 As Salaamu Alaikum, welcome to lecture number 34 of the course on statistics and probability. Students, you will recall that in the last lecture, I began the concept of point estimation and having given the basic definition of the concept, I discussed with you some desirable qualities of a good point estimator. You will recall that I discussed with you the concepts of unbiasedness and consistency. Unbiasedness k havali se aapko yaad hoga ke bunyadi concept ye hai that the sampling distribution of your statistic is centered on the parameter that you are trying to estimate. For example, if you are trying to estimate the population mean mu, then the mean of the sampling distribution of x bar lies exactly at the same point where mu itself lies. Our distribution is neither to the right nor to the left as far as the center of our distribution is concerned. It is neither to the right nor to the left of mu, but it is on mu. And students, what about consistency? Aapko yaad hain haa boh bada complicated sa expression lekin uska jo lube-lubab hai, wo itna mushkil nahi hai. Aur wo kya tha? An estimator is said to be consistent. If the probability that the estimator is very close to the parameter, if this probability increases and tends to 1 as the sample size increases, then we say that our estimator is a consistent estimator. So, wali pehda hota hai students ke consistent estimator ko hani hum iss baat ko ek desirable property kyun kahitthein. Dekhe iss pe thora sa gaur kiye. Sarai jo estimation ho re hai that is on the basis of the data that you have collected from the population, but you have not collected the data from the entire population. You have only collected data from a sample. To ye to hum intuitively understand kr satte hain aur accept kr satte hain that the larger the sample, the more representative it should be of the population. To issi hawale se hum ye bhi sochenge ke if the sample size is large, our estimator should be closer to the parameter than if it was an estimator based on a sample that was relatively small. Issi hawale se hum we have this concept and you will see that it tellies with what we would expect intuitively. Yani, to repeat what I just said once more, as the sample size increases, if the probability increases that my estimator will be very close to the parameter, then I say that my estimator is consistent and this is a desirable quality. Students, you will recall that I mentioned to you that another very important desirable property of a point estimator is efficiency. And let me now define this one for you formally. As you now see on the screen, an unbiased estimator is defined to be efficient if the variance of its sampling distribution is smaller than that of the sampling distribution of any other unbiased estimator of the same parameter. In other words, suppose that there are two unbiased estimators, T1 and T2 of the same parameter theta, then the estimator T1 will be said to be more efficient than T2 if the variance of T1 is less than the variance of T2 and students, I would like you to concentrate on the graphical representation of this concept that you now see on the screen. So, I am going to show you that both of them are unbiased estimators, T1 and T2 are unbiased estimators, T1 and T2 are unbiased estimators, T1 and T2 are unbiased estimators. So, let us say that our sampling distribution is centered on the parameter theta, students, you can see that one distribution is relatively tight and the other one is relatively wide. Now, we said that the one whose variance is smaller is the one which is more efficient. I hope that you will all have said that T1 is more efficient. That is tighter than that of T2 and we all know that if the variance is small, then of course, the standard deviation will also be small and if the standard deviation is small, then the spread of the distribution will be less and it will be a tighter distribution. Students, you should note that we are always looking for efficiency as far as point estimation is concerned. Most of them are closer to the parameter that we are trying to estimate. In other words, in a real life situation, the probability is higher that our estimator will be close to the parameter. So, this is a very important concept and one that is linked with it is that of relative efficiency. As you now see on the slide, the relative efficiency of T1 compared to T2, where both T1 and T2 are unbiased estimators, this relative efficiency is given by the ratio variance of T2 divided by variance of T1. And if we multiply this expression by 100, then of course, we can represent the relative efficiency in percentage form. And thus, we have a means of comparing different unbiased estimators of one particular parameter. Students, let us talk about sample mean and sample median. I had already told you that sample mean is an unbiased estimator and a consistent estimator of mu. Or if our distribution is normal, then the sample median is also an unbiased and consistent estimator of mu. But you will be interested to note that as far as efficiency is concerned, the sample mean is more efficient than the sample median. So if we want to choose one of the two in order to estimate mu, which one will be preferred? It is obvious that x bar is to be preferred over x tilde because of this reason that it is more efficient than x tilde. Alright, now that I have discussed with you some basic desirable properties of a good point estimator, students, I would like to discuss with you briefly the various methods of estimation that have been developed. As you now see on the screen, some of the more