 कोर्स on dealing with materials data presently we are going through the sessions on hypothesis testing we have so far considered the sessions our way which deals with so if we quickly write down what we have seen we have dealt with the hypothesis H0 or H0 null hypothesis that mu is equal to mu 0 we have assumed that population is normal with mean mu and variance sigma square and we consider two possibilities possibility A when sigma 0 sorry sigma square is known and other possibility we deal with is when sigma square is unknown in either case we would like to discuss or we have already discussed the three alternates the first alternative which is a two sided alternative that mu is not equal to mu 0 second alternative is that mu is greater than mu 0 and the third alternative we have discussed is mu is less than mu 0 under both the cases that sigma square is known and sigma square is unknown we found when sigma square is known we found that test statistic is z which is x bar minus mu 0 over sigma square root n here we have assumed that we have a sample x1 x2 x3 xn is a random sample from normal population with mean mu and variance sigma square so x bar is an average of x1 x2 x3 xn and we found that this is distributed as normal 0 1 it has no unknown parameters it is completely known distribution and therefore if we consider the alternative that mu is not equal to mu 0 we found that the critical region can be defined as probability or the probability of critical region can be defined as z in absolute value is greater than some value z which should be alpha where alpha is fixed is the type 1 error this is under H0 that is when the null hypothesis is true we call this as the this is the critical region and this is the probability of critical region which is equal to alpha and then we find that actually in that case probability of z greater than small z 1 minus alpha by 2 is equal to alpha by 2 and this is the in that case z 1 minus alpha by 2 is a critical value means that the decision is going to be reject H0 if this statistics z is greater than z 1 minus alpha by 2 if we take the alternate of mu greater than mu 0 then again the test statistic is same is z and the critical region let us properly define it is x1 x2 xn such that z is greater than some value z small z this is your critical region under H0 so you can say that probability of critical region under H0 is equal to probability that z is greater than small z should be alpha and therefore what we find is that it has to be probability that z greater than z 1 minus alpha is equal to alpha if this 1 minus alpha by 2 and 1 minus alpha is confusing you I would like to remind you that you take a standard normal population this is normal 0 1 this is your mean 0 then z of z sub a is defined as this probability so that this probability is a so in this case if we want this probability to be alpha then a has to be 1 minus alpha and therefore we are taking 1 minus alpha or similarly we are taking this because this is 2 sided so then when we come to the next possibility is that alternate of H that mu is less than mu 0 then our critical region is going to be x1 x2 xn such that z is less than small z sorry it should be this way curled bracket so probability of c under H0 is equal to probability of z less than small z which should be alpha and therefore we find that probability of z less than z alpha is equal to alpha and therefore the critical value this is the critical value this is the critical value in this case the decision is in the previous case the decision is if reject H0 if z is greater than z small z1 minus alpha and in this case the decision would be reject H0 if z is smaller than small z alpha so this is what we have done so far next thing what we have made change is that suppose we are taking the sigma square as unknown then please recall that the z statistics turns into a t statistic which is same as x bar minus mu 0 divided by s over square root n where s is the sample standard deviation and now this is distributed as t with n minus 1 degrees of freedom so again there is no parametric dependence on it we already know n so we know n minus 1 so we know the distribution this should be small t sorry it is a distribution that we are already aware of and therefore now you can start that your H0 which says that mu is equal to mu 0 under H1 that mu is not equal to mu 0 your critical region is going to be all x1 x2 x3 xn such that the absolute value of t is greater than some t that is if you take a t distribution which is a symmetric distribution and we say that if t lies in any of this area it is our rejection region within this area we accept it so when with a fixed alpha from the beginning we would like to have probability of critical region given H0 which is probability of absolute t greater than small t which has to be alpha and therefore just as we did in the previous case capital T greater than 1 minus alpha by 2 has to be alpha by 2 and this becomes your criteria for decision where you will say that H0 to be rejected if t is greater than small t 1 minus alpha by 2 sorry here I should write n minus 1 and here also I should write n minus 1 because that specifies this degrees of freedom so on the same line quickly we can say that if we consider the testing of hypothesis H1 I mean the sorry the alternate of H2 which says that mu is greater than mu 0 then do I need to give much explanation please recall our decision is reject H0 if the same t is greater than small t 1 minus alpha with n minus 1 degree of freedom and in the case of the third hypothesis mu less than mu 0 third alternate hypothesis the the decision making is going to be reject H0 if t is less than t alpha n minus 1 n minus 1 in degrees of freedom so this is what we have learnt so far in the hypothesis testing the important two statistics we have found so far is z which is x bar minus mu 0 over sigma square root n distributed as a normal 0 1 and in the other case our test statistic is t which is x bar minus mu 0 over sample standard deviation square root n which is distributed as a students tree distribution with n minus 1 degrees of freedom in the today's case what we would like to do is consider a case when you want to test that sigma square itself is equal to some predecided value sigma 0 so today what we want to consider is a case of sigma square is equal to sigma naught square where sigma square is population variance and sigma naught square is given when this is the case when this is the case we start out that we have a sample x1 x2 x3 xn from again a normal population with mean mu and sigma square and we want to sigma square is unknown and we want to test the hypothesis that sigma square is a given value sigma naught square the situation can arise just as it would arise in a any testing of hypothesis for mu is equal to mu 0 again we consider the alternate hypothesis h1 first as sigma square is not equal to sigma naught square it is a two sided alternative ok let us follow the six steps let us follow the six steps of hypothesis testing so first step is that fix alpha ok so we have fixed up alpha number two clearly write your hypothesis null hypothesis which is sigma square is equal to sigma naught square and your alternative verses you are testing against the alternative h1 which says that sigma square is not equal to sigma 1 sigma 0 square ok what