 As-Salaam-Alaikum. Welcome to lecture number 35 of the course on statistics and probability. Students, you will recall that in the last lecture, I discussed with you various methods of point estimation. You will recall the method of moments, the method of least squares and the very important method of maximum likelihood. After a brief discussion of these methods students, we started the other very important area of estimation and that is interval estimation. You will recall that towards the end of the last lecture, I derived for you the 95 percent confidence interval for mu. You remember that I had told you that generally, we will not do derivations in this course, but I will give you this derivation so that you can realize the lower and the upper limit of the confidence interval. Otherwise, if I would have just given you the formula, it would not have made much sense to you. Students, you will recall that the derivation led us to the formula x bar plus minus 1.96 sigma over square root of n plus and minus meaning that if you minus, you obtain the lower limit and if you plus, you obtain the upper limit. And you will remember that I had told you that in many real life situations, sigma, the standard deviation of the population is not known and then what do we do? The best we can do is to estimate sigma. If we apply the formula summation x minus x bar whole square over n minus 1 then of course, we are talking about small s square which is an unbiased estimator of sigma square. And hence, our formula for the confidence interval now becomes x bar plus minus 1.96 s over square root of n. Let us apply this now to an example. As you now see on the screen, consider a car assembly plant employing something over 25,000 men. In planning its future labour requirements, the management wants an estimate of the number of days lost per man each year due to illness or absenteeism. A random sample of 500 employment records shows the following situation. Now, we have two columns. The first is number of days lost and we have the values none, 1 or 2, 3 or 4, 5 or 6, 7 or 8, 9 to 12 and 13 to 20. And the second column is of the number of employees who had that many absences. 48 persons were not absent even once, 43 were absent either once or twice and 90 either three times or four times. And in this way, we have all the figures in the second column which add to 500 the sample size that we drew. Construct a 95 percent confidence interval for the mean number of days lost per man each year due to illness or absenteeism. Students, you saw that we had a frequency distribution available. But, importantly keep this in mind that this frequency distribution pertains to the sample that we have drawn. As you have seen, there are more than 25,000 employees of that company. But, sample that is of only 500 employees and those records pertain to just 500 people. But, since 500 is a big figure, the absences of the figures have been compiled into a frequency distribution. Our aim is that we do not just want to know for this sample that what is the mean number of absences. In fact, we would like to draw an inference for the whole population. Or if we construct this confidence interval, we can estimate in the form of an interval with which we can attach a level of confidence. So, how do we tackle the situation? So, as you now see on the screen, the computations which are exactly the same as we have learned earlier lead to x bar equal to 5.38 days and s equal to 3.53 days. Substituting these values in the formula, our 95 percent confidence interval for mu comes out to be 5.38 minus 1.96 into 3.53 over the square root of 500. This is the lower limit and for the upper limit, we have exactly the same numbers, the only difference being that the minus sign is replaced by the plus sign. Obviously, we have put the number 500 in the square root sign, because the sample size in this particular problem is 500. And students, upon solving the two expressions we obtain the lower limit as 5.07 days and the upper limit as 5.69 days. Students, aaye is result ko interpret karne ki ek martabha phir koshish karte hain, dekhye level of confidence tha 95 percent. Haam ye kye rahe hain ke agar haam is krisam ke, bohot se confidence intervals construct karte based on the different samples that we could have drawn from this population, to 95 percent of these intervals would have contained the true parameter value. Yani jo true mean number of absences hain haa, wo inme se pachan ve fisad me lai karna tha. Likin of course, ek real life situation mit haam ek hi draw karenge aur ye joh me draw kia uski basis pe jo interval banaa that is 5.07 days to 5.69 days. Ye range kis chees ko represent kar rahi hain, ke haam ye kye rahe hain ke mean number of absences jo hain haa that may be lying in this range and we are saying this on the basis of that procedure in which the level of confidence is 95 percent. Baat zara lambi hogei, likin baat ahem hain. Ek bada hi interesting point aur important point me aapko right now convey kar dena chatiun. Dekhi ye jo derivation last time ki thi uske tahit. The equation that we obtained at the end was that the probability