commonly and widely known methods are the method of moments, the method of least squares and the method of maximum likelihood. Let us begin with the method of moments. The method of moments, which is due to Carl Pearson consists of calculating a few moments of the sample values and equating them to the corresponding moments of a population. Thus, getting as many equations as are needed in order to solve for the unknown parameters. Come, let us try to understand this concept in some detail. Look, students, you will remember that we talked about that x bar is an estimator of mu, s square is an estimator of sigma square. So, similarly, higher moments, they would be estimating the corresponding population moments. For example, m3 is an estimator of mu3, m4 is an estimator of mu4. After all, x bar and s square are also moments. x bar is the first moment about 0 and s square is the second moment about the mean. So, this method of moments, the concept is that sample moments are estimating the corresponding population moments. So, we can match them. Thus, we will get equations, which can be solved simultaneously to obtain the values of the parameters that we would like to estimate. Let me explain this to you with the help of an example. As you now see on the screen, let x be uniformly distributed on the interval 0 theta. Find an estimator of theta by the method of moments. Now, in order to solve this problem, the first point to note is that the probability density function of the given uniform distribution is f of x is equal to 1 over theta such that x lies between 0 and theta. This equation, I have just presented to you, students, you may be thinking that first, when we read the uniform distribution, its formula was different. Actually, if you pay attention, this is a special case of that. You will remember that we said that f of x is equal to 1 over b minus a, where x itself lies between a and b. Now, if you keep a 0, that is, we are starting from 0 and we are going up to b, then what will my f of x become? Then I will have f of x is equal to 1 over b minus 0 such that my x goes from 0 to b. And if you keep b as theta, then what will be your equation? f of x is equal to 1 over theta minus 0 such that x lies between 0 and theta. And students, this is exactly what I have presented to you. As you now once again see on the screen, this distribution is given by f of x is equal to 1 over theta minus 0, which is of course, equal to 1 over theta such that x goes from 0 to theta. Students, this particular uniform distribution has only one parameter. If I want to estimate this one lone parameter using the method of momentum estimates, what should I do? According to what I just explained, I should compute a sample moment and match it with the corresponding population moment. And by doing so, I will obtain one equation which when I solve, I will get an estimator of theta. So, as you now see on the screen, the first sample moment about 0 that is m 1 dash is sigma x minus 0 raised to 1 over n. In other words, sigma x over n of course, that is the sample mean. And the first population moment about 0 that is mu 1 dash is given by the interval of integral of x minus 0 raised to 1 into f of x. And this integration is being done with respect to x and the limits are 0 to theta. And on simplifying, this can be written as the integral of x into f of x. And since f of x is equal to 1 over theta therefore, this integral is the integral of x into 1 over theta with respect to x with the limits from 0 to theta. Upon solving, we obtain 1 over theta multiplied by x square over 2 within the limits 0 to theta. And upon further simplification, the final answer is that mu 1 dash is equal to theta over 2. Now, matching the first sample moment with the first population moment, we obtain x bar is equal to theta over 2 or in other words, theta is equal to 2 times x bar. Hence, the moment estimator of theta in this particular problem is equal to twice of x bar. Students, aiyes mamale pe zara zyada deft me jaate gaur karte. This is a uniform distribution. Iski ek edge 0 pe hai, dosri edge theta pe hai. Aur iska jumeen hai mu that is equal to theta over 2. Isle ke wo bilkul chunke symmetric hai. Isle uska mean johai wo exact center me hoega. To is baat pe gaur ki je mu is equal to theta over 2. Aur khud theta jo hai uska aur mu ka aapis me kya relation hai. Obviously, theta itself is the double of theta by 2, yani, the double of mu. In other words, our equation is that theta is equal to 2 times mu. Ye to hai wo situation jo actually really exist karte hai for this probability distribution. Theta is equal to 2 mu. Lekkin haam agar sample data leh rahe hain aur mu hum nahi compute kr sate, because hum ne poori population ya poori probability distribution encompass hum nahi kr sate. Aur sample data leha, to hum to mu nahi nikal sate, hum to sirf x bar compute kr sate, which is an estimator of mu. To ye jo equation banti hai, theta is equal to 2 mu. Isme agar aap mu ki jaga x bar rakhdein to kya equation banti hai. Theta is equal to 2 times x bar. And this is exactly what you obtained through the method of moments. Aap ne rekhah students ke agar aap in cheezon pe hoar karein, to it is very interesting aur aap ko subject ki philosophy ko samajne mein madad miltiye. And it is so interesting ke jo cheez intuitively aap accept kareinge, mathematics ki through bhi aap ko bohi cheez mil jaati. Let me now apply this concept to another example. As you now see on the screen, let x 1, x 2, so on up to x n be a random sample of size n from a normal population with parameters mu and sigma square. Find