 यागा द astronauts लोनी भा duri bhile आपकत further और मे conditioning वआब रेकon pure रेकon तो �しくझब नाँम el रीए on रीए on रीन यिस देल इस तो आपन आपन tsos अद िी ने कप Inte , और आ़उयारत मर तरушे चब लाग anecd isämोग खन stitch äd बया थी किए हैं बशमाँ डंका इसनी में। धद दी यी, महच़ एखतिंने करे abayi Path-in sankítis eaten ganda तर आ़उयार वे से, दै लगता कर्टिक ucz clinicaleth Arsenal falls in that region we say that hypothesis is to be rejected then that region we are going to call critical region. When we do this, when we make such a decision that null hypothesis is true or null hypothesis is not true in any such situation we commit 2 kinds of errors the first error is called अल of an book me अल six आल अल अल आल तो उझा अल अल अल अल आल औज तै या घरatorio आँन चौटला, के शस्ते सो ब फ энब ईगृटे, मेंग कता, से पँउज आणल हमन dumbbell, ke खिल वह का shelves खिल, के脱़ह काौना तिरट की वच्छ Nom शल्लें,िरट के षाट मने काई तेश्तसी हूँ, स्त्रेस के कारा ती ओचा दू की जीग, डव्य च् lyn strategies थो अपहास्वाय सह्वास्वूपड़ी से णंजा, नू मर्माधा loan क से yeah already , को आए, � boost Hmm, because withНजद engaged, the , बग्द, अदशता के लँवागते मे bishar , आदशता की लेकता मैंत oil, इ�HARE Birthdays are , अदशम �ход, फ Hahaha is थध तश कल काछ he Anся दें कर सम्तेंत् करinda ज़ ہےで ब स्या साथ तच不過 kasha publata तो बस्या स्या Granny के वेर � 및 मिंसे ज़न है। liyeh थद land रहाेooo Develop लगे ofsor very carefully the procedure that we are following right now assumes normality, बड लोक at the details. I will bring it up to your notice that what we call test statistic. आँ वे we will define it pretty soon. that actually is a stand alone it is stands by Its self and it does not really need a normal assumption. We will have one such case also in not in this session, but in future sessions. So, let us start with the classical approach of 6 steps quickly, first, we fix a level अवाज़ा क्य़ मेंवाजे़। God is a small value typically chosen at 1% 5% 10% क्य़ मेंवाजे़। 10% क्य़ मेंवाज़। Then we clearly state what is an null hypothesis and what is its alternate hypothesis that is critical region false for what condition is what we have to define. दे उगे से ज़़ा मैं आपका वार चीन के सब दिया ठा है. कि रहाद़ सब गोरूज रहाड लिए मैं एक शीथ करने व्र्टान का वज रहे है। तो तो, तो तो तो तो तो आपका लिए मैं ग़ाद़ के सब उग़िपूना मैं जाग्से नहीं!! then we chose an appropriate estimator for theta. where H naught or the null hypothesis can be rejected. Then we calculate the error probability that the data falls in critical region when actually null hypothesis is true and we fix it to a level alpha. This way we find out exact nature of critical regency. Because we have a nature of critical regency its final point is to make a decision whether the data you have taken, the sample value that you have got falls within the critical region or not. If it does your decision is hypothesis is to be rejected, if it does not then you say that you do not have sufficient evidence to reject the null hypothesis. As I said we are going to take a model case of normal distribution with mean mu variance sigma square and here we are going to assume that sigma square is known. So we have a data which is a sample random sample independent and identically distributed normal mu sigma square sample x1, x2, x3, xn where sigma square is known but mu is unknown and we want to test the hypothesis that mu is a prefixed value or pre decided or known value mu 0 versus a two sided alternative which says that mu is not equal to mu naught. So it could be less than mu naught, it could be larger than mu naught but it is not mu naught. So first we have to fix an alpha so let us say that alpha is fixed. We state the null hypothesis and the alternate hypothesis. In the third step we have to find a test statistic. We are going to perform all the test to test whether the mean is equal to a fixed value mu naught. Therefore mean is our point of focus and therefore we find a test statistic in terms of the sample value x1, x2, x3, xn which is x bar which says that expected value of x bar is mean mu and therefore x bar is a good estimator. Now when we say that we want to reject the null hypothesis if it is not in close vicinity of mu 0 what do we mean by that say this is your mu 0 and you have some normal distribution because that is what we have assumed and some one sigma limit is this much this is one sigma limit this is unknown to you I mean presently it is known to us. Now what I would like to say is that I should once I find an x bar once I find an x bar the question is if my x bar is not exactly mu 0 I cannot expect it to be exactly equal to mu 0 but even if it is in some vicinity of mu 0 I should be happy but if it is way beyond the mu 0 then I can say that my null hypothesis does not seem to be true. So this region where I said that it is way beyond the mu 0 value is called a critical region so that critical region I am defining is that if somewhere here I have x bar and if I take x bar minus mu 0 and I take its absolute value positive or negative it should be larger than some pre-decided value some value not pre-decided but some value C if it is larger than C this is what I have written if my absolute value of x bar minus mu 0 is larger than some value C which says that it is not in vicinity of mu 0 then that should be my critical region. Now what I find is that if I take x bar minus mu 0 and divided by appropriately with sigma divided by square root n remember x bar is distributed as normal with mean 0 and variance sigma square by n. So if I take a z is equal to x bar minus mu 0 divided by sigma over square root n then it is distributed as normal 0 1 if h 0 is 0 that is if mu 0 is mu is equal to mu 0. So if this is the case so then I actually change my test statistic to z because now it is distributed in a area where it neither depends on mu nor depends on sigma square it is a parametric independent distribution that I have got and therefore z is what I choose as my test statistic or x bar minus mu 0 divided by sigma over square root n is what I choose as my test statistic and I move forward. So this step I have just repeated here for the continuity