 मैंग्लदीखादा हमाने तादशे हैं, अर्खेरा बाशाँ मैंग्लगी लूचे क्यारी वूँँणा कूँऍर, बत ज्डूष और पूँभ्टिस्टीं आर्बशाँ क्यारी बाशाँ मैंगुए लूँँँँँँँँँँँँँँ... All the while, we have worked with the case in which we have considered a normal distribution as the population distribution. So, if we review it quickly, what we have seen is we have tested the null hypothesis if the population is normal with mean mu and variance sigma square and then we have tested the hypothesis that mu is equal to sum given value mu naught against the 3 alternative, the 2 sided alternative that mu is not equal to mu naught and the other 2 are mu is greater than mu naught or mu is less than mu naught. In both the cases we found that if the population variance sigma square is known, then it reduces down to testing a hypothesis using a test statistic z which is a standard normal variate. It is a normal variate with mean 0 and variance 1. In case we assume that sigma square is unknown, then it reduces to a test statistic which is a students t distribution statistic and it is distributed as students t distribution with n minus 1 degrees of freedom. The advantage of the 2 is that none of the 2 statistics distribution under H0, when you assume that null hypothesis is true, it does not depend on any other parameter than the value n or in case of z, it does not depend even in the size of the sample. Something I have not shown here but we also went through a testing of hypothesis process for testing that the population variance is equal to a given value sigma 0 square and we found that chi square that is n minus 1 sample variance divided by the sigma naught square is a chi square variate and that variate if we call it w then or we have called it y probably. That is the variate which is a test statistic. It becomes a chi square distribution with n minus 1 degree of freedom. So, it does not have any other parameter. Only parameter is a known value depending on the sample size and again we tested the 3 alternatives and we found 3 critical regions in which we can reject the hypothesis when we assume that null hypothesis is true. So, then comes what we want to do now. As I said before, we already made an assumption that the underlying population distribution is normal but that may not be always the case and as it was mentioned earlier also that in the material science and metallurgical engineering, number of times the data do not follow a nice symmetric distribution like normal and we cannot really apply central limit theorem because the data size is not sufficiently large. So, in this presentation today, in this session today, we would like to derive a test statistic to test the hypothesis, the same two hypothesis that the population mean is equal to a given value versus its 3 alternatives and population variance is equal to given value versus its 3 alternative but the underlying population distribution will not be assumed to be normal. In this case, as a case, we have assumed it to be viable distribution. We had mentioned it earlier. You just observe the test statistic that we found in testing the hypothesis that mu is equal to mu 0 when sigma square or the population variance is known is z which depends only on x bar and rest of them are known parameters. If you say sigma square is unknown, then sigma square gets replaced by the sample variance. So, t statistic is depends on x bar and s and otherwise mu 0 and n which are all the known quantity. This x bar and s are very special statistics because expected value of x bar is always mu and variance of x bar is always sigma square over n no matter what the underlying population distribution. Also if you look at the expected value of s square is also sigma square independent of what distribution underlying distribution is it need not be normal. So, this actually indicates that the same z and t can well be used as test statistics to test the same hypothesis mu is equal to mu 0 when sigma square is known and mu is equal to mu 0 when sigma square is unknown and we can follow the practice only thing is assumption of normality makes life easy because the distributional aspect of the test statistics z and t become very obvious. One is a standard normal distribution and the other is a t distribution with n minus 1 degree of freedom which may not be the case when you deviate from the assumption of normality. But well in that case with the advent of computers and so much of computing facility and software there always should be possible to find a numeric solution through either the numerical analysis or through Monte Carlo simulation and that is exactly what we are going to demonstrate or rather show it here. So, let us consider a case that we want to test the hypothesis mu is equal to mu 0 under viable distribution. So, here I have given you a situation in which such a problem can arise. Suppose there is an indigenous engine component and this component is now need to be replaced where we earlier the manufacturer was using an imported component. Now the developers of the indigenous component has to show that it is performance or the performance criteria that the imported component meets is are also met by the indigenous component and suppose the criteria is matching the low cycle fatigue life, LCF life is the property we would like to compare. So, assume that we have a data sample data from the indigenous component x1, x2, x3, xn and we would like to from this data we would like to show that the mean LCF value and the standard deviation of LCF value from this sample is same as what you would get from the imported component. Remember it is one can show that LCF values can follow closely viable distribution. So, let us set up the problem. We want to test if LCF of indigenous component is same as that of an imported component. So, for comparison we have two pieces of information. We have an LCF data x1, x2, x3, xn which is a random sample of