 As-salamu alaikum. Welcome to lecture number 45 of the course on statistics and probability. Yes, the last lecture of this particular course. Students, you will recall that in the last lecture I discussed with you the chi-square test of goodness or fit. And we applied this concept to the example in which we fitted the binomial distribution to real data. In today's lecture, I will be applying this concept to the case of the Poisson distribution. And as you now see on the slide, the platform manager of an airlines terminal ticket counter wants to determine whether customer arrivals can be modeled by using a Poisson distribution. The manager is especially interested in late night traffic. Accordingly, data for the time period of interest have been collected as follows. Number of arrivals per minute are 0, 1, 2, 3 and so on. And the frequencies are 84, 114, 70 and so on. And the sum of the frequency column is 400. The question is, is the distribution Poisson students, let us try to understand this problem. That is the number of arrivals per minute. That means 400 minutes and they should have been selected at random if this distribution, if this whole method is to be properly applied. Now, the 400 minutes we have randomly selected, we have seen per minute how many people are arriving and that data is in front of you. As you once again see on the screen, corresponding to the number of arrivals per minute equal to 0, the frequency is 84. What does it mean? It means 84 minutes is kisham ke the, when we did not have even one person arriving. Similarly, 114 minutes were searched when only one person arrived at the counter and 70 was searched when there were two persons arriving in that one minute. Students, the question is, what is this distribution Poisson? You remember when I had discussed the Poisson process, how was it defined? A process where events occur randomly, either over a time scale or over a distance scale. Obviously, we are talking about the time scale and one minute is being regarded as one unit of time. So, we are understanding that customer arrival is a random process. So, we would like to think that this data should be fitted by a Poisson distribution. If I do want to fit a Poisson distribution to this data, students, what will be the procedure? You remember that in the case of binomial, we first computed x bar and then we equated x bar to n p. n was already known and so p was found using this equation. Now, in this particular situation, students, the one lone parameter of the Poisson distribution is mu, where mu is the mean itself of the Poisson distribution. We will have to replace it by its estimate and that of course, is x bar. So, as you now see on the screen, in order to compute x bar, we construct the column of f x, which is the product obviously of the x column with the f column. Sigma f x is equal to 800 and dividing by 400, x bar comes out to be equal to 2. Now, the formula of the Poisson probabilities is e raise to minus mu, mu raise to x over x factorial, but since we have to replace mu by x bar, our formula becomes e raise to minus 2, 2 raise to x over x factorial. Applying this formula, students, we obtain the Poisson probabilities as you now see in the third column on the slide in front of you. You will put the values of the Zahire 0, 1, 2, 3, 4 and so on and you obtain probabilities as 0.1353, 0.2707 and so on such that the sum of the probabilities is 1. Multiplying the probabilities by 400, the sample size we obtain the expected frequencies which are 54.12, 108.28 and so on. Students, now that we have computed the expected frequencies, of course, now is the time for us to apply the chi-square test of goodness of fit. So, as you once again see on the slide, given the column of the observed frequencies O i and the column of the expected frequencies E i, we will be constructing and the column of O i minus E i whole square divided by E i. Adding this last column, chi-square comes out to be 78.88, but students, please before you go on with the thing, please note that we have combined the last four values, the last four categories that we have. If I add 0.08 to 0.36, the sum is still less than 5. When I add 1.36, the sum is not yet 5 and then when I add 4.80 to the sum of the first three, I obtain 6.60 as the expected frequency. Since I have combined the expected frequencies, I must also combine the observed ones and thus I obtain 24 corresponding to 6.60 and students, these are the two numbers that will be used in order to find O i minus E i and the other quantities in that particular part of the table. This sorry bath, you may have to start key students, I hope you remember that even in the last lecture I mentioned to you, that this is a very basic requirement of this particular test, the chi-square test of goodness of it, that none of the expected frequencies should be less than 5 and therefore, if we do obtain sum which are less than 5, then we will add in the manner that I just explained, so that the thing that we get ultimately is 5 or more. Let us have a look at the slide one more time. The sum of the last column is 78.88 and this particular chi-square statistic follows the chi-square distribution having k minus 1 minus r degrees of freedom, where k is the number of categories that we have after having combined any classes that we needed to combine. Therefore, if you count now after having combined the last four, we have 7 classes and not 10. You will recall from the last lecture that r is the number of parameters that you estimate from the sample data. As I just said there is only one parameter of the Poisson distribution mu and we are estimating