sake. So the next step is that probability that I have to find the type 1 error. So probability that x 1, x 2, x 3, x n is such that z value is greater than c dash because here I have used c I am calling it c dash because it is not exactly the same it should be alpha okay alpha is a pre-decided value. So this is absolute value of z. So I can say you remember that this is now the distribution of z I have shown see here this is the this is the distribution of z okay. So it is mean and its 1 sigma value is here this is 1 sigma limits of it this is 1 sigma limit and its mean value is 0. Now when I say that absolute z has to be greater than c dash it means that I am looking at some c dash here and minus c dash here and the value should lie here and this total should be alpha total value should be alpha. So it means that I am saying that that if you take only this much area which says that it is a z greater than c dash. So this is z greater than c dash this area is going to be alpha by 2. So now it says that your c dash has to be c dash has to be such that probability of z greater than c dash has to be alpha by 2 and therefore c dash is small z 1 minus alpha by 2. Now these are the small z values if I put the values remember z is the capital Z is a random variable small z is the values that it takes. So if I take any value what is the value of c dash well it has to be this has to be alpha by 2 it means that all the area on this side of the curve has to be 1 minus alpha by 2 and remember the z a value is defined as minus infinity 2 z a sorry is defined as z a such that 1 over square root 2 pi exponential minus 1 half x square dx is equal to a. So you have this probability is 1 minus alpha by 2 that also brings you to the value of z sub alpha by 1 minus alpha by 2 and therefore it says that c dash is z 1 minus alpha by 2 and therefore we get the critical region that if x bar minus mu 0 divided by sigma over square root n remember that now this is a number because x bar is known mu 0 is known sigma is known and n is known if this number is larger than 1 minus z 1 minus alpha by 2 then we reject the null hypothesis in this once again I say that the test statistic z x bar not as a value but as an estimator this is called a test statistic as I have mentioned here and small z 1 minus alpha by 2 is called the critical value of the test because that is the value which decides the critical region. So now to understand it better let better let us have go through a example. So here I have an example it is completely a made up example and a row space industry is interested in buying certain super alloy rods from a foundry and the industry has been told that the super alloy would have a yield strength of about 1110 MPa with a standard deviation of 110 MPa. The industry takes the aerospace industry takes a random sample of 100 from the supplied lot now it wants to find out whether this lot whatever the mean it has does it satisfy does it fall in the vicinity of 1110 MPa. So it wants to what it does it finds that actually the average yield strength is 1129 MPa now company has to take a decision whether it should accept the supply. So let us first understand in our terminology what all has been given to us number one mu 0 is what the average company has claimed foundry has claimed. So it is 1110 MPa sigma square variance is known standard deviation is 110 MPa the sample of size 100 is taken so n is equal to 100 and it has found a mean yield strength of 1129 MPa. So let us follow the classical steps let us fix the alpha at 5% so it is alpha is 0.05 null hypothesis is that the mean value should be 1110 MPa and alternate hypothesis is we are taking only one sided alternate and we say that it has to be it is not equal to 1110 MPa because it is 1129 MPa well 10 and 1029 how much difference does it make we do not know the statistic of interest is x bar which takes a value of 1129 MPa I am sorry I forgot to write the unit let us write it down it should be m p a ok it is very important to realize that I just thought that when we say that mu is equal to 1110 MPa what we really mean is that mean lies in the vicinity of 1110 MPa the reason being it has a variance of 110 sorry standard deviation of 110 MPa so it can vary within that range and therefore the question comes whether this is a outside the range or not and that range is what we are trying to find and that is the range which is acceptance region what we are looking for is the rejection region so here we take the next step we say that the test statistic which is x bar minus mu 0 divided by sigma over square root n should be greater than some value z when h 0 is true so we find that this value should be equal to if you take the probability then that is the error if you put mu 0 there you are accepting the null hypothesis but in reality it is not true and therefore this is the rejection region when null hypothesis is true therefore you have written mu is equal to mu 0 that should be alpha which is 0.05 so and z is as we said before it is a parameter less distribution where distributed variable it is distributed as normal 0 1 not even parameter less it is a unit less because we have already divided MPa with MPa so it is a unit less parameter less random variable which has a normal 0 1 it is fully defined random variable so then probability of that z greater than we know that 1 z of small z sub 1 minus alpha by 2 has to be 0.025 this has to be alpha by 2 and if you look into it as we said please this is something which we must remember so I am going to repeat it several times in today's and pre other sessions this and other sessions we are looking for z value here so that this region is alpha but we always calculate z of any a is such that minus infinity to z of a 1 over square root 2 pi exponential minus 1 half x square dx is equal to a so it takes this region as its probability this interval is this region and therefore this value sorry this is alpha by 2 so this side is 1 minus alpha by 2 and therefore z like a it has to be 1 minus alpha by 2 so that we find that if alpha is alpha by 2 is 0.2 0.025 z 0 is 0.975 