is 0.95 that is 95 percent that x bar minus 1.96 sigma hour square root of n is less than equal to mu is less than equal to x bar plus 1.96 sigma hour square root of n. Ye statement bilkul sehi hai. Jab takke x bar ki value specify nahi hoti, tap tak ye statement sehi hai. Ke is baat ki probability pachan ve feesad hai ke ye jo interval hai that will contain mu. But students, the moment you have actually drawn a sample, actually found in x bar and actually computed limits the way we have just done. Aap agar me is kusam ki statement doon, it is incorrect. Agar me ye statement doon is example me that the probability is 0.95 that 5.07 is less than or equal to mu is less than or equal to 5.69. Yehani mu lies in this range is baat kain kaan pachan ve feesad hai. Mathematically speaking it is wrong. Deke mu jo true population mean hai nahi uski koi ek particular value hai. Ya vo value is interval ke andar hai aur ya vo value is interval ke bahar hai. If it is inside this interval then we should say that the probability of it being in this interval is cent percent that is 1. Ine misal ke tor pe agar aap fars ki je that the true mean is 5.25. To 5.25 to hamesha hamesha aur hamesha 5.07 or 5.69 ke darme anhi lai kare gana, this is cent percent probability. So, in this case we should say that the probability of this happening is 1 aur agar fars ki je that the actual true value of mu is 6.25. So, 6.25 to kabhi kabhi bhi 5.07 or 5.69 ke darme anh lai nahi karshat tana. So, the probability of 6.25 lying in that range is 0. It is neither 1 and nor is it 0.95. Aap ne dekhah ki yeh bahot lambi chauri discussion, main ek mathematical intricacy ke upar kar di, but students this is vitally important. Aap yeh statement nah kabhi dein that the probability that mu is between 5. something and 6. something or whatever, ke a number hain agar aur uske bhi chme mu lai kar rahe hain aur aap kahe ki iss baat ki probability 0.95 hain. Once the sample has be drawn and it has materialized and you have a proper some actual numerical value for x bar and you have correspondingly some actual numerical lower and upper limits, uske baat either that interval contains mu or it does not contain mu. Either the probability of mu lying in that interval is 1 or it is 0. Tuphe sabaal yeh ki sab kuch hum kar kya rahe hain? Main phir repeat karthi hu students ke ham ne yeh kya, ke hamara procedure iss khusam kahe that if we had done this, this process again and again and again, then 95 percent of our intervals of this type would have contained mu. To iss baat pe hum khus hain. Iss baat pe hum khus hain ke hamara yeh process iss type kahe. Aur issi basis pe students, numerous intervals are constructed. Not only for mu, but also for p, the proportion of successes in a binomial population, for sigma square, the variance of any particular population and for other quantities. And in this course as we go along, you will find that I will be constructing intervals for you of this type for mu, mu 1 minus mu 2, p, p 1 minus p 2 and later also for sigma square. And also students for sigma 1 square over sigma 2 square, yani do alag alag population hai, unke alag alag variances. And we are interested ke un variances ka apis me kya relation hai, usko estimate kar na chahate. So, this is a detailed area, a vast area, but yeh jo khus me ne abhi pichle 10 minute ke dharan kaha that is the crux of the matter. Aur issi liye main issko itni detail me explain kya, kyunke from my point of view it is absolutely useless to apply formulas without understanding the concept. Hamesha koshish kijiye ke aap pehle uski, uski rooh tat pachin aur uske baad usko real life situations me appropriately apply kering. Alright, now that we have had a detailed discussion regarding the core concept involved students, let me now bring your attention to another important point. Yeh jo me ne aap ke saath saari discussion pehle bhi ki aur aaj bhi ki, this was pertaining to 95 percent confidence. This is the generally used level of confidence aksar aqad hum yehi use karthe. But this is not the only one that we can use, we can be 99 percent confident, we would like to be more confident why 95 and in some situations we might find that we might be happy with even only 90 percent confidence. To issi liye main, it means that there should be a general formula jis me a saari levels of confidence accommodate hojain. And for that let us now look at the slide that we have. In general the lower and upper limits of the confidence interval for mu are given by x bar plus minus z alpha by 2 s over square root of n where the value of z alpha by 2 depends on how much confidence we want to have in our interval estimate. Ab saawali pehda hota hai ke z alpha by 2 se hammari kya muradhe. If you have a look at the diagram that is now in front of you students, you will find that the two points that we have towards the tails of our standard normal distribution, they are being called z alpha by 2 the one on the right and minus z alpha by 2 the one on the left. Pehlae right side valet point per concentrate ke j this is z alpha by 2 and students the subscript alpha by 2 implies that the area to the right