these parameters by the method of moments. Aap ne rekhah ke statement mein hum nahi yon parha ke find these parameters by the method of moments. Lekin saap zahe rahe aap ki agar method of moments ki baat ho rahe hai, to find ka matlab hai ke estimate kareinge hum because we are collecting data on sample basis and then only we will apply the method of moments. So, students aap iss case mein how will we estimate these two parameters? Jaise ke pehle kaha tha ke we will be matching the sample moments with the corresponding population moments. But how many equations will I have in this particular case? I should have two equations. Why? Because the normal distribution has two parameters that we are trying to estimate mu and sigma square. So, as you now see on the screen, the first two sample moments about 0 are m 1 dash is equal to x bar just as before and m 2 dash is equal to sigma x minus 0 whole square over n, which upon simplification becomes sigma x square over n. What about the corresponding moments of the normal distribution? These are mu 1 dash is equal to mu and mu 2 dash is equal to mu 2 dash is equal to sigma square plus mu square. Now, this second equation is particularly important and you will realize that we obtain this equation from the basic equation that we have already discussed in this course. The one that relates the moments about the mean to the moments about arbitrary origin A or the moments about 0. We have already studied that mu 2, which is the same thing as the variance is equal to mu 2 dash minus mu 1 dash square and since mu 1 dash the first moment about 0 is the same thing as the mean mu. Hence, our equation becomes sigma square is equal to mu 2 dash minus mu square and taking mu square to the other side, we obtain mu 2 dash is equal to sigma square plus mu square. Now, in order to obtain the estimators of mu and sigma square by the method of moments we match the sample moments with the corresponding population moments and thus we obtain the following two equations. Equation number 1 mu is equal to sigma x over n and equation number 2 sigma square plus mu square is equal to sigma x square over n. Solving these two equations simultaneously, students we obtain mu hat is equal to x bar and sigma square hat is equal to summation x minus x bar whole square over n, which is nothing but capital S square the variance of our sample. You have seen that the method of moments through estimators, they are none other than what we have been discussing and we are familiar with and that is the simple mean of the sample x bar and the simple variance S square. This is the method, very, very simple. If you have only one equation, you need just one equation. If you have to estimate two as in the case of the normal distribution then you should form two equations and solve them simultaneously and so on and so forth. So, students although it is very simple, but there is one point that is to be noted that moment estimators are generally inefficient, that is not generally not very tight. So, that is a shortcoming of this method. The method that I will discuss with you next very briefly is the method of least squares. This is a method by Gauss and Markov and it is based on the theory of linear estimation. As you now see on the slide, an estimator found by minimizing the sum of the squared deviations of the sample values from some function that has been hypothesized as a fit for the data, such an estimator is called the least squares estimator. Students, I think you are trying to confuse it completely, but there is no need to do that because you will recall that in lecture number 15, I have already discussed with you this concept when we talked about regression. You remember that first of all we made a scatter diagram or our points, our graph cupor I, if we note that this seems to be a linear pattern then we decided that we will fit a straight line to this particular data set. That particular line, which is the one for which the sum of the squares of the distances between the points and the line that sum is minimum. Exactly this is the method and the principle of least squares and what you just saw on the screen is telling with what I just explained. Students, why is this being called a method of estimation? With reference to regression that I just discussed, you will be interested to know that the line that we are fitting to our sample data, actually this is an estimate of a line that we would have obtained, if we had not collected only a small data set on sample basis, but if we had access to the entire population of data points for that particular situation. That is, if we do not have 5, 7, 10, 12 or 20 points, but if we have 1 crore or half points regarding those two variables that we are correlating with each other and regressing one on the other. If we have 1 crore or half points, then the line that we get in terms of the means of the y values corresponding to various x values, that true line would be our estimate of this line and we have established this line by the method of least squares. In this course, I will not be going into too much mathematical detail, but I would like to convey to you a few points, so that if you are so interested, you may study in further detail on your own. The next method that I would like to discuss with you is a very, very important method of estimation. This is known as the method of maximum likelihood. This was introduced by the great statistician Sir R. A. Fisher in 1922. And although the technique and the concept is a bit advanced and it involves the concept of the likelihood function, which I will