# en/01D9UwYi1M4v.xml.gz
# ht/01D9UwYi1M4v.xml.gz
(src)="1">
(trg)="1"> ...
(src)="2"> This is the same problem that we had in the last video .
(trg)="2"> Sa se pwoblèm menm sa nou te genyen nan videyo dènye a .
(src)="3"> But instead of trying to figure out whether the data supplies sufficient evidence to conclude that the engines meet the actual emissions requirement , and all of the hypothesis testing , I thought I would also use the same data that we had in the last video to actually come up with a 95 % confidence interval .
(trg)="3"> Men olye ke yo ap eseye pou evalye si done fournitures ase evidans , dezyem prèv pou fèmen sa motè ranpli kondisyon émissions aktyèl la , ak tout de la ipozisyon sonde , mwen te panse mwen tou ta pwal itilize menm data nou menm ki te nan videyo dènye a aktyèlman vini ak yon 95 % entèval konfyans .
(src)="4"> So you could ignore the question right here .
(trg)="4"> Se konsa ou te kapab inyore kesyon isit .
(src)="5"> You can ignore all of this .
(trg)="5"> Ou ka meprize tout sa .
(src)="6"> I 'm just using that same data to come up with a 95 % confidence interval for the actual mean emission for this new engine design .
(trg)="6"> Mwen sèlman ap itilize sa menm done pou vini ak yon 95 % entèval konfidans pou réel emisyon mechan pou sa nouvo motè conception .
(src)="7"> So we want to find a 95 % confidence interval .
(trg)="7"> Se konsa nou vle jwenn yon entèval konfidans 95 % .
(src)="8"> And as you could imagine , because we only have 10 samples right here , we 're going to want to use a
(trg)="8"> Menm jan ou te kapab kwè , paske nou gen sèlman 10 échantillons dwat isit la , nou pral ta vle itilize yon
(src)="9"> T- distribution .
(trg)="9"> T- distribisyon .
(src)="10"> And right down here I have a T- table .
(trg)="10"> Et droit desann isit la mwen gen yon tab T .
(src)="11"> And we want a 95 % confidence interval .
(trg)="11"> Apre sa , nou vle yon entèval konfidans 95 % .
(src)="12"> So we want to think about the range of T- values that 95 -- or the range that 95 % of T- values will fall under .
(trg)="12"> Se poutèt sa , nou vle pou panse osijè de gamme de T- valè ke 95 - ou ranje sa 95 % nan T- valè ap tonbe anba a .
(src)="13"> So let 's think about this way .
(trg)="13"> Se konsa nou panse osijè de chemen sa a .
(src)="14"> So let me draw a
(trg)="14"> Se pou m´ fè yon
(src)="15"> T- distribution right over here .
(trg)="15"> T- distribisyon droit sou isit la .
(src)="16"> So a T- distribution looks very similar to a normal distribution but it has fatter tails .
(trg)="16"> Se konsa yon distribisyon T parèt trè sanble ak yon nòmal distribisyon men li gen pi gra queue .
(src)="17"> This end and this end will be fatter than in a normal distribution .
(trg)="17"> Fin sa a ak nan fen sa a y´ a fre pase yon nòmal distribisyon .
(src)="18"> And then we want to find an interval , so if this is a normalized T- distribution the mean is going to be 0 .
(trg)="18"> E lè sa a nou vle pou jwenn yon entèval , sa si sa se yon nòmalizasyon T- distribisyon : la pwal 0 .
(src)="19"> And we want to find interval of T- values between some negative value here and some positive value here that contains 95 % of the probability .
(trg)="19"> E nou vle pou konnen entèval de T- valè ant kèk negatif valè isit la ak kèk pozitif valeur isit la ki contient 95 % nan pwobabilite a .
(src)="20"> So this right here has to be 95 % .
(trg)="20"> Se poutèt sa a isit- menm gen pou être 95 % .
(src)="21"> And to figure what these critical T- values are at this end and this end , we can just use a T- table .
(trg)="21"> Pou figure sa kritik T- valè sa yo se lè sa a fen ak nan fen sa a , nou kapab sèvi jis ak yon tab T .
(src)="22"> And we 're going to use the two- sided version of this because we 're symmetric around the center .
(trg)="22"> E nou pral sèvi ak vèsyon pipiti de la de kote youn de sa paske nou gen simetrik nan mitan .
(src)="23"> So you look at the two- sided , we want a 95 % confidence interval , so we 're going to look right over here , 95 % confidence interval .