it by x bar, therefore r is equal to 1. So, the degrees of freedom for this particular statistic are k minus 1 minus r that is 7 minus 1 minus 1 that is 5 degrees of freedom. And students, this is a right tail test, the entire level of significance, the entire area has to be in the right tail. So, if we set alpha equal to 0.05, then we will look in the chi-square table under 0.05 against 5 degrees of freedom. And as you now see on the slide, the critical value is 11.07. Since our computed value 78.88 is much, much larger than 11.07, therefore it is evident that in this particular problem, we reject the null hypothesis which said that the fit is good and we conclude that it is not a good fit. In fact, students, here 78.88, which is our result, it is so much larger than 11.07 that we are inclined to think that it might even be highly significant. You remember that if your level of significance is 5 percent, if it is only 1 percent or if your computed value is further than that, then of course we say that it is not only just significant but highly significant. So, I would like to encourage you to have a look at the chi-square table for 5 degrees of freedom under 1 percent alpha and then find out for yourself. The next point I would like to convey to you is, what is the rationale of this point that I said? So, it will always be the right tail test students. OI minus EIE whole square over EIE and then you add these up. It is the difference or the distance between the observed and the expected frequency. If those distances are small, then their squares will be small and when you add all those quantities, whatever you do after that when you add that thing, that will also be relatively big. In fact, do not you see students that if your expected frequencies, your observed frequencies with exact tally, then OI minus EIE will be equal to 0 or if they are all 0, what will be chi-square equal to? Obviously, 0. So, this is an ideal situation that your expected frequencies are exactly matching with your observed frequencies. Of course, we cannot expect such an ideal situation in the real life world where we have so much randomness, but I hope that you have got the essence of the whole discussion that the smaller the value of chi-square, the more evidence there is that it is a good fit. If it is so large that it falls in the tail, the right tail of the distribution, then we say that this is not a good fit and we should reject H naught. Then we say that this is not a good fit and we should reject H naught. Students, now let us start the discussion of the chi-square test of independence. You are interested in the formula that is quite similar to the one that we just had and actually this can also be regarded as a kind of chi-square test of goodness of fit. Let me explain this concept with the help of an example. A random sample of 250 men and 250 women were polled as to their desire concerning the ownership of personal computers. The following data resulted. In the top row, you have the men and the women and in the first column, you have the two situations, either they want to own a PC or they do not want a PC. Now, the figures in the body of the table are 120, 130, 80 and 170. Now, the figures in the body of the table are 120, 130, 80 and 170. You can see that out of 250 men and 120 women, they want to own a personal computer, whereas the women we chose 250 men and 80 women are of this type. We would like to test the hypothesis that the desire to own a PC is independent of the sex of the person and we would like to conduct this test at the 5 percent level of significance. This test involves exactly the same kind of a pattern as we have had before. The first step is the formulation of the null and the alternative hypotheses and the null hypothesis students in any such situation will always be that the two variables of classification are independent, whereas the alternative hypothesis would always be that the two variables of classification are not independent. The underlying mathematics of this procedure is such that as you know, we always begin by assuming that H naught is true, that the two variables of classification are independent. Then, we conduct all the steps under this hypothesis and in that case, our expected frequencies and our chi-square will be computed in a way in which the testing can be done. After having set the hypotheses, of course, the second step is the level of significance and we can as usual set it at 5 percent. The third step is the test statistic and as you now see on the screen, the test statistic in this situation is chi-square is equal to sigma sigma oij minus eij whole square over eij and please note that the first summation is over i and the second is over j. Students, I had told you that the pattern is very similar to the one that we had before. The difference is that we are saying sigma sigma oij minus eij, but I am not saying oij in Punjabi. I am saying it in English and why is it that we have a double summation students? The reason is that as you know, in this case, we have a bivariate table. So, I represents rows and j represents the columns. That is over all the values and therefore, we have the double summation. What is the next step? Of course, the fourth step is to compute our test statistic. Until here, students, pehle hame eij joke is particular problem, there are four e11, e12, e21 and e22. These have to be computed and as you now see on the slide, e11 that is the expected frequency of the first cell. The cell lying in the first row as well as the first column, students, this expected frequency e11 is obtained by multiplying the marginal total directly