not be going into the details of, I would like to give you a basic idea about the core of this concept. As you now see on the screen, according to this method, we would like to consider every possible value that the parameter might have and for each value compute the probability that the given sample would have occurred if that were the true value of the parameter. That value of the parameter for which the probability of a given sample is greatest is chosen as an estimate. Students, once again, of course, you want to get confused, but again I would like to say that step by step, step by step, step by step. Now, one example is that if we want to estimate the proportion of success in the population for any particular population. Now, you know that if we say that every possible value of parameter to consider, then in this case, P that can have any value from 0 to 1, you will agree that proportion of success or probability of success cannot be less than 1 or not more than 1 because 1 means 100 percent. So, we have determined that our parameter, which we are going to estimate, will be brought somewhere between 0 and 1. Now, the question arises which value between 0 and 1 which we can regard as the value. Now, we have drawn a sample, a sample materialized. So, if the information in that sample says that this information has a particular value of this particular parameter, you see, the concept is that of maximum likelihood. As this given sample is present, what is the value of P that is more likely? You see, there is an example that if in reality P is 0.9, then my sample that has come in, if P hat is coming in 0.3, then this is not happening with it. If the proportion of success in the population is 90 percent, then why did my sample come in 30 percent? It should have been relatively high in the sample. This is intuitively understandable. So, the discussion is that once again without going into the mathematical details which are quite advanced, I just wanted to convey to you a basic idea. We are trying to say that given this particular sample, what is the value of the parameter that is most likely in the light of this information? Having said all this, I will make it clear here that this is a very non-mathematical way of trying to explain it. The mathematics involves differentiation because as you have studied in pure mathematics, when you are trying to find the maxima or the minima of a function, you do take derivatives. So, we will talk in that way, which is evident that scientifically speaking that is the best way. So, we will talk in different ways. But once again, the object of my discussion is to give you a preliminary intuitive idea that is involved in this particular method. Students, let me give you a few examples of the estimators that we obtain by the method of maximum likelihood in various situations. As you now see on the screen, for the Poisson distribution given by e raise to minus mu, mu raise to x over x factorial, the maximum likelihood estimate of mu comes out to be x bar for the geometric distribution given by p into q raise to x minus 1 such that x is equal to 1, 2, 3 and so on. The MLE that is the maximum likelihood estimator of p comes out to be 1 over x bar. In other words, the MLE of p for this particular distribution is equal to the reciprocal of the sample mean. For the Bernoulli distribution given by p raise to x, q raise to 1 minus x such that x can take either of the two values 0 and 1, the MLE of p comes out to be x bar. And having discussed these discrete situations, we now go to some examples of continuous probability distributions. So, for the exponential distribution given by theta into e raise to minus theta x such that x is greater than 0 and theta also is greater than 0, the MLE of theta is 1 over x bar, the reciprocal of the sample mean. For the normal distribution with parameters mu and sigma square, the joint maximum likelihood estimators of mu and sigma square come out to be x bar and s square. Capital s square which is not an unbiased estimator but which is consistent. Students, in this course we have discussed many times that the normal distribution is one of the most frequently encountered distributions in practice. So, you have seen that for this frequently encountered distribution, the simplest things that we compute, the sample mean and the ordinary sample variance, they are the maximum likelihood estimators of mu and sigma square. That means, the simplest quantities we compute without thinking of all these advanced techniques and these very advanced criteria for good methods of estimation. It is so interesting that these simplest quantities in very, very criteria fulfill. X bar is unbiased, consistent, efficient and capital s square although it is not unbiased but it is consistent and efficient. So, now we will apply this on a practical example. As you now see on the screen, it is well known that human weight is an approximately normally distributed variable. Suppose that we are interested in estimating the mean and the variance of the weights of adult males in one particular province of a particular country. A random sample of 15 adult males from this particular population yields the following weights, 131.5 pounds, 136.9, 133.8 and so on. Find the maximum likelihood estimates for theta 1 equal to mu and theta 2 equal to sigma square. Now, students in order to solve this problem, the first thing that we note is that this sample, very small sample of size 15 has been drawn from a normal population with mean mu and variance sigma square. It has been mathematically proved as already discussed that