(trg)="23"> Se konsa , ou gade nan a de kote nou vle yon konfidans 95 % entèval , se poutèt sa , nou pral voye je dwat sou isit la , 95 % entèval konfyans .
(src)="24"> We have 10 data points , which means we have 9 degrees of freedom .
(trg)="24"> Nou gen 10 done pwen , ki vle di nou pa gen 9 degre de libète .
(src)="25"> So 9 degrees of freedom for our 10 data points .
(trg)="25"> Se konsa 9 degre de libète pou pwen 10 enfòmasyon nou yo .
(src)="26"> We just took 10 minus 1 .
(trg)="26"> Nou sèlman pran 10 moins 1 .
(src)="27"> So if we look over here , so for a T- distribution with 9 degrees of freedom , you 're going to have 95 % of the probability is going to be contained within a T- value of -- so the T- value is going to be between negative , so this value right here is 2 . 262 , and this value right here is negative 2 . 262 .
(trg)="27"> Si nou voye je sou isit la , se konsa pou yon T separe ak 9 degre de libète yo , nou pral dwe 95 % de la pwobabilite pwal être contenue nan yon valè T de - se konsa valè T a pwal ant négatif , se konsa valè dwa sa a isit la , se 2 . 262 , Et sa a valè dwat isit la se negatif 2 . 262 .
(src)="28"> That 's what this right here tells us .
(trg)="28"> Sa se sa sa a isit- menm di nou .
(src)="29"> That if you contain all the values that are less than 2 . 262 away from the center of your T- distribution , you will contain 95 % of the probability .
(trg)="29"> Si sa nou genyen tout valè sa yo mwens ke 2 . 262 kite sant de distribisyon T ou , ou p' ap genyen 95 % nan pwobabilite a .
(src)="30"> So that is our T- distribution right there .
(trg)="30"> Se poutèt sa se nou T- distribisyon la a .
(src)="31"> Let me make it very clear .
(trg)="31"> Pou m´ fè l´ trè klè .
(src)="32"> This is our T- distribution .
(trg)="32"> Sa se distribisyon T nou . ...
(src)="33"> So if you randomly pick a T- value from this
(trg)="33"> Si ou reponn au yon T valè de sa
(src)="34"> T- distribution , it has a 95 % chance of being within this far from the mean .
(trg)="34"> T- distribisyon , li gen yon chans 95 % Des cours nan sa byen lwen a vle di .
(src)="35"> Or maybe we should write this way .
(trg)="35"> Ou gen dwa nou ta dwe ekri wout sa a .
(src)="36"> If I pick a random T- value , if I take a random T- statistic --
(src)="37"> let me write it this way -- there 's a 95 % chance that a random T- statistic is going to be less than 2 . 262 , and greater than negative 2 . 262 .
(trg)="36"> Si mwen chwazi yon o aza T- valè , si m´ pran yon o aza T- estatistik - kite m´ ekri l´ men ki jan- pa gen yon 95 % chance sa yon o aza T- estatistik k pou mwens pase 2 . 262 , Et pi gran pase negatif 2 . 262 .
(src)="38"> 95 % percent chance .
(trg)="37"> 95 % pousan chans .
(src)="39"> Now when we took this sample , we could also derive a random
(trg)="38"> Koulye a lè nou te pran echantiyon sa a , nou te kapab tou tirer yon o aza
(src)="40"> T- statistic from this .
(trg)="39"> T- estatistik de sa .
(src)="41"> We have our sample mean and our sample standard deviation , our sample mean here is 17 . 17 -- figured that out in the last video , just add these up , divide by 10 -- and our sample standard deviation here is 2 . 98 .
(trg)="40"> Nou dwe nou mwayen echantiyon Et devyasyon estanda echantiyon nou , nou echantiyon isit la se 17 . 17 - sipoze sa soti nan videyo dènye a , jis ajoute sa
(trg)="41"> leve , divize 10 - ak nou estanda echantiyon devyasyon isit la se 2 . 98 .
(src)="42"> So the T- statistic that we can derive from this information right over here -- so let me write it over here -- the
(trg)="42"> Se konsa a T- estatistik sa nou ka tirer de enfòmasyon sa a droit sou isit la ki se pou m´ ekri l´ sou ici - a
(src)="43"> T- statistic that we could derive from this , and you can view this T- statistic as being a random sample from a
(trg)="43"> T- estatistik sa yo nou te kapab tirer de sa , e ou kapab wè sa a T estatistik menm jan yo te yon ti aléatoire de yon
(src)="44"> T- distribution .
(trg)="44"> T- distribisyon .
(src)="45"> A T- distribution with 9 degrees of freedom .