to the right of this cell by the marginal total directly below this cell and dividing this product by the grand total 500. Hence, e11 is equal to 200 into 250 over 500 and that is equal to 100. Students, ejo formula henna, this is according to what you learnt quite a few lectures ago, that if two events A and B are independent, then the probability of A and B is equal to the probability of A into the probability of B. Usi ki ek modified form abhi istimal ki and this is the way you will compute all the expected frequencies. Me pehle repeat karthi ho jis cell ki expected frequency chahiye uske saamne wala total uske niche wale total se multiply karthi je and divide this product by the grand total of all the observations. Next, of course, we will construct the columns of oij minus eij and the square and so on. And hence, as you now see on the slide, constructing all those columns, chi square comes out to be 13.33. Students, ye jo columns humne banai hai ki ab aap ki apni marzi hai ke aap kis order me oij ki values iss column me likhe, but the only thing that you have to keep in mind is that whichever order you choose for the oij, the same order must be maintained for eij. Students, what is the fifth step of any hypothesis test in procedure? The critical region. Now, in this particular situation, it can be mathematically shown that this particular statistic chi square follows the chi square distribution having r minus 1 into c minus 1 degrees of freedom, where r is the number of rows and c is the number of columns. This particular example may r is equal to 2 and c is also equal to 2. Therefore, we have 2 minus 1 into 2 minus 1 degrees of freedom, that is 1 degree of freedom. So, looking at the chi square table, against 1 degree of freedom under 0.05, we obtain, as you now see on the slide, 3.84 as the critical value. The last step is the conclusion and since our computed value 13.33 is larger than 3.84, therefore, we reject H naught. Students, our null hypothesis is that the two variables of classification are independent, that is, the sex of the person and the desire to own a personal computer, these are independent. So, we have rejected this hypothesis. That means that we are saying that the two variables of classification are associated. So, naturally, the question arises, which gender is there in which it is a higher proportion of persons who would like to have a personal computer? If there is a relationship with sex of this thing, then we should study it, that in which category this thing is more prevalent. So, let us have a look at the data once again. As you now see on the slide, out of the 250 men, 120 are such who would like to have a PC whereas, out of the 250 women, only 80 are of this type. Now, since our chi-square has come out to be significant, therefore, we can say that this difference that we can see in the data, this indicates a significantly higher proportion of men wanting to have a personal computer as compared with women. Students, with reference to the chi-square, always keep in mind the very important point that for both the chi-square test of goodness of fit and the chi-square test of independence, the number of observations in our sample should be at least 50. Otherwise, this procedure of hypothesis testing will not be suitable. Students, this example we have done, this was the simplest case when our bivariate table, which is technically called a contingency table, it was only a 2 by 2 table, instead of the rows or the columns. But of course, we can apply this procedure when we have more than 2 rows or more than 2 columns. So, let us consider one more example as you now see on the screen. A national survey was conducted in a country to obtain information regarding the smoking patterns of the adult males by marital status. A random sample of 1772 citizens, 18 years old and over, yielded the following data. Now, students, you can see that we have 4 categories for marital status, single, married, widowed and divorced and we have 3 categories for smoking pattern. Total abstinence, only at times, and regular smoker. We would like to use this data to decide whether there is an association between marital status and smoking pattern. Students, I would like to throw this as a challenge to you. The steps are absolutely similar to the ones that we just did. So, please conduct it yourself and find out whether there is an association between the marital status and the smoking pattern or whether you do not find such an association and you accept the null hypothesis that the 2 variables of classification in this case are independent students. This is to determine whether or not there is an association between 2 qualitative variables. Jase, sex of the person and whether or not he or she wants to own a personal computer. Obviously, these are qualitative variables. If we want to know the relationship between quantitative variables, then what we will do? You will recall lecture number 15, in which I discussed with you in detail the concepts of regression and correlation. Students, where we talked about R or regression line fit, that was in the area of descriptive statistics, but then we also have methods by which we can do inferential statistics regarding regression and correlation. We can test hypotheses about rho, the population correlation coefficient and beta, the population regression coefficient. Which we have R or B, they are of course the sample values, but we would not have a chance to discuss those in detail in this particular course, QK students. If we look at it this way, there are many more things which could be discussed, but the basic purpose of the course