the joint maximum likelihood estimators of mu and sigma square are X bar and capital S square. So, all we need to do is to compute X bar and S square for this particular data set and doing so, we obtain X bar is equal to 133.43 pounds and S square is equal to 5.10 pounds squared. Students, of course, we can now very confidently say that these are the maximum likelihood estimates of the mean and variance of the population of weights in this particular example. Students, we have discussed at some length the concept of point estimation. Point estimation kya hai? A single quantity that we compute from the sample in order to estimate the corresponding parameter. Lekin aap gaur karin to iss method me ye limitation hai that in any particular real situation when we have a particular sample and we have computed this quantity. We have no way of knowing how close or how far this particular estimate is from the parameter that we are trying to estimate. We have discussed in detail that X bar is an unbiased estimator of mu. Iska mafum kya tha? That if you do not compute just one X bar from one sample, but if you draw all possible samples of a particular size from that population and you find X bar for every one of them. To bo jo all possible X bars honge unki jo mean value hai that is identical with mu. But in a real life scenario this does not help us tremendously. Kyuke real life me aap all possible samples draw nahin kar rahe. You are just drawing one and you are just having one X bar that you compute. Ho sakta hai ki ye X bar mu can iss deek hi ho in which case you are lucky. Lekin ye bhi to mumkin hai na, ke it is quite far and it is towards the edge of your sampling distribution of X bar. And if it is towards the edge then it is quite far away and the distance is quite large from mu. To ye shortcoming hai of point estimation that you do give a quantity as an estimate of your parameter. But you are not able to attach a statement with it as to how confident you are that it is close to the parameter. Internal estimation is that procedure in which you are able to give not one value, but a range of values within which the true population parameter value is expected to lie. And you are able to give a level of confidence along with this interval. As you now see on the slide a confidence interval is an interval computed from the sample observations X1, X2 so on up to Xn with a statement of how confident we can be that the interval does contain the true parameter. Let us develop this concept with the help of the example of the Ministry of Transport test to which all cars irrespective of age have to be submitted. You will recall that this test looks for faulty brakes, staring lights and suspension and it was discovered after the first year that approximately the same number of cars had 0, 1, 2, 3 or 4 faults. Students, aapko yakinan ye example yad aagya hoga. Or ye baat importantly yad ki jhe when we drew all possible samples of size 2 from this uniform distribution our sampling distribution of X bar was triangular. But when we increase the sample size to 3 our sampling distribution was like something like a normal distribution rather than triangular. And when we increase the sample size to 4 the shape of the distribution became even more like a normal distribution. This example ki baat me dubara ish liye kar rahi hoon ke ye jho normality hai of the sampling distribution which of course also is presented formally by way of the central limit theorem which we have already discussed. I wanted to draw your attention to this fact once again. Ke aksar okaat in real life situations. Hamara sample size 2, 3 or 4 to nahi hoega. It will be much larger. Aur agar 4 size pe hum dekh rahe hain ke it is something like a normal. To zahere hai ke size 30, 50, 100, 500. Ispe to wo usse bhi bohot zyada normal hogi. So, the concept of confidence interval that I am now going to present to you is based on this basic point that in many situations when we are trying to estimate mu by computing X bar our sampling distribution of X bar can be regarded as a normal distribution and hence I will now derive for you the confidence interval for mu based on this normal distribution. Students aap kahenge ke derivation karne lage hain. All through I have been saying that in this particular course I am not going to do derivations but I will be simply giving you the basic idea and the basic concept and then we will be applying these concepts to real life situations. Yes true. That is exactly what I have been doing most of the time and I will be doing in the next lectures too but students believe you me if you concentrate on this one derivation that I am going to do now for you. You will be at ease with the concept of interval estimation. As you now see on the slide when sampling from an infinite population such that the sample size n is large X bar is normally distributed with mean mu and variance sigma square over n. Hence the standardized version of X bar that is Z equal to X bar minus mu over sigma over square root of n this is normally distributed with mean 0 and variance 1. Now for the standard normal distribution we have area between 0 and 1.96 equal to 0.4750 as indicated in the area table of the standard normal curve and therefore the area from 0 to minus 1.96 is also 0.4750. This implies that the area between minus 1.96 and 1.96 is 0.95 and since area represents probability therefore we can say that the probability that Z lies between minus 1.96 and plus 1.96 this probability is equal to 0.95. Now substituting X bar minus mu over sigma over square root of n in place of Z in this