(trg)="45"> Yon T- separe ak 9 degre de libète .
(src)="46"> So the T- statistic that we could derive from that is going to be our mean , 17 . 17 minus the true mean of our population .
(trg)="46"> Se konsa , a T- estatistik sa nou te kapab tirer de sa se va pou nou : , 17 . 17 moins a : tout bon de nou popilasyon an .
(src)="47"> Or actually you would say the true mean of our sampling distribution , which is also going to be the same as the true mean of our population , because that 's our population mean over there , divided by s , which is 2 . 98 over the square root of our number of samples .
(trg)="47"> Ou ou ta di ke tout bon vle di nou D´ distribisyon , ki tou pwal tou a vre : nan popilasyon an , paske se pou sa popilasyon an mechan pase la a , divize pa s , ki se 2 . 98 sou gwo lakou a rasin nou kantite echantiyon yo genyen .
(src)="48"> We 've seen this multiple times .
(trg)="48"> Nou te wè sa a plizyè fwa .
(src)="49"> This right here is the T- statistic .
(trg)="49"> Isit- menm se T- estatistik .
(src)="50"> So by taking this sample you can say that we 've randomly sampled a T- statistic from this 9 degree of freedom
(trg)="50"> Se konsa , t´ ap fè echantiyon sa a ou kapab di ke nou te gen au
(trg)="51"> lot yon T estatistik de 9 degre libète sa a
(src)="51"> T- distribution .
(trg)="52"> T- distribisyon .
(src)="52"> So there 's a 95 % chance that this thing right over here is going to be between -- is going to be less than 2 . 262 and greater than negative 2 . 262 .
(trg)="53"> Se konsa pa gen yon chans 95 % . se yon bagay byen sou isit la va gen ant - ap gen mwens ke 2 . 262 Et pi gran pase negatif 2 . 262 .
(src)="53"> So the 95 % probability still applies to this right here .
(trg)="54"> Se konsa , pwobabilite 95 % a toujou la pou dwa sa a isit la .
(src)="54"> Now we just have to do some math , calculate these things .
(trg)="55"> Koulye a , nou jis gen pou fè kèk matematik , kalkile bagay sa yo .
(src)="55"> So let me get my calculator out .
(trg)="56"> Se pou m´ mete m´ calculatrice yo deyò . ...
(src)="56"> And so let me just calculate this denominator right over here .
(trg)="57"> Se konsa se pou m´ kalkile sa a sèlman denominatè droit sou isit la .
(src)="57"> So we have 2 . 98 divided by the square root of 10 .
(trg)="58"> Se konsa nou dwe 2 . 98 ki te divize pa rasin kare de 10 .
(src)="58"> So that 's 0 . 9423 .
(trg)="59"> Se poutèt sa se 0 . 9423 .
(src)="59"> So what I 'm going to do is I 'm going to multiply both sides of this equation by this expression right over here .
(trg)="60"> Se poutèt sa , se sa mwen pral fè m ap grennen timoun nan tou de kote yo ekwasyon sa a pa espresyon sa a droit sou isit la .
(src)="60"> So if I do that -- so let me just do that right over -- so if I multiply this entire -- this is really two equations or two inequalities I should say .
(trg)="61"> Se konsa si m´ fè sa ki se pou m´ sèlman fè sa byen Sur - sa
(trg)="62"> Si mwen eple kou tout sa a - sa se vrèman ekwasyon de ou de inegalite mwen ta dwe di .
(src)="61"> That this quantity is greater than this quantity and that this quantity 's greater than that quantity .
(trg)="63"> Quantité sa a gen plis pouvwa anpil pase kantite sa a e ke quantité sa a pi konsekan pase sa quantité .
(src)="62"> But we can operate on all of them at the same time , this entire inequality .
(trg)="64"> Men , nou ka opere sou yo tout nan menm moman an , sa inekwasyon ak tout antye .
(src)="63"> So what we want to do is multiply this entire inequality by this value right over here .
(trg)="65"> Se poutèt sa nou vle fè se sa a tout inekwasyon ak valè sa a droit sou isit la .
(src)="64"> And we just calculated it at that value -- let me write it over here -- that 2 . 98 -- I 'll write it right over here -- 2 . 98 over the square root of 10 is equal to 0 . 942 .
(trg)="66"> E nou jis calculé li nan valè sa ki fè m´ ekri l isit- la ke 2 . 98 - m´ ap ekri li dwat sou isit la ki sou 2 . 98 sou rasin kare de 10 egal a 0 . 942 .
