 अत्वाज बवाज इत्वाज रवेष्वेज नदा्ज मेरा अगिताई. सब वी वो थवाज नडवाई। विएद्वाज विपाष शंश्ठै की जबाज बगी उद्वेख। परता वाज अद्वाज लिख। विदवाज आप देना करी बिवया, about the population that we are trying to understand. So far, we introduced the concept of hypothesis testing, we gave six steps of classical hypothesis process and we followed these steps to derive the hypothesis testing procedures under the assumption that the population is normal. We consider two cases, one is when the variance of the population is known and the other when the variance for population is unknown and both the times we try to test the hypothesis, the null hypothesis that the mean of the sample mean of the population, I am sorry mean of the population is equal to a fixed value mu 0. In both the cases, we also demonstrated it through an example and we found that when variance is known, it is the standard normal deviate z which is equal to the sample mean minus mu 0 divided by standard deviation over square root n is the test statistic and when sigma square is unknown, x bar that is the sample mean minus the mu 0 divided by sample standard deviation over square root n which is distributed as t with n minus 1 degrees of freedom is the test statistic. In this particular session, we would like to talk about the type 1 error in testing of hypothesis process and talk about type 2 error as a function of parameter under testing. So far what we have done is we have set up the procedure, the 6 steps of classical testing. We define the critical region by equating it to the fixed value type 1 error alpha. But if you look at the classical approach, it says that fix the level alpha at minimum possible level and then the probability of type 1 error should be actually less than or equal to alpha. So far we have equated it with the alpha, it does not make any difference but we would like to set up a test procedure in which we actually calculate the probability of type 1 error and we test it with the alpha. So you remember we had a critical value in the previous approach and we were comparing the test statistic with the critical value. Here we would like to compare the probability with the alpha value that is with the fixed value for which we want to have your type 1 error smaller than that fixed value. So let us start, we again take the same two cases, we assume that the population is normal, mean is unknown, sigma square that is the variance of the population is known, this is one case, when it is unknown it is a second case and on the sidelines I would like you to tell you that this assumption of normality gives a very beautiful closed form solution. So it is very easy to understand this procedure but we will also consider a case in which it is not a normal distribution and still the same statistic can also be utilized with some more calculations to arrive at the similar testing of hypothesis procedure. So we start our test statistic is z which is a sample mean minus mu 0 divided by standard deviation over square root n, it is a parameter without any unknown parameter it is a normal distribution without known parameters 0 and 1, let us call z sub 0 as an actual value of this z when x bar is known. So if known value of x bar is small x bar is available from the data and mu 0 and sigma are given to you then this is a known quantity it is a number therefore type 1 error we actually says that your z 0 should be smaller than the absolute value of z, this defines your critical region and this critical region will be then we will say that reject h 0 if probability that z 0 is smaller than absolute random variate z is less than or equal to alpha or equivalently we can say that probability of z 0 less than z the random variable is less than or equal to alpha by 2. Shall we repeat this once again why it becomes alpha by 2 because here is where I find many people tend to get confused so I am sorry if it is being too much of a repetition but my experience says that it is worth repeating it. So we will have z is a standard normal distribution with a mean 0 and here is what we are looking for this area and this area to be the critical region. Since this is symmetric if the probability of critical region under h 0 is alpha then each of this region has to be alpha by 2 this is first part. So that is why we have come that if you want to take only z greater than that is the random variable z greater than small z 0 this is what we would like to have alpha or smaller. Remember that it will have alpha value if this is z alpha by 1 minus alpha by 2 if you recall the previous session. So this whole area may be I should change the color of the pen let us make it green. So this whole area is alpha by 2 and this is the area we are saying that should be smaller than alpha by 2. So then we continue so if we take the example I am not going to repeat the statement of the example you can take it up from the previous session but then z 0 turns out to be 1.72 and I am sorry this is the miserror I do not want to write this so please remove this. And so we have probability of z 0 smaller than z which is probability of 1.72 smaller than z and that probability turns out to be 0.04727 which is much greater than 0.025 which is alpha by 2 and therefore we cannot reject the lot we have not enough evidence to reject the lot we have to accept it. This comes out from the previous slide that if you have to accept it this probability has to be greater than alpha by 2 and that is what we find here. If sigma square is unknown it is the same thing except that it becomes a t statistic because the sigma the way standard deviation gets replaced by sample standard deviation then it follows a t distribution and it becomes a t random variable and therefore if you take a t 0 or a w 0 as it is a known quantity given so type 1 error becomes that w 0 because we have already assumed h to be 0 I mean sorry null hypothesis to be true. So we have already put a mu 0 here and therefore this is the type 1 error probability and that has to be is equal to alpha or this is the type 1 probability where t is distributed as a t distribution with n minus 1 degrees of freedom. So we say that reject the null hypothesis if the w 0 is less than absolute value of