of this particular point is alpha by 2. First ke j hammari z alpha yaha pehla hota to uska matlab yeh hota ke is point ke right side peh joe area hai that is equal to alpha. Acha aap saawali yeh ke yeh hammari yeh jo quantities likhi hui hai one minus alpha alpha by 2 yeh inka kya matlab hai. This sils leh meh the first thing to understand is that the area in the middle of the diagram represents the level of confidence. So, if the level of confidence is 95 percent it means that 1 minus alpha is equal to 0.95 which in turn means that alpha is equal to 0.05 means that alpha is equal to 0.05. Now, when I divide alpha by 2 in other words if I divide 0.05 by 2 obviously I will get 0.025 that is 2.5 percent and this is exactly the area that we would like to have in the right tail as well as the left tail of the distribution. In other words the area on the right as well as on the left is 0.025 and the one in the middle equal to 0.95. So, that when we add all these areas we get 1 the total area under the normal distribution. Generally speaking I repeat in the middle we will have 1 minus alpha which will denote the level of confidence and on the right side as well as on the left side we will have alpha by 2. This Z alpha by 2 here students this Z value we will get from the area table as we have seen before when we were studying the normal distribution in its own right. Whenever you solve a problem of normal distribution then Z value you will get from the area table. Now, as you see on the next slide if 1 minus alpha is equal to 0.95 then the area table gives us the Z value as 1.96, but if the level of confidence is 99 percent then Z alpha by 2 comes out to be 2.575 which can be rounded to say 2.58. Also if the level of confidence is 90 percent then Z alpha by 2 comes out to be 1.644 5 students. I am sure that you should have no problem in finding these values because you have already quite a bit of practice I am sure by now of how to consult the area table of the standard normal distribution. Now, these 3, 4 values I have shared with you actually these you know if you just memorize them then you do not need to read the area table as I have already said 95 percent confidence is the most commonly used level of confidence and for this 1.96 value we get from the area table. So, that can be memorized very easily and the other 2 I have told you that is also very easy to remember. So, you do not have to look at the area table again and again and again 1.645 is the value when we have 90 percent confidence and 2.58 is the value which we will have if we want to have a very high level of confidence and that is 99 percent. Next point is that the formula which we have developed now is X bar plus minus Z alpha by 2 S over square root of n you have seen that I have said S. So, this means that we are assuming that sigma the way the standard deviation of the population is unknown and so we have to replace it by S. This formula this is valid for large n or was central limit theorem be I have to remind you again that if your sample size large so, then your sampling distribution that has mean equal to mu and standard deviation equal to sigma over square root of n, we have derived this interval from that and now finally, we have put S at the place of sigma. Large n is the practical means how large it is means practically how large it is for students. The rule of thumb is that if your n is greater than or equal to 30, then you can apply this formula quite adequately. So, in reality it is only 30 which you can apply. It should have been very large, but we find that even if it is only 30 it is quite adequate. So, as you now see on the slide for large n that is n greater than equal to 30 a sample of this size drawn from an infinite population. The confidence interval for mu is given by X bar plus minus Z alpha by 2 into S over square root of n and S is equal to the square root of sigma X minus X bar whole square over n minus 1. Let us now apply this concept to a few more examples. The Punjab highway department is studying the traffic pattern on the GT road near Lahore. As part of the study the department needs to estimate the average number of vehicles that pass the Ravi bridge each day. A random sample of 64 days gives X bar equal to 5,410 and S is equal to 680. Find the 90 percent confidence interval estimate for mu, the average number of vehicles that pass the Ravi bridge per day. So, let us now look at some of the information available to us. You have seen that sample size is 64 and the rule of thumb given to you is greater than 30. So, we can apply this formula that I have been discussing with you. Next is that X bar is available. We have taken the data of these days. That is, record is kept every day. You know that the tool takes and so on. So, of course, a record is maintained. So, we found that the record of 64 days is X bar is equal to 5,410. On the average, in those 64 days, X bar is equal to 5,410 days. 5,410 vehicles were passing the Ravi bridge each day. On the average, standard deviation a measure of variation. So, that variation the measure is S and as you notice that is equal to 680. So, applying these