is to give you an insight into the fundamentals of the subject. I think that what we have done up till now would suffice for that, inshallah. This brings us to the end of the series of topics that I wanted to convey to you in some detail in this course. For the remaining part of today's lecture, students, I will be conveying to you a few very interesting points. The very first of which is the degrees of freedom. I said to you that the parameters that exist in the mathematical equations of those distributions, they are called degrees of freedom. But the question is why on earth are they called degrees of freedom? What do we mean by that? Let me try to explain this to you with the help of an illustration. This is the line segment. This thing is still free to move in this plane. It can be rotated and otherwise it can move. So, we have one degree of freedom. Students, if I fix both the points, now I do not have any degree of freedom for the movement of this pencil or the line segment. But if we step into the three-dimensional world, the one that you and I live in, then you will note that this plane, this whole plane is fixed here. This plane itself is able to move in this three-dimensional world. And in the three dimensions, I do have one degree of freedom. So, this is the kind of concept that is involved in this discussion. Now, coming to our statistical work. You remember that when we tested H naught mu is equal to something against H alternative that mu is not equal to that quantity. And if we were using the t test, we said that the t statistic x bar minus mu naught over s over square root of n follows the t distribution having n minus 1 degrees of freedom. So, this is n minus 1 q. Let me explain this with the help of an example, which you now see on the slide. Suppose that we have a sample of size n equal to 6 and suppose that the sum of the sample values is 20. That is, we have the following situation. . . . . . . . . . . . . . . . . the sixth one, because the sum is 20. If you freely choose the first 5 values and the sum that comes out to be 17, then Lamohala sixth value has to be equal to 3. Hence, we do not have 6 degrees of freedom to choose values, but we do have 6 minus 1, that is 5 degrees of freedom. To dekhah aapne students, it is quite an interesting concept. Ishi baatko ek aur tarasi sumjne ki koshish karte. I have 6 observations. I find their sum and then I throw away one of those 6. Students, I can regenerate that one that I have thrown away, because I already know the sum and I know I have the other 5 values. To degrees of freedom jahena, they are 6 minus 1 wo ek jo mai regenerate kar sati hu, that is the one for which I do not have the degree of freedom, that that can also be chosen freely. Wo to un baaki 5 values se hi regenerate ho jaegi. So, this is the kind of concept why we say that we are dealing with degrees of freedom. Coming back to the T distribution, the chi-square distribution, the F distribution students. Generally, we can define degrees of freedom in this manner that the degrees of freedom are the total number of observations in our sample minus the number of parameters that you estimate from the sample data. Ye jo T distribution ki baat hai hi kare lete sab se pehle. As you now see on the slide, in testing H naught mu is equal to mu naught against H alternative that mu is not equal to mu naught, our statistic X bar minus mu naught over s over square root of n follows the T distribution having n minus 1 degrees of freedom. Here, the point to be noted is that s is the estimate of sigma, the population standard deviation. Goya hum ek parameter jo hai population ka, that is the standard deviation usko hum estimate kare hain by the sample standard deviation small s. Chuke hum ek parameter chahe aap usko standard deviation kehle je, yaa population variance kehle je. Chuke ye ek parameter estimate ki aaja raha hai from the sample data therefore the degrees of freedom are n minus 1. Ishi tara f statistic ki baat kare lete hai you will recall that when we were trying to test H naught sigma 1 square is equal to sigma 2 square, our test statistic was f is equal to s 1 square over s 2 square aur hum ne kaha tha ye statistic n 1 minus 1 comma n 2 minus 1 degrees of freedom vali f distribution ko follow karte hai. Why do we have n 1 minus 1? Because s 1 square that we have in the numerator is an estimator of sigma 1 square to hum ne goya ek parameter estimate kia and hence n 1 minus 1 degrees of freedom for the numerator and a similar situation holds for the denominator aur jo abhi hum ne thori der pehle chi square test of independence conduct kia usme to bohati interesting situation hai. I said to you that we have r minus 1 into c minus 1 degrees of freedom to ye kistrase ho gaya boh jo example tha ke jisme either man or woman and either you want a p c or you do not want a p c to usme 2 rows and 2 columns and therefore we have 2 minus 1 into 2 minus 1 that is 1 into 1 and that is 1 degree of freedom. Ye kaisin hua? Let us look at that table once again. Now I am presenting to you the same table in such a way that I do have information regarding the marginal totals and the grand total, but I have removed the four data values that I had in the body of the table. For example, suppose that the value of the first cell is 100 then automatically the value in the second cell of the top row will also be 100 because the sum of the row is 200 and the value in the second cell of the first column will be 150 because the sum of that column is 250 and then of course we can also find the value in the last cell that is the cell in the second row and the second column. To