random variable t that probability has to be less than alpha equivalently in the same argument you can say that the probability of w 0 less than the random or the let us put it other way round. The random variable t has to be greater than w 0 so probability that random variable t is greater than w 0 is 1 minus alpha that is the correct way we put it otherwise you accept the hypothesis once again if we go to the super alloy rods example now you know that in the second case we had taken sample variance or sample standard deviation as 112 mpa so this value turns out to be 1.69 so probability that a t t random variable with n minus 1 degree of freedom is larger than 1.69 is 0.047 which is definitely greater than 0.025 and therefore we cannot reject the lot we do not have sufficient evidence to reject the lot. Let us talk about type 2 error type 2 error is accepting the null hypothesis when in reality it is not true so probability that type 1 sorry it should be type 2 let us correct it here and we will change the color of the pen to black this has to be type 2 error please correct it it has to be a type 2 error so that then we again consider the case of normal population with sigma square known so we want to test the hypothesis that null hypothesis that mu is equal to mu 0 verses alternate that mu is not equal to mu 0 so beta is probability of acceptance of h 0 when population mean is not mu 0 it is some other mu it means that beta is a function of mu because please remember mu is equal to mu 0 completely defines everything here when you say that mu is not equal to mu 0 it does not completely define and therefore it is it becomes the type 2 error becomes a function of mean value which is not equal to mu 0 and therefore it can be expressed at beta of mu as probability when you take mu is equal to mu not mu 0 of the test statistic because our critical region is coming through test statistic and you are making it acceptance so it is less than or equal to z alpha by 2 which is the critical value of the test and therefore this becomes minus z alpha by 2 less than x bar minus mu 0 over sigma over square root n less than or equal to z alpha by 2 this is called an operating characteristic curve which has argument mu and the response beta if you look at this function beta of mu when h 0 is not true z is equal to x bar minus mu over sigma over square root n is distributed where mu is not equal to mu 0 it is distributed as a standard normal variate so we can derive it from here this is the probability as derived earlier we have added or rather we have subtracted from every side the mu divided by sigma square root n and I think there is an error here and let me correct it yes there is an error here there are lots of errors so let us start correcting it here we have to subtract it in all the sides so this has to be minus so we have subtracted everywhere mu over sigma square root n this is subtracted from all the sides so this has to be minus and then what then what we find is I think this step there is an error this step there is an error let us do it here so therefore what you have to do is z minus mu 0 over sigma square root n okay this is correct and then what we are doing is yes this z is this no this is correct this z is this and therefore it is z minus mu 0 over sigma square root n and this should be minus and therefore you take mu 0 you add mu 0 over sigma square root n on both the sides so you get this so this is correct this is correct with this correction this is correct with this correction and therefore it is phi remember phi is phi of A is another notation where minus infinity to A 1 over square root 2 pi exponential 1 half x square dx is called phi of A it is a function it is a cumulative distribution function of standard normal variate so this shows the cumulative distribution function of standard normal variate so what it says is that it says that the area under this curve can be shown as area in this curve minus the area in this curve so that is what it is being shown here it takes the full area here minus this area is this central area is what this equation shows so if you take the super alloy rods area I mean the case and you say that suppose the actual mean yield strength is 100 and 1120 MPa and but we accepted the null hypothesis you remember in the previous case we said that the null hypothesis is acceptable it means that you have accepted that mu 0 is 1110 MPa I have forgotten to put the unit here please make sure this is MPa so in this case what is the type 1 error we have committed so we calculate it out beta of this mean value which is the difference between these z alpha by 2 is 1.96 I am mistaken I think right the mistake has happened here it is z 1 minus alpha by 2 and the other is alpha by 2 so here alpha by 2 what we mean is this area okay so in that case it is 1.96 so actually we are taking z 1 minus alpha by 2 please make correction here that here I am sorry for so many corrections but it is z 1 minus alpha by 2 okay and therefore we get the type 2 error as 0.85 which is actually is very high one of the reasons could be you will see that the yield strength is not near really normal and we are comparing it with the standard normal variate and therefore it might be giving us this value anyway this was just an example to demonstrate how the type 2 error is to be calculated so let us summarize we carried out the classical hypothesis testing procedure by fixing the type 1 error to alpha in this case what we did is we actually calculated out the type 1 error and we showed it that if we set up the decision procedure that you reject the null hypothesis if the type 1 error is less than or equal to alpha and you accept the null hypothesis if it is greater than alpha we introduced also I mean reintroduce the concept of type 2 error as a function of mean of normal population when variance is unknown variance is known please note that this becomes this mean is not the same as mu 0 this is under the alternate hypothesis and the same process you can extend it for the case when the population variance is unknown instead of dealing with a standard normal deviates z you will be working with a t deviates with a degrees of freedom n minus 1 thank you