kwisikali sokwaneliseka kwabathengi esinamanqaku ama-5 ... Masiyijonge le ngxaki sisebenzisa ingqikelelo ephezulu.
Ukuvavanya i-hypothesis ephezulu, siya kubala ubungakanani besampulu ngolu hlobo lulandelayo:
Oko kukuthi, ukuba unemibono ye-2103 kwiqela, unokuqiniseka nge-90% ukuba uya kufumana umphumo we-0,10 okanye ngaphezulu. Kodwa ukuba i-0,10 iphezulu kakhulu kuwe, isenokungafanelanga ukuvavanya i-hypothesis yokuphakama kwayo. Unokufuna ukuqiniseka ukuba uqhuba isifundo ngesayizi encinci yesiphumo, esifana ne-0,05. Kule meko, uya kufuna ukuqwalaselwa kwe-8407, oko kukuthi, isampuli iya kwanda ngamaxesha angama-4. Kodwa kuthekani ukuba sinamathela kubungakanani bethu bokuqala besampulu kodwa sonyuse amandla aye ku-0,99 ukuze singathandabuzi ukuba sifumana iziphumo ezilungileyo? Kule meko, n kwiqela elinye liya kuba yi-3676, esele ingcono, kodwa inyusa ubungakanani besampuli ngaphezu kwe-50%. Kwaye ngenxa yoko, asisayi kuba nakho ukuphikisa i-hypothesis engekhoyo, kwaye asiyi kufumana mpendulo kumbuzo wethu.
Kuthekani ukuba endaweni yoko sivavanya i-hypothesis yokungasebenzi kakuhle?
Ubungakanani besampulu buya kubalwa kusetyenziswa ifomula efanayo ngaphandle kwedinomineyitha.
Umahluko kwifomula esetyenziswayo ekuvavanyeni ingqikelelo ephezulu yolu hlobo lulandelayo:
- Z1−α/2 ithathelwe indawo nguZ1−α, kodwa ukuba wenza yonke into ngokwemigaqo, uthatha indawo α = 0,05 nge α = 0,025, oko kukuthi, eli linani elifanayo (1,96)
- ibonakala kwidinomineyitha (μB−μA)
- θ (ubungakanani besiphumo) indawo yayo ithathwa ngu Δ (umda wempumelelo engekho ngaphantsi)
Ukuba sicingela ukuba µB = µA, ngoko (µB − µA) = 0 kunye nokubala ubungakanani besampulu yomda ongekho ngaphantsi koko besiya kukufumana kanye xa sibala ukongama kumlinganiselo wesiphumo sika-0,1, mkhulu! Sinokwenza uphononongo lwesikali esifanayo ngeengcamango ezahlukeneyo kunye nendlela eyahlukileyo yokufikelela kwizigqibo kwaye siya kufumana impendulo yombuzo esifuna ukuwuphendula ngokwenene.
Ngoku cinga ukuba asikholelwa ngokwenene ukuba µB = µA kunye
sicinga ukuba µB mbi ngakumbi, mhlawumbi ngo-0,01 iiyunithi. Oku kwandisa i-denominator yethu, ukunciphisa ubungakanani besampulu kwiqela ngalinye ukuya kwi-1737.
Kwenzeka ntoni ukuba uguqulelo B ngenene lungcono kunohlelo A? Siyayikhaba i-hypothesis engekhoyo yokuba i-B imbi kuno-A ngaphezu kwe-∆ kwaye samkele enye i-hypothesis yokuba i-B, ukuba imbi kakhulu, ayikho mbi kune-∆ kwaye ingaba ngcono. Zama ukubeka eso sigqibo kwinkcazo-ntetho enqamlezileyo kwaye ubone ukuba kwenzeka ntoni (ngokungqongqo, yizame). Kwimeko apho kufuneka uqhelane nekamva, akukho mntu ufuna ukuzinza "kubi ngaphezu kwe-Δ kwaye mhlawumbi ngcono."
Kule meko, sinokuqhuba isifundo esibizwa ngokuba ngokufutshane kakhulu "ukuvavanya ingcamango yokuba enye yeenketho iphezulu okanye ingaphantsi kwenye." Isebenzisa iiseti ezimbini zeengqikelelo:
Iseti yokuqala (efanayo naxa kuhlolwa ingqikelelo yokusebenza kakuhle okungaphantsi):
Iseti yesibini (efanayo naxa uvavanya ingqikelelo ephezulu):
Sivavanya i-hypothesis yesibini kuphela xa eyokuqala yaliwe. Kuvavanyo olulandelelanayo, sigcina inqanaba elipheleleyo leempazamo zohlobo lwe-I (α). Ngokwenza oku, oku kunokufezekiswa ngokudala i-95% yexesha lokuzithemba kumahluko phakathi kweendlela kunye nokujonga ukuba isithuba esipheleleyo sikhulu kuno-Δ. Ukuba ikhefu alidluli -Δ, asinakulahla ixabiso elinguziro kwaye siyeke. Ukuba lonke ixesha lokuphumla ngenene likhulu kuno-−Δ, siyakuqhubela phambili kwaye sibone ukuba isithuba sokuphumla siqulathe u-0.
Kukho olunye uhlobo lophando esingakhange sixoxe ngalo - izifundo zokulingana.
Izifundo zolu hlobo zinokuthi zitshintshwe zizifundo zokuvavanya i-hypothesis yokusebenza ngaphantsi kwaye ngokuphambene noko, kodwa zona ngokwazo zinomohluko obalulekileyo. Uvavanyo olungekho ngaphantsi lujolise ekuboniseni ukuba ukhetho B lulungile njengo-A. Kwaye uphando lokulinganisa lujolise ukubonisa ukuba ukhetho B lulungile njengo-A, kwaye ukhetho A lulungile njengo-B, olunzima ngakumbi. . Ngokwenyani, sizama ukugqiba ukuba ngaba lonke ixesha lokuzithemba lomahluko phakathi kweendlela liphakathi −∆ kunye no-∆. Amaphononongo anjalo afuna iisayizi ezinkulu zeesampulu kwaye aqhutywa rhoqo. Ke kwixesha elizayo xa usenza uphononongo apho eyona nkxalabo yakho iphambili ikukuqinisekisa ukuba inguqulelo entsha ilungile, musa ukuzinzisa "ukungaphumeleli ukuphikisa i-hypothesis engekhoyo." Ukuba ufuna ukuvavanya i-hypothesis ebaluleke ngokwenene, qwalasela iindlela ezahlukeneyo zokukhetha.
umthombo: www.habr.com