lubbe lubab ye students that you can choose only one value freely. Baak ki tino ki tino juhena wo to un marginal totals ki wajah se determine ho jayengi. Aap freely choose nei kar satte. The only one that you can choose freely that may be in the first cell or in the second or third or fourth usse farak nei patta. The point to note is that you have only one degree of freedom and one can be written as 1 into 1 and that can be written as 2 minus 1 into 2 minus 1 in other words R minus 1 into C minus 1. Of course, if you have a 2 by 3 table or a 3 by 4 table to ye hi baat jo me ne present ki it will be in that context aur me aap ko encourage karungi ki aap ek 2 by 3 table banaye aur marginal total likh le aur uske baat khud iss pe work karein ke aaya ye jo formula R minus 1 into C minus 1 wala hai jo ke uske iss me 2 minus 1 into 3 minus 1, yehani 1 into 2, yehani 2 degrees of freedom ke mutta radif ho ga kya waak hi uss table me which has 2 rows and 3 columns. Do you really have only 2 degrees of freedom iss pe aap khud work ki jay. Students, the next concept is that of the P value aap ne dekha ke humne bohat saare hypothesis test kye we discussed various procedures, but in every situation what we were doing was that we determined the critical region and we decided whether our computed value falls in the acceptance region or in the critical region, but students we can also draw a conclusion regarding our hypothesis by means of what is called the P value. As you now see on the slide the probability of observing a sample value as extreme as or more extreme than the value that has been observed, given that the null hypothesis is true, this probability is called the P value. Students I would like to explain this to you with the help of the example in which we considered the hourly wages of computer analysts and registered nurses. As you will recall the statement of the example was a survey conducted by a market research organization 5 years ago showed that the estimated hourly wage for temporary computer analysts was essentially the same as the hourly wage for registered nurses. This year a random sample of 32 temporary computer analysts from across the country is taken. The analysts are contacted by telephone and asked what rates they are currently able to obtain in the marketplace. A similar random sample of 34 registered nurses is taken and the resulting wage figures are listed in the following table. As you can see the wages are 24.10, 23.75 and so on for the computer analysts and similar figures for registered nurses. We would like to conduct a hypothesis test at the 2 percent level of significance to determine whether the hourly wages of the computer analysts are still the same as those of the registered nurses. Students you will recall that our null hypothesis was mu 1 minus mu 2 is equal to 0 against the alternative that mu 1 minus mu 2 is not equal to 0. We carried out various steps and the computed value of our test statistic came out to be 3.43 which was greater than the tabulated value, the critical value 2.33. Hence we rejected H naught. The figure that you now see on the screen illustrates this point Abdeh Krenke, our value 3.43 johair that is lying in the right tail of the sampling distribution and it is to the right of 2.33 and therefore we reject the null hypothesis. Students yeh joh result humne obtain kia we could also have got the same result by the p-value method. Ab p-value kya thise? Subse pehla step johair wo yeh hai ke aap compute keejiye how much area lies to the right of our computed value. So, as you now see on the slide the value 3.43 in the right tail of the standard normal distribution is in front of you and as you can see the area beyond this point is 0.0004. Of course, we get this area by looking up the area table of the standard normal distribution. Now, since this is a two-tailed test therefore we are interested not only in this area which is on the right tail, but also the area which is to the left of minus 3.43 and obviously that is also equal to 0.0004 students in this situation the p-value is given by 0.0004 plus 0.0004 and that is equal to 0.0008. That means, we have computed the probability of this point students that our value actually is greater than that of the p-value or greater than that of the p-value and since this is a two-tailed test, we will also say that the negative counterpart of the p-value is greater than that of the p-value or greater than that of the p-value. We have found that it is only 0.0008. Now, since this is extremely small and much smaller than the level of significance that is 2 percent in this problem therefore we reject H naught. Students we always begin by assuming that H naught is true. Point is that if H naught is true, then what is the probability of extreme value? 0.0008 is highly improbable. So, if we have such a low probability for getting this value that we have actually got then we say that there is something wrong and we cannot accept the null hypothesis in this situation. So, therefore, we will reject it. I would like to encourage you to work on this problem on your own. You will simply take the area in that one tail. You will compute the area beyond that value which you have actually got as your computed value of your test statistic and once you have computed it, you will see that is that large or is it small? As you can now see on the slide, the p-value is a property of your data and it indicates how improbable your obtained result really is. So, if it is a very very small figure then we say that this result is very improbable under the null hypothesis and therefore, we reject H naught. A very simple way