values in the formula, we obtain 5,410 plus minus z alpha by 2, 680 divided by the square root of 64. Now, since the level of confidence interval that we want to have is 90 percent. Hence, z alpha by 2 is equal to 1.645 and substituting this value, we obtain 5,410 plus minus 1.645 into 680 divided by 8 and solving this expression, the lower limit of our confidence interval is 5,270.2 and the upper limit is 5549.8. Rounding these numbers to the nearest whole number, we obtain 5,270 to 5550. This means that on the basis of 90 percent level of confidence, we are saying that the average number of vehicles that pass the Ravi bridge each day, this average number lies somewhere between 5,270 and 5,550. Here, a very important number is 5,270. So, we are saying that the average number of vehicles that pass this bridge each day, this number lies somewhere between 5,270 and 5,550. No, this is very important. I have said that the average number of vehicles that pass the main number of vehicles lies in this range. I have not said that the minimum number of vehicles that can pass is 5,270. It is possible that the day when only 4000 cars pass, it is possible that the day when 6500 cars pass, it is possible that it is possible that it is possible that it is possible. Let us now have a look at another example. Suppose a car rental firm wants to estimate the average number of kilometers traveled per day by each of its cars rented in a particular city. A random sample of 110 cars rented in this city reveals that the mean travel distance per day is 85.5 kilometers with a standard deviation of 19.3 kilometers. Compute a 99 percent confidence interval to estimate mu. Now, to solve this problem, the first thing to note is that n is equal to 110 much larger than 30 and x bar is equal to 85.5 s equal to 19.3. So, applying these values in the formula, we obtain 85.5 minus z alpha by 2 into 19.3 over square root of 110 for the lower limit and for the upper limit, the minus sign is replaced by the plus sign. Now, since we wish to have 99 percent confidence, therefore, z alpha by 2 is equal to 2.575 and substituting this value, our limits come out to be 80.8 and 90.2. So, we are saying with a very high level of confidence that is 99 percent that the mean distance, travel distance, this mean value lies somewhere between 80.8 or 81 if we round it, 81 kilometers and 90.2 that is 90 kilometers. So, this is the way say students, you have seen that we give a range of values and we say that our parameter lies in this range. We say this with a certain level of confidence. Now, I would like to draw your attention to another very interesting and important way of interpreting a confidence interval. As you now see on the screen, because of the fact that sigma x bar is equal to sigma over square root of n, our confidence interval can be conveniently written as x bar plus minus z alpha by 2 into sigma x bar, where sigma x bar of course, represents the standard error of x bar. The confidence interval for mu can therefore be defined as x bar plus minus a certain number of standard errors of x bar. In other words, we can say that our confidence interval is a point estimate plus minus a few times the standard error of that estimate. Now, the question arises how many times and the answer is that this thing depends on the level of confidence that we wish to have. In the case of 99 percent confidence, the in order to estimate mu our z alpha by 2 is equal to 2.58 which is approximately equal to 2.5 or in other words 2.5. So, we can define our confidence interval as x bar plus minus 2.5 times standard error of x bar. Similarly, in the case of 95 percent confidence z alpha by 2 is equal to 1.96 which is approximately equal to 2. So, in this case we can define our confidence interval as x bar plus minus 2 times the standard error of x bar and so on and so forth. In the case of 99 percent confidence, the standard error of x bar plus minus a certain number of standard errors. In the case of 99 percent confidence, the standard error of x bar plus or 95 percent confidence level chahiye, to those standard errors add or subtract kalli jai. Agar aap usko visualize kare, to it is very simple x bar us mesi agar minus karenge you get the lower limit and agar aap plus karenge you get the upper limit and you get the two limits of your confidence interval. So this is an important and very interesting way of interpreting and remembering the concept of a confidence interval. Or students aap me aapki tavat jo ek bohot hi important point ki tarab dilana chati hu. Aap kahange aaj to sabhi cheezein bohot zyada important hai, well let it be so. Baat hi hai ke aapne dekhah, ke level of confidence is variable aap chahiye to nawe fisad confidence pehi you can be quite happy. But obviously it is intuitively understandable that the higher the level of confidence the better it is. I mean why should it be 90 percent? Why not 99 percent or even 99.99 percent? The problem is that as you raise the level of confidence given any particular sample of a particular size, students your confidence interval widens. Abhi abhi aapne dekhah tha, ke main kaha ke agar pachan ve