(src)="65"> So if I multiplied this entire inequality by 0 . 942 I get , on this left- hand side over here I have negative 2 . 262 times 0 . 942 -- and it 's a positive number that we 're multiplying the whole inequality by , so the inequality signs are still going to be in the same direction -- is less than -- we 're multiplying this whole expression by the same expression in the denominator so it 'll cancel out .
(trg)="67"> Se konsa , si mwen miltiye inekwasyon tout sa a pa 0 . 942 mwen jwenn , sou sa a sou bò gòch bò isit la sou mwen te négatif 2 . 262 fwa 0 . 942 - Et li a yon nonb pozitif ki nou ap multipliant inekwasyon tout a bò la , se konsa mirak inekwasyon sont toujou ale nan menm direksyon an ki se mwens ke - nou ap multipliant espresyon tout sa a pa menm bagay la tou espresyon nan piti denominatè se poutèt li ap annuler .
(src)="66"> So we 're just going to be less than 17 . 17 minus our population mean , which is going to be less than 2 . 262 times , once again , 0 . 942 .
(trg)="68"> Se konsa nou jis pral gen mwens ke 17 . 17 moins nou
(trg)="69"> Popilasyon : , ki pwal gen mwens ke 2 . 262 toujou , yon fwa ankò , 0 . 942 .
(src)="67"> Let me scroll over to the right a little bit .
(trg)="70"> Kite m´ faire sou bò dwat yon ti jan .
(src)="68"> 0 . 942 .
(trg)="71"> 0 . 942 .
(src)="69"> Just be clear , I 'm just multiplying all three sides of this inequality by this number right over here .
(trg)="72"> Jis pou konnen , se mwen menm ki jis multipliant tout twa kote yo de inekwasyon sa a pa dwat sou ici nonm sa a .
(src)="70"> In the middle this cancels out .
(trg)="73"> Nan mitan sa annuler deyò .
(src)="71"> So if I multiply -- I 'll just write it over here -- 0 . 942 , 0 . 942 , 0 . 942 .
(trg)="74"> Se konsa , si mwen eple kou - m ap jis ekri li sou ici- 0 . 942 , 0 . 942 , 0 . 942 .
(src)="72"> This and this is the same number so that 's why those cancel out .
(trg)="75"> Sa e sa se menm nonb se poutèt sa , se poutèt sa annuler .
(src)="73"> And now let 's get the calculator to figure out what these numbers are .
(trg)="76"> Koulye a Ann obtenir la calculatrice pou evalye sa anpil nan moun sa yo ye .
(src)="74"> So if we have the 0 . 942 times 2 . 262 .
(trg)="77"> Si nou gen 0 . 942 a tan 2 . 262 .
(src)="75"> So we 're going to say times 2 . 262 is 2 . 13 .
(trg)="78"> Se poutèt sa , nou pral di tan 2 . 262 se 2 . 13 .
(src)="76"> So this number right over here on the right- hand side is 2 . 13 .
(trg)="79"> Se poutèt sa a anpil byen sou isit la sou a
(trg)="80"> Angle yo bò kote pa 2 . 13 .
(src)="77"> This number on the left is just the negative of that .
(trg)="81"> Nonm sa a sou bò gòch se jis négative de sa .
(src)="78"> So it 's negative 2 . 13 .
(trg)="82"> Se konsa li se negatif 2 . 13 .
(src)="79"> And then we still have our inequalities -- is going to be
(src)="80"> less than 17 . 17 minus the mean , which is less than 2 . 13 .
(trg)="83"> E nou toujou gen inegalite nou yo ki pwal mwens ke 17 . 17 moins vle di , ki se mwens ke 2 . 13 .
(src)="81"> Now what I want to do is I actually want to solve for this mean .
(trg)="84"> Koulye a , kisa mwen vle fè se mwen aktyèlman vle bay solisyon pou sa a vle di .
(src)="82"> And I don 't like that negative sign in the mean .
(trg)="85"> Apre sa , m pa renmen sa negatif siy nan a vle di .
(src)="83"> I 'd rather have this swapped around .
(trg)="86"> Non mwen sa remplacez nan .
(src)="84"> I 'd rather have the mean minus 17 . 17 .
(trg)="87"> Mwen ta gen plutôt a : moins 17 . 17 .
(src)="85"> So what I 'm going to do is multiply this entire inequality by negative 1 .
(trg)="88"> Se poutèt sa mwen pral fè se sa a tout inekwasyon pa negatif 1 .
(src)="86"> If you do that , if you multiply the entire thing times negative 1 , this quantity right here , this negative 2 . 13 will become a positive 2 . 13 .