of understanding what interpretation to make is that if your p-value is less than the level of significance, then you will reject H naught and if it is greater then you would accept H naught. But I would like to encourage you to work on this problem on your own and to study quite a few text books and other books in this regard. What I would like to now discuss with you students is the relationship between interval estimation and hypothesis testing. Students, that there seems to be a relationship between interval estimation and hypothesis testing. After all, do you not remember that when we derived the confidence interval for mu, dharmyanme area rakhata 1 minus alpha is side way alpha by 2 or dhusri side may be alpha by 2 and when we did hypothesis testing, in case of a two tailed test, if our level of significance is alpha, then we have alpha by 2 area on this side and alpha by 2 on the other side. They definitely does seem to be some kind of a relationship between the two. So, let me throw a proposition to you. As you now see on the slide, let L comma u be the 95 percent confidence interval for the parameter theta, where L stands for the lower limit and u for the upper limit of the interval. Then we will accept the null hypothesis H naught theta is equal to theta naught against H 1 theta is unequal to theta naught at 5 percent level of significance. If theta naught falls inside the confidence interval, but if theta naught falls outside the interval L u, then we will reject H naught. In the language of hypothesis testing, the 95 percent confidence interval is known as the acceptance region and the region outside the confidence interval is called the rejection region. Also, in this way we can say that the end points of the confidence interval represent the critical values. Students, I would like you to work on this on your own. You might arrive at the conclusion that what kind of a relationship exists between interval estimation and hypothesis testing. You might arrive at the conclusion that what was just presented is correct or you might arrive at something slightly different. So, that is a challenge for you. Students, we are drawing close to the end of today's lecture and to the end of this course. But before I close, I would like to convey to you that statistics is a vast discipline. In this course, I have presented to you some of the very basic and fundamental concepts. But of course, there are numerous other concepts that could be discussed. For example, with reference to experimental design, we could talk about the Latin square design, factorial designs and so many other very interesting and important designs. With reference to statistical inference students, we could have talked about non-parametric statistics, non-parametric inference and in a similar way, there are so many other topics that we could talk about. This particular course should be regarded as the beginning of a long journey in the science of statistics. And if you are interested, you would definitely want to pursue studies in this area at some stage or the other. Another very interesting and important point, students, that I would like to convey to you is that I have presented so many different problems and so many different techniques and most of the time I have been presenting to you the solutions of the problems step by step in a lot of detail. The purpose was to give you an idea of the fundamental concept involved in that particular technique. But having said this, students, I would now like to convey to you that there are numerous computer packages that you can use and you can arrive at the solution within seconds. You are a computer science student. Who can know this better than you? And the packages are numerous, as I said, SPSS, SAS, Stata, S plus, Minitab and so on and so forth. So, I would like to encourage you to attempt quite a few problems on these packages. You will simply be clicking a button and you will be arriving at all your regression coefficients and all kinds of things that you would like to do. And speaking of latest trends, students, I would now like to give to you one of the latest definitions of statistics. Statistics is a science of decision making for governing the state affairs. It collects, analyzes, manages, monitors, interprets, evaluates and validates information. Statistics is information science and information science is statistics. It is an applicable science as its tools are applied to all sciences including humanities and social sciences. And students, at this point, I would like to repeat what I said in the very first lecture of this course. It has been said that statistical thinking will one day be as necessary for efficient citizenship as the ability to read and write. Yes, this is the importance of statistics. Education is a wholesome process. The essence of effective teaching is effective and inspiring communication on the part of the teacher and concentrated attention and effort on the part of the student. As far as the teacher is concerned, the teacher is concerned, as far as students' responsibility is concerned, as I said, it is concentrated attention and a lot of effort. And as far as students' responsibility is concerned, as I said, it is concentrated attention and a lot of effort. Students, if this course has been able to inculcate in you the fundamentals of statistical and probabilistic thinking, then it has served its purpose. I wish you the very best in your pursuit of knowledge. May Allah bless you with the wealth of knowledge and with the capability to utilize this knowledge for the betterment of humanity. Salih al-Nakhmi. Ejazat dichi. Allah hafiz.