size n and we have a mean value of imported component mu 0 and the standard deviation of imported component sigma 0. Let mu denote the mean value of LCF of indigenous component and sigma denote the standard deviation of the indigenous component. Then in the present case we would like to test the hypothesis that mu is equal to mu 0 where the population distribution is viable. In reality we have to assume three parameter viable distribution. In the case of special random variable we have given a very brief introduction to three parameter viable distribution. So, we will revisit it here. The probability density function of three parameter viable distribution is given in this format where psi is called the location parameter, beta is called the scale parameter and c is called the shape parameter. Now, let us take a transformed variable w which is equal to x minus psi over beta. Then it turns out to be that is also a viable distribution with location parameter 0, scale parameter 1 and the shape parameter c. So, it means that this depends only now on one parameter which is called c, it is a shape parameter. This is also called a standard viable distribution. Remember standard normal distribution does not depend on any parameter, standard viable distribution depends on one parameter which is a shape parameter. The mean of standard viable distribution mu w, mu sub w can be given as a gamma function of 1 over c plus 1. Now, if we transform the variable the viable variable into w then x can be written as psi plus beta plus w and therefore expected value of x is expected is psi plus beta times expected value of w and therefore it is psi plus beta m sub w and similarly the variance of the three parameter distribution viable distribution x variance of x is equal to sigma square can be written as beta square sigma sub w whole square. Let x1, x2, xn be a random sample with viable psi beta and c then w1, w2, wn becomes a random sample from a viable distribution with 0, 1 and c and x bar is nothing but psi plus b beta w bar and s square is beta square sw square. Let us define viable t statistic that is our t statistic that we already know x bar minus mu divided by s over square root n and which can be written as square root n w minus mu w divided by sw and this I call it a t but it will not follow the students t distribution. Whatever it may follow I call it a viable t distribution and I call this a viable t statistic. t is called a viable t statistics and unlike t distribution viable t statistic also depends on one parameter unknown parameter c which is the shape of viable distribution. So, it depends on of course the degrees of freedom should be n minus 1 and because we have estimated x bar so if the degrees of freedom will be n minus 1 but it will also have another parameter along with it which is c. Now we consider the testing of hypothesis mu is equal to mu 0 against the 3 alternatives h1 sorry it should be h1, h2, h3 but there is a mistake here. Let us correct it so that the mistake does not continue. So, it is h1, h2 and h3 so it is mu is not equal to mu 0 mu is greater than mu 0 and mu less than mu 0. We straight away assume that sigma square is unknown then the t statistic as defined above can be a t statistic for testing the hypothesis that mu is equal to mu 0. Let us define what will be the critical region. You recall what we did in the past and we follow the same steps. So, if we fix the alpha error the type 1 error at alpha then for the testing the null hypothesis against the 3 alternatives the first alternative reject h0 reject h0 if absolute value of t is greater than I call it t Weibull 1 minus alpha by 2 with n minus 1 degrees of freedom and of course there will be a parameter I am sorry of course there will be a parameter c which I have missed out. So, this is going to be your critical value the alpha 1 minus alpha by 2 comes under the same argument we are assuming that it is going to be symmetric if it is not symmetric then it has to have two value I think it is it would be more appropriate if we write it that this can be written as t less than t w 1 minus sorry alpha by 2 n minus 1 c or t is greater than t sub w 1 minus alpha by 2 n minus 1 c this is the correct way this needs to be corrected this has to be either this or that then if we consider the rejection I mean the alternate hypothesis as mu is equal to mu 0 then we say that reject h0 if t is greater than t w 1 minus alpha n minus 1 c if we take the alternate that mu is equal to is less than mu 0 then we have to say rejected 0 if t is less than t w alpha n minus 1 c please make sure that this correction is made this is the full correction this is incorrect it has to be this t has to be less than t w alpha by 2 n minus 1 or t has to be greater than so in other words like a chi square distribution we are assuming that this distribution will also be kind of a symmetric distribution and then we are taking two values this is t alpha by 2 and this is t 1 minus alpha by 2 with the rest of the parameters so there is n minus 1 and c here also there is n minus 1 and c this is the case with respect to this so if you are looking for this situation then you are going to look into this area to be alpha and therefore this value is going to be t 1 minus alpha n minus 1 c and if you are looking for the fourth case if you are looking for this fourth case then you will have to take some value here where this value this area is going to be alpha and therefore this is going to be t sub w alpha n minus 1 and c I hope this is clear so these are going to be the critical region so how do we find any critical value and again I have to add a c here I have to add a c here because it is actually so this should be written as t sub w alpha n minus 1 and c so it is a weibull t distribution depends on degrees of freedom n minus 1 and shape parameter of a weibull distribution c therefore weibull t distribution