fisad confidence hai to 1.96 ya 2 standard errors aap piche jayenge, 2 standard errors aage jayenge. So you will get a certain interval, but when you increase the level of confidence to 99 percent aap ku yaad hain aap 2.58 yaani 2.5 standard errors piche jayenge aap 2.5 standard errors aage jayenge and it is a wider interval. So this is the problem, ke agar aap level of confidence ko raise karte hain to aap ka interval widen ho jaata hai to aap kahange ki problem kya hai hone de. No that is not very wonderful students isliye ke the narrower the interval the better. After all aap zara ghor kije aap main statement deine lagi ho zara sochi hai. If I say that I am 99.99999 percent confident that the mean height of the adult males in this particular city, this mean height lies somewhere between 4 feet and 12 feet, will it make any sense? Yaani itna wide interval main dia aur mujhe koi kaini sakta ke ma ghalat kaini hui isliye ke bo mean height vaak hi 4 feet or 12 feet ke dar main hi kaini hogi and I am saying it with 99.99999 percent level of confidence, it does not make much sense. The narrower your interval the better. For example, if I say that I am this much confident that the mean height of these males lies somewhere between 5 feet 6 inches and 5 feet 7 inches students, aap ne dekha immediately aap ka zahan jo hai yeh sko accept karta hai as something to be accepted and to be happy about dekhye kis kadar narrow interval main aapko dia ki bai hi 5 feet 6 inches or 5 feet 7 inches ke dar maian mean height lie karthi. Actually the narrower your interval the better and students what can be the ideal width of your interval, will it not be zero ke bilkul hi koi uski width na ho and you are able to give just one figure. Then you are back to point estimation, the x bar value itself that one lone answer that you have obtained that one single point that is an estimate of your parameter. But as stated in the last lecture the problem with point estimation is that once you have drawn one particular sample and found this point estimate you have no way of saying how close this particular value is to the true parameter value, how close is it or how far is it. agar aap ek interval construct kar leten to uske saath you are able to attach this level of confidence, yani that probability that we have been talking about earlier. aap issari discussion ka jist kya hai hum chahti yeh ke interval narrow bhi ho aur level of confidence bhi hi ho. But students as I just indicated this is not possible for any given particular size in the in because of this reason that as you increase your level of confidence your interval widens iska ilaaj yeh ke aap sample size bada karne because if you look at the formula it is x bar plus minus z alpha by 2 sigma over square root of n jab yeh n uske denominator me aara hai to iska matlab hai ke agar n increase karega to wo puri jo cheese hai which is being added and subtracted from your x bar that entire expression will become smaller obviously as you increase the thing in the denominator that whole thing becomes smaller aur agar aap bo ek choti cheese add or subtract kareenge from your x bar then obviously you are achieving a narrower interval. The problem is that in many situations you cannot afford to increase the sample size beyond a certain value aap ke itne resources nahi hote itne paise aap ke paise available nahi hote budget ke point of view se. Ke jiski bhajase you are able to increase the sample size to as large size that you might wish to have you may have time constraints the smaller your sample the less time you will need to collect the data and to analyze the data and so on and so forth. So, agar aap sample size nahi brhaasakte then you have to strike a compromise between how high a level you wish to have and how wide the interval you can tolerate alright students. Now that we have had a discussion regarding the confidence interval for mu let us proceed to the confidence interval for mu 1 minus mu 2. aap haam ye baat kar rahe ke we have two populations and we are interested in determining the difference between the means of the two populations. Istra ke situation kab arais karegi for example, if there is a boys college and there is a girls college and we are talking about the marks of the students to end munkin hain ke agar vo dono college compete karthe hain to unke jo principles hain unne interest ho is baat me ke determine ki a jai ke dono colleges ke jo marks hain unne kitna fark hain on the average how much is the difference between the marks obtained by the students of the boys college and the marks obtained by the students of the girls college. To aaye is concept ko is apply karthe hain two different examples prior to which of course we first look at the formula for this particular interval. As you now see on the screen the confidence interval for the difference between the means of two populations that is mu 1 minus mu 2 is given by x 1 bar minus x 2 bar plus minus z alpha by 2 into the square root of s 1 square over n 1 plus s 2 square over n 2 and this formula is valid when we are drawing large samples independently from two infinite populations. This formula