is the best statistic to represent this sigma 0 well we know that s square which is equal to 1 over n minus 1 summation i is equal to 1 to n xi minus x bar whole square this is just to recall is a good statistic because expected value of s square is equal to sigma square we have shown it in the past and therefore the fourth step we can define a critical region such that x1 x2 x3 xn should be such that the s square value is larger than some let us call it x value ok this is going to be our critical region so in that case fifth step is to identify the critical region so we find that n minus 1 times s square divided by sigma square is distributed as chi square with n minus 1 degrees of freedom please recall this n minus 1 times the sample variance divided by the population variance is distributed as chi square n minus 1 and chi square n minus 1 does not depend on any parameter it is n minus 1 which is already known because you already have a sample of size n so you already know n ok so therefore your probability is that n minus 1 s square over sigma square which I call it an x this will create a confusion so let us call it a y such under the condition h0 this probability should sorry this greater than x should be greater should be equal to alpha I will rewrite this so that there is a very clarity in it y which is equal to n minus 1 s square over sigma square is greater than x under h0 this should probability should be alpha if h0 is correct this means that you are looking for probability of y which is n minus 1 s square over sigma naught square greater than x should be alpha now because the alternate is on both sides actually this is insufficient we have to say that for alternate which is both sides that which says that sigma square is not equal to sigma naught square in a chi square distribution it should lie somewhere between two limits a and b where this is also alpha by 2 and this is alpha by 2 so considering that we modify our critical region as y less than a union y greater than b such that this is a mathematical sign for such that such that probability of y less than a under h0 is equal to alpha by 2 and probability of y greater than b I think it is better that I use an eraser because it is creating a problem so here we are y greater than b under h0 should be alpha by 2 so here please remember you cannot have a single critical value because chi square distribution is not a symmetric distribution it is a skewed distribution and therefore you cannot have a solution like you had it with respect to the standard normal distribution and the t distribution with n minus 1 degrees of freedom and therefore actually you are going to have to find two critical values and these two critical values can be given as chi square n minus 1 alpha by 2 and b is going to be chi square n minus 1 1 minus alpha by 2 please remember we are always considering the probability on the left hand side of the value so these are your critical value and your decision is going to be reject h0 if either p sorry either y which is n minus 1 a square over sigma naught square is less than chi square n minus 1 alpha by 2 or y is greater than chi square n minus 1 1 minus alpha by 2 I hope this is clear that we are what we find let us quickly go through it if we follow the sixth step this is the sixth step so if we follow the sixth step we fix the first alpha value so first we fix an alpha value this is our null hypothesis and the alternate hypothesis we find that s square is the correct statistic to take to find a test statistic for this or to find a critical region for this test and therefore this test statistic we chose as a sample variance because expected value of sample variance is the population variance then we define a critical region x1, x2, x3, xN such that x square should be greater than x and then we take appropriate statistic which would be devoid of any unknown parameter so we say that it is n minus 1 I better write a parenthesis here so it is n minus 1 square over sigma square actually here itself I should have used this kind of an arrangement but we have actually continued with the same arrangement but we find that when you simplify this and you want to find the critical value we realize that this is a two sided hypothesis and therefore we must have the two values a and b and therefore we find that the critical region can be modified to this and finally we come up with the test now I believe it is not very difficult to understand as to what would be the case if you take h is 0 as sigma square is equal to sigma naught square versus h2 which says that sigma square is greater than sigma naught square then naturally you will say that your test statistic is y which is n minus 1 s square over sigma square and this is distributed as chi square with n minus 1 degrees of freedom under h0 and therefore a probability of y greater than x has to be alpha now we are correct because if you look at the region we want the sigma square to be greater than sigma naught square this is sigma naught square then you would like to have it greater than some value which is a critical value which will be a critical value which I am calling x but now you can make out that in that case x is nothing but chi square with n minus 1 degrees of freedom and 1 minus alpha please remember again if you take the alpha chi square distribution and if you wish to have this value such that this is alpha in that case this area is 1 minus alpha and therefore this value is I call it sorry chi square n minus 1 1 minus alpha so that is the value which will come so the decision is going to be reject h0 if n minus 1 s square over sigma naught square is greater than chi square value of chi square with n minus 1 degrees of freedom at 1 minus alpha probability similarly you can easily show that if you want to test the hypothesis h0 is equal to sigma square is equal to sigma naught square versus the hypothesis h3 which says that sigma square is less than sigma naught square then your test statistic is same which is the test statistic is going to be n minus 1 s square divided by sigma 0 square which will be distributed as chi square with n minus 1 degrees of freedom and we are looking for a probability that y is less than x has to be alpha under h0 so if you look at the probability curve you are looking at this value of x so that this is alpha and this is given by chi square n minus 1 alpha and therefore your decision is going to be reject h0 if n minus 1 s square over sigma naught square is less than chi square with n minus 1 degrees of freedom and alpha probability so with this we conclude this session on testing of hypothesis for testing of hypothesis for sample variance so sample variance sorry not sample variance it should be population variance sigma square the hypothesis we tested is sigma square is a given value sigma naught square and we considered three alternative one is a two sided alternative which says that sigma square is not equal to sigma naught square second one say that sigma square is greater than sigma naught square and the third hypothesis say that sigma square is less than sigma naught square in all the cases we found that test statistic y is nothing but n minus 1 sample variance divided by sigma naught square which is distributed as chi square with n minus 1 degrees of freedom and then we derived the three tests thank you