 to the course on dealing with materials data. In the process of statistical inference, we are undergoing the sessions on hypothesis testing. So far we have considered the testing of hypothesis with the two sided alternative that hypothesis that mu is equal to mu 0 with the alternate that mu is not equal to mu 0. We found that when the population variance is known, the test statistic is Z which is sample mean minus mu 0 divided by standard deviation over square root n. If it follows, if the population is normal, this follows a normal distribution. If sigma square is unknown, then the test statistic is defined as T which is X bar minus mu 0 divided by sample standard deviation divided by square root n. And under the assumption of the normal population, this T statistic follows a T distribution with n minus 1 degrees of freedom. Then we also found that the sixth step of classical hypothesis testing procedure can also be carried out by working out the probability of type 1 error. And we also went through how to derive the type 2 error as a function of alternate hypothesis mean mu when it is not mu 0. In this session, we are going to consider the case of two sided alternatives. There can be two of them, mu is greater than or it should not be greater than or equal to, it has to be only greater than. So, let me correct myself. This should be only greater than mu 0 and they should be only less than mu 0. So, there are again we are taking the two possibilities under normal population assumption that variance is known and variance is unknown. One sided alternative hypothesis, you have either this or this and we will follow the six steps of classical hypothesis testing. So, let us take the case when sigma square is known, mu is unknown and we are assuming normality that is the population is normal with mean mu and variance sigma square. We have the null hypothesis as mu is equal to mu 0 and mu or the alternate hypothesis as mu is greater than again the error has occurred. So, I stand corrected, mu is greater than mu 0. Here also it will be mu is greater than mu 0. So, then we have the test statistic like before it is going to be x bar because expected value of x bar is mu and then we say that we can reject the null hypothesis that is the critical region has to be such that now I have removed the absolute value x bar minus mu 0 has to be greater than c. So, if you look at the, sorry, okay, if you look at the normal population, this is your mu 0 and there is some sigma here, some value sigma is there. What we are saying is that x bar has to be in the vicinity, but you have to reject it if it is very far away on the positive side. So, actually if it is anywhere far away from mu 0, you have to reject it, but to some extent we are going to accept it. So, we are going to make some limit if it is very far away and this value is what I call c. If it is very far away, then this is the region where I am going to reject it. So, I am saying that x bar minus mu 0 has to be greater than c. So, this is, if I plot this x bar minus mu 0, it has to be greater than somewhere here. So, if your x bar value itself is some larger than this value, then you are going to reject it. So, again our test statistic is standard normal devi at z which is x bar minus mu 0 over standard deviation over square root n which is a normal with a known population mean and standard deviation when H naught is true. And therefore, we come to the test statistic. This is where the difference is, so I have shown it in red. The absolute value is gone. It has to be greater than c dash which is equal to alpha and therefore, probability of z greater than c dash is alpha, then c dash has to be a small z alpha. And we will say that the decision is if the z value is greater than z alpha, then we reject at 0. Let us quickly plot it up. This is a standard normal barrier distribution with mean 0. And we say that we want this area to be alpha. So, z alpha. Actually I should have said z 1 minus alpha because again in the table as I said, z alpha is defined as probability of minus infinity to z alpha is 1 over square root 2 pi exponential minus 1 half x bar x square dx. This has to be alpha. So, if you want this area to be alpha, z has to be z 1 minus alpha and therefore, it is here. Then you reject the null hypothesis, please make this correction. So, anyway I have already shown in the plot. So, if we take the same example again, exactly the same example as before. And now we do the one sided testing in which we take a beam is equal to mu 0 is that is the hypothesis that the population mean is 1110 MPa versus alternate that mean is greater than 1110 MPa. And at present we have found our from our sample the mean value. The sample mean is 1129 MPa, sorry for this mistakes in the slide. So, this is MPa. So, accordingly we work out the procedure. The next step is that what is the critical region has to be x, the capital z greater than small z if H naught is true and then we would like it to be alpha. So, we say that z is distributed as normal 0 1. I think now by now you must have realized what the procedure we are following. So, this becomes alpha which is this has to be 1 minus alpha. It has been calculated as 1 minus alpha I know, but it has not been written correctly. So, we find that this particular value is 1.72 and alpha as I have already written as 1 minus alpha. So, the z value is 1.645 and 1.72 is indeed greater than 1.645 and therefore the hypothesis is rejected. Remember in the previous case we had accepted the null hypothesis in this case with the changed alternate we are going to reject the null hypothesis. I think the mean value has also been changed, this value x bar has also been changed. So, let us take the other alternate. I think you would understand in the other alternate what has to happen before going into the detail. If you are looking at the population distribution and this is mu 0, you are going to reject when this has to be small. So, mu is less than mu 0. So, up to some value you are going to accept it, below that you are going to reject it. If it is very very small you are going to reject it and therefore this value is going to be z alpha because this is what we would like to have alpha and this will be exactly z alpha. So, coming to the point you see that that is exactly z alpha, here I have made a mistake I have totally mixed up the two things this has to be alpha, this is also alpha and then you reject the null hypothesis. If you take the example you can work out that it is also you can work out the example and you will find whether you take a mean value to be say 1108 which is smaller than 1110 and then see if the hypothesis, how the hypothesis procedure works and do you reject the null hypothesis or you accept it. The two sided case when sigma square is unknown, I do not think we need to go through the whole procedure by now because we have clearly seen that under the null hypothesis of mu is equal to mu 0, the underlying test statistic is this, when sigma square is known. If sigma square is unknown then sigma square gets replaced by the sample variance and therefore it becomes a t statistic. So, this is when sigma square is known and this is when sigma square is unknown then it gets replaced by sample variance square. Thus the two sided case will accordingly change to a t distribution with n minus 1 degrees of freedom while this becomes a normal distribution with zero mean and one standard deviation. I do not think we need to repeat this in a totality because it is a very obvious case. So, let us summarize it. We discussed one sided alternate of mean value to be greater than means mu 0 or less than mu 0. In both the cases we found that when sigma square is known or rather in all the cases, we found that when sigma square is known, z which is standard normal deviate which is x bar minus mu 0 over sigma square root n is the test statistic and the test will procedure will follow normal distribution. If sigma square is unknown then the test statistic takes a value t because you replace sigma by sample standard deviation and it follows the t distribution with n minus 1 degrees of freedom. Now before we close the session I would like to repeat or bring out one particular feature here. Remember that x bar that is the sample mean is a good estimator. It is an unbiased estimator of population mean no matter whether the distribution is normal or non-normal or whatever. Similarly, sample standard deviation or sample variance is an unbiased estimator of a population variance. It does not depend on the distribution and therefore these two statistic the z statistic and the t statistic play a central role when you want to test a hypothesis with respect to mean. Similarly, in the next session we will say that if you want to have a hypothesis testing for the variance of the population. So, you want to test a hypothesis which is something like sigma square is equal to sigma not square and what do you think would be the central statistic? It has to be s square over sigma not square and it would be distributed as chi square with n minus 1 degrees of freedom and just as t and z this is going to be the test statistic. We will look into it in the next session. Thank you.