which is 1.96 if you plug in the value of x bar now we are taking actual values of x bar then this comes to 1.72 which is not greater than 1.96 and therefore we cannot reject the null hypothesis and so the industry should accept the supplied lot this is the decision this is how the six steps are to be followed what is to be remembered is that up to this point you are talking about this value as a random variable and at this point in reality it is actual value maybe we should make a change here we should instead of calling x bar we should call it a small x bar and this x bar may be replaced by small x bar that would make the that would bring more clarity I believe so we find our final decision in this case is that we accept the lot suppose sigma square is not known here I have shown both the cases parallely we have already derived the six steps for the sigma known you will find that they are going absolutely parallely when sigma square is unknown your x1 x2 x3 xn is same sigma square is unknown u is also unknown your null hypothesis and alternate hypothesis is same as what we had before alpha is to be fixed which is same null hypothesis are same third step v is the same that expected value of x bar is mu so x bar is the estimator therefore h naught can be rejected if x bar minus mu is greater than c once again we are saying that x bar has to be in the vicinity of mu naught may not be a bad idea to repeat what we had said before so let us do it that we have this is mu naught and this is the distribution of the population and the variance is sigma sorry standard deviation is sigma and then we are saying that the x bar has to be in some vicinity of mu naught so that we can say that if it is beyond if it is too large it is not acceptable please remember that we we are accepting the fact that it can never be dot mu zero we are never saying that it should be dot mu zero we are saying that it will fall in the vicinity of mu zero and when it is very far away when it is too far away from mu zero that is when we define our rejection region so here we have to define a rejection region that x bar minus mu is greater than some quantity c and therefore now we have a test statistic just as we did what we did we neutralize we remove the units and we brought a test statistic to a level which had a distribution without any unknown parameters so here also what we find is it is a t which is equal to x bar minus mu zero instead of sigma we have a sample standard deviation divided by square root n and we know that under normality this is distributed as t n minus one it is a t distribution with n minus one degrees of freedom when h naught is true and therefore on the similar line instead of a normal table you will be looking into the t table please remember that the t table t distribution is also a symmetric distribution but it has a thicker tails it does not have a thin tail like normal here also the value zero is in the center and therefore here you are looking once again if the t value is defined t of a is defined if this is a then this is the probability defined ta then this is called a probability a we would like to have this to be alpha by two then this has to be one minus alpha by two your a has to be one minus alpha by two so accordingly it is shown here and we say that if your t statistic which is x bar minus mu zero divided by sample standard deviation by square root n is greater than t one minus alpha by two for the n minus one degrees of freedom of t distribution then you reject h zero let us again take an example the same example i am taking now i am not saying that it has a variance of 110 mpa we do not know the variance 100 samples have been taken the mean yield strength is found to be 1129 mpa and the standard deviation is found to be 112 mpa should the company accept the supply so here we have mu zero is equal to 1110 mpa which is what the company has claimed n is 100 x bar is 1129 mpa and sample variance sample standard deviation is 112 mpa so we follow the six steps fix alpha sale 0.05 we have mean is equal to 1110 mpa versus mean mean is not equal to 1110 mpa the statistic of interest or i think i have made a mistake it should have been i don't know why it has come here this should have been 29 it is 1129 next we take the test statistic because sigma square is unknown as a t statistic which is x bar minus mu zero over standard deviation sample standard deviation by square root n and therefore probability of the absolute value of t statistic i call this a t statistic this is equal to t so this is absolute value of t is greater than c should be alpha which is 0.05 then you find again you have to take the value i repeat again because this is very important to understand this is t distribution and generally t of a is defined as when this probability of a so if you have this side probability of alpha by 2 you would like to have a to be 1 minus alpha by 2 and therefore your t a should be t 1 minus alpha by 2 and alpha by 2 is 0.025 if you go to the table you will find that t 99 because 100 is the data point so it is t 99 with alpha by 2 1 minus alpha by 2 it comes to 1.98 and our test statistic value is 1.69 which is not greater than 1.98 so again we cannot reject the hypothesis and therefore we company has to accept industry has to accept the lot so let us summarize it in this session we explained two examples or two methods assuming normal population we followed six steps of classical hypothesis testing process and both the methods where sigma square is known and sigma square is unknown we gave an example to demonstrate what really we have to do the few things to remember is that please follow the six steps one after the other to avoid any kind of confusion second thing please notice that the underlying statistic is actually x bar minus mu 0 over sigma square root n or x bar minus mu 0 over sample standard deviation square root n please remember that these are the two central statistic we are working with and they are very important they do not really require a normal assumption you can deviate from it only the procedure changes thank you