(trg)="89"> Si ou fè sa , si ou eple kou tout bagay negatif 1 , fwa sa a quantité dwa isit la , sa 2 . 13 negatif pral tounen yon 2 . 13 pozitif .
(src)="87"> But since we are multiplying an inequality by a negative number you have to swap the inequality sign .
(trg)="90"> Men , depi nou sont multipliant inekwasyon ak yon pa yon négatif numéro ou gen pou twoke siy inekwasyon .
(src)="88"> So this less than will become a greater than .
(trg)="91"> Se konsa sa a mwens ke sa pral tounen yon pi gwo pase .
(src)="89"> This negative mu will become a positive mu .
(trg)="92"> Fò negatif sa a ap tounen yon fò pozitif .
(src)="90"> This positive 17 . 17 will become a negative 17 . 17 .
(trg)="93"> 17 . 17 Pozitif sa a ap tounen yon negatif 17 . 17 .
(src)="91"> We 're going to have to swap this inequality sign as well , and this positive 2 . 13 will become a negative 2 . 13 .
(trg)="94"> Nou pwal gen pou twoke inekwasyon mirak sa a tou ,
(trg)="95"> Et 2 . 13 pozitif sa a ap tounen yon negatif 2 . 13 .
(src)="92"> And we 're almost there .
(trg)="96"> Apre sa , nou gen prèt pou genyen .
(src)="93"> We just want to solve for mu .
(trg)="97"> Nou jis vle pou rezoud pou fò .
(src)="94"> Have this inequality expressed in terms of mu .
(trg)="98"> Gen inekwasyon sa a te eksprime fò tèm .
(src)="95"> So what we can do is now just add 17 . 17 to all three sides of this inequality , and we are left with 2 . 13 plus 17 . 17 is greater than mu minus 17 . 17 plus 17 . 17 is just going to be mu , which is greater than -- so this is greater than mu , which is greater than negative 2 . 13 plus 17 . 17 .
(trg)="99"> Se konsa , kisa nou kapab fè se koulye a jis ajoute 17 . 17 tout twa kote yo de sa inekwasyon , e nou ki rete ak 2 . 13 plus 17 . 17 se plis pase fò moins 17 . 17 plis 17 . 17 jis pwal fò , ki gen plis pouvwa pase - se poutèt sa pi gwo pase fò , ki gen plis pouvwa pase negatif 2 . 13 plis 17 . 17 .
(src)="96"> Or a more natural way to write it since we actually have a bunch of greater than signs , that this is actually the
(src)="97"> largest number and this -- oh sorry , this is actually the smallest number and this over here is actually the largest number , is actually flipped -- you can just re- write this inequality the other way .
(trg)="100"> Ou yon jan sa fèt toupatou plis pou yo ekri l´ depi lè nou gen aktyèlman yon pakèt moun sa pi gran pase mirak , ki aktyèlman se a pi gwo kantite ak sa - o padon , sa se aktyèlman a pi piti nombre Et sa a sou isit la se aktyèlman la plus anpil , aktyèlman retournée- ou ka sèlman re- ekri sa inekwasyon lòt bò .
(src)="98"> So now we can write -- actually let 's just figure out what these values are .
(trg)="101"> Men koulye a , nou kapab écrire - aktyèlman an nou jis kèk figi soti sa valè sa yo ye .
(src)="99"> So we have 2 . 13 plus 17 . 17 .
(trg)="102"> Se poutèt sa , nou dwe 2 . 13 plis 17 . 17 .
(src)="100"> So that is the high end of our range .
(trg)="103"> Sa se menm bagay la haut de gamme nou .
(src)="101"> So that is 19 . 3 .
(trg)="104"> Se poutèt sa se 19 . 3 .
(src)="102"> So this value right over here , so this is 19 -- let me do it in that same color -- this value right here is 19 . 3 is going to be greater than mu , which is going to be greater than -- and this is negative 2 . 13 plus 17 . 17 .
(trg)="105"> Se konsa , valè sa a droit sou isit la , se poutèt sa 19 - pou m´ fè l sa a menm koulè - dwat isit la se 19 . 3 valè sa a se va gen plis pouvwa anpil pase fò , ki pwal pi bèl lontan pase - Et sa se negatif 2 . 13 plis 17 . 17 .
(src)="103"> Or we could have 17 . 17 minus 2 . 13 , which gives us 15 . 04 .
(trg)="106"> Ou nou te kapab gen 17 . 17 moins 2 . 13 , ki ban nou 15 . 04 .