does not have any closed form solution the way we have defined it it does not have a closed form solution hence the critical values of weibull t are generally simulated we will give you the here how to simulate it so this montecarlo simulation can be carried out that first you take mu 0 because that is already given to you the next is you convert this c 0 is equal to 1 over gamma inverse mu 0 gamma inverse is the inverse gamma function mu 0 minus 1 you remember why we are doing this because mu 0 is actually defined as gamma of 1 over c 0 plus 1 please recall so if you take the inverse function and any mathematical programming language will have an excess how to find a gamma inverse function so you can as well use this then you generate capital M number of samples of size small n capital M number of samples of size n M is typically 5000 10000 lakh i mean depending on computer capability and the random generators capability you can decide what should be the size of m so you generate this M number of samples of size small n from weibull 0 that is standard weibull distribution with a c 0 which you have already calculated in the previous step with that you generate the weibull random variables this is your sample so you call it wji i runs from 1 to n it represents the sample and j represents the number of simulations these j are m populations and therefore you have j is equal to 1 to m wj bar and swj square you calculate the weibull t statistic using this formula once you have M capital M of tj you sort them out from smallest to the largest and if you take 1 minus alpha time mth value of tj that will be the alpha level critical value of weibull t distribution depending on c 0 so what you really need to do is for different values of c 0 you have to simulate this or as and when needed you take what is your m 0 you convert it into your c 0 and write a nice program so that every time it generates this critical value and gives you please remember when we do Monte Carlo simulation it is very important to see the stability of it make sure that the seed that is given into it does not conflict with your Monte Carlo simulation process if suppose now we want to test sigma square is equal to sigma naught square well under normal population if you want to test our test statistics w is n minus 1 s square over sigma 0 square which is a chi square n minus 1 degrees of freedom just as in the previous case of testing mu is equal to mu 0 we find rather we observe that expected value of s square is sigma naught square or sigma square independent of what is underlying population hence w can also be a test statistic for the present case and when population under concern is weibull such a statistic is called a weibull chi square statistics and the distribution is called weibull chi square distribution again you can find the critical region here also please remember the statistic will depend on c so let me write it down everywhere so we will have this with c here also there will be a c so if you are taking a null hypothesis sigma 0 square is equal to sigma square is equal to sigma 0 square and your alternate is sigma square is not equal to sigma 0 square then your test is divided in two region reject null hypothesis if your test statistic w is smaller than chi square w that is weibull chi square with alpha by 2 n minus 1 degrees of freedom and parameter c by the way this parameter such a parameter c is generally called a nuisance parameter or w is greater than chi square w square that is the weibull chi square at probability 1 minus alpha by 2 with n minus 1 degrees of freedom and nuisance parameter c your alternate is sigma square is greater than sigma 0 square then we say that rejected 0 if chi your statistics w is greater than the weibull chi square with probability 1 minus alpha n minus 1 degrees of freedom and nuisance parameter c and if sigma square is less than sigma 0 square is your alternative then the critical value is weibull chi square at alpha probability n minus 1 degrees of freedom and nuisance parameter c so you find the critical values of weibull chi square distribution with n minus 1 degrees of freedom and nuisance parameter c well we have to take the same steps as given to simulate the critical values instead of calculating a test statistic which is test statistic which is tj you calculate wj and sort wj from smallest to the largest and take 1 minus alpha times mth value which will give you weibull chi square critical value please remember here the value given is sigma 0 square so there will be a slight change in the beginning you will take sigma 0 square and from that you will calculate there is you have to find out the formula for sigma 0 square which is given earlier and then you have to calculate the value of c0 and then do the simulation so let us summarize it we defined the testing of hypothesis process for testing mu is equal to mu 0 and sigma square is equal to sigma 0 square under non-normal distribution in particular we took the case of weibull we did this because we found that the test statistic z and t which we have chosen under normal population their properties of choosing them for statistic is independent of what is the underlying population so we decided that the same statistic can also be useful we felt that it can also be useful to test the hypothesis under different population distribution we did that with the weibull distribution we found weibull t and weibull chi square statistic we found the critical region and we gave the steps through Monte Carlo simulation to simulate the critical values please remember there is nothing holy about weibull this is just a case has been given to you because when you say that distribution is non-normal there are too many probability possibilities come up so we have shown you one possibility you can change it and see how the test procedure can develop as and when the need arises thank you