equation we have probability that minus 1.96 is less than or equal to X bar minus mu over sigma over square root of n less than or equal to 1.96 this probability is equal to 0.95. Let us rewrite this equation in such a way that we take the term sigma over square root of n which is in the denominator of the central expression we take this term to the expression on the left and the expression on the right and by when we do that of course it will no longer remain in the denominator but it will be attached to minus and plus 1.96 respectively in the numerator. Hence our equation now becomes the probability that minus 1.96 into sigma over square root of n is less than or equal to X bar minus mu is less than or equal to 1.96 sigma over square root of n is equal to 0.95. Students it is extremely simple. The moment you do that the term does not remain in the denominator of course it will attach in the numerator or in the main expression in the left hand side term as well as in the right hand side term. Now there are just a few more steps to this derivation and then we will arrive at our confidence interval. As you now see on the slide we have probability minus 1.96 sigma over square root of n less than equal to X bar minus mu less than equal to 1.96 sigma over square root of n this whole probability equal to 0.95. Abhi thori der pehle ham sigma over square root of n ko left or right sides pehle gaythe ab hum jo term X bar hai in the middle usko bhi lekar jaana chahthe to the left hand side and to the right hand side. Chunke middle expression me X bar is being added to minus mu. Isliye jab hum usko left hand side pehle gaye to the left hand side pehle gaye to the left hand side pehle ham. So, we obtain the probability that minus X bar minus 1.96 sigma over square root of n is less than or equal to minus mu is less than or equal to minus X bar plus 1.96 into sigma over square root of n this whole probability is equal to 0.95. Chunke ab hamare central expression me minus sign aah rahe which is not extremely wonderful for our purposes. Therefore, we would like to multiply the entire expression inside the brackets all the expressions by the minus sign, but we all know that the moment we will do that the less than equal sign will interchange with the greater than equal to sign and hence students we now obtain the probability that X bar plus 1.96 sigma over square root of n is greater than or equal to mu is greater than or equal to X bar minus 1.96 sigma over square root of n this whole probability is equal to 0.95 students. This derivation we have just done actually we have reached our confidence interval but we want to do one more step and what is that we have said that something is greater than mu and mu is greater than something instead we want to do another reverse order. We would like to say that something is less than mu and mu is less than something. So, what should we do? Just do this, reverse it. Write the term on the right hand side and write it on the left hand side and bring it on the right hand side. Naturally what you will do is that your greater than sign will become less than sign again. For example, if you say that 3 is less than 4 is less than 7 this is equivalent to saying that 7 is greater than 4 is greater than 3. So, as you now see on the screen the probability that X bar plus 1.96 into sigma over square root of n is greater than or equal to mu is greater than or equal to X bar minus 1.96 into sigma over square root of n this probability is equal to 0.95. This statement is equivalent to the statement that the probability that X bar minus 1.96 into sigma over square root of n is less than or equal to mu and that is less than or equal to X bar plus 1.96 sigma over square root of n this probability is 0.95 that X bar minus 1.96 into sigma over square root of n is less than or equal to mu and that is less than or equal to X bar plus 1.96 sigma over square root of n. These two quantities that appear inside the bracket on the left hand side and the right hand side they are known as the lower and upper limits of our 95 percent confidence interval for mu. Students, that the probability is 95 percent that the interval X bar minus 1.96 sigma over square root of n to X bar plus 1.96 sigma over square root of n this interval will actually contain mu the parameter that we are trying to estimate and that is why students this is called a 95 percent confidence interval. Of course, we can have 99 percent confidence or sometimes we may even be happy with 90 percent confidence, but then as you saw in the derivation then those numbers 1.96 and minus 1.96 will be replaced by other numbers which we will get from the area table of the standard normal distribution. A or point note or in a real life situation many times of course, sigma is unknown. So, what do we do? We replace sigma square by its unbiased estimator small s square which you now see on the screen as indicated last time it is summation X minus X bar whole square over n minus 1 and in this case our interval becomes X bar minus 1.96 s over square root of n to X bar plus 1.96 s over square root of n. Students, in today's lecture I discussed with you the desirable property of efficiency then we went on to various methods of point estimation last but not the least we have started the concept of interval estimation. In the next lecture I will be discussing with you various real life applications of the concept of interval estimation. In the meantime I would like to encourage you to practice and to study the concepts involved in point estimation. My best wishes to you and until next time Allah Hafiz.