is being presented in that situation when the population variances sigma 1 square and sigma 2 square are unknown as is generally the case and therefore, we will replace them by their estimates s 1 square and s 2 square. Students ab is course me ek derivation to me kar hi chuki ho regarding confidence intervals. Lekin ab is formulae ki derivation mai nahi karna cha hongi, but I would like to draw your attention to the fact that the basic pattern is exactly the same as before. It is the point estimate plus minus z alpha by 2 into the standard error of that particular point estimate. This situation me jo parameter hon estimate karna chahti hain ya jo quantity hon estimate karna chahti hain that is mu 1 minus mu 2 aur iska jo point estimate hain that is x 1 bar minus x 2 bar aur aapne dekha ke yehi quantity hain jis ke baad hon kar rahe hain plus minus z alpha by 2 into the square root of some big expression. Now, that expression the square root of s 1 square over n 1 plus s 2 square over n 2 this is the standard error or in other words the standard deviation of the sampling distribution of x 1 bar minus x 2 bar our point estimate. Again repeating myself ke asal me to sigma 1 square or sigma 2 square hon hain. Lekin agar wo available nahi hain to hon ne replace kar dete hain by their point estimates. So, let us now apply this formula to an example as you now see on the screen the means and variances of the daily incomes in rupees of two samples of workers are given in the following table the samples being randomly drawn from two different factories. Now, the data is for factory A and factory B the sample sizes are 160 and 220 the mean values are rupees 12.80 and rupees 11.25 and the variances are 64 and 47. Calculate the 90 percent confidence interval for the real difference in the incomes of the workers from the two factories. In order to solve this question the first thing to do is to denote factory A by the subscript 1 and factory B by the subscript 2. Of course, we can reverse this if we want and in that case factory B will be denoted by 1 and factory A by 2, but if we use 1 for A and 2 for B then we find that x 1 bar is equal to 12.80 and x 2 bar is equal to 11.25, n 1 is 160, n 2 is 220 and s 1 square is 64 and s 2 square is 47. Applying all these values in the formula we obtain the lower limit as 0.27 and the upper limit as 2.83. Of course, in this formula we have put z alpha by 2 is equal to 1.645 because we are interested in a 90 percent confidence interval. Students you have seen that our interval is 0.27 and upper limit is 2.83. Now, how we will interpret this? We are saying that on the basis of 90 percent confidence we can say that the difference between the incomes of the workers of factory A and the workers of factory B, the mean difference between their wages lies in this range. On the average, this difference lies in this range. I would like to encourage you to think about this on your own. If you understand that this is a better picture than narrowing, then what you would have to do? You would have had to reduce the level of confidence further. already we are only 90 percent confident. Then we might have had to come down to 80 percent confidence. How foolish it might seem if I were to say that the difference between these two quantities, this difference lies in the range 1.11 and 1.31 and I am saying this with 20 percent confidence. Students, let us apply this very interesting concept to another example. Suppose a study is conducted in a developed country to estimate the difference between middle income shoppers and low income shoppers in terms of the average amount saved on grocery bills per week by using coupons. Random samples of 60 middle income shoppers and 80 low income shoppers are taken and their purchases are monitored for one week. The average amount saved with coupons as well as the sample sizes and sample standard deviations are given in the following table. For the middle income shoppers, we are using the subscript 1 and for the low income shoppers, we use the subscript 2, so that n 1 is 60, x 1 bar is 5.84 dollars and s 1 is 1.41 dollars. Similarly, n 2 is 80, x 2 bar is 2.67 dollars and s 2 is 0.54 dollars. Now, we wish to use this information in order to construct a 98 percent confidence interval to estimate the difference between the mean amounts saved with coupons by the middle income shoppers and the low income shoppers. In order to solve this question, the first thing we do is to compute z alpha by 2 corresponding to level of confidence equal to 98 percent. And if we look at the area table of the standard normal distribution, we find that z alpha by 2 is equal to 2.33 against this particular level of confidence. Substituting all the available values in the formula, we obtain 5.84 minus 2.67 minus 2.33 into the square root of 1.41 square over 60 plus 0.54 square over 80, 2.72 as the lower limit and 3.62 as the upper limit of our confidence interval. In other words, we are saying that on the basis of 98 percent confidence, we estimate that the mean difference in the savings of the two categories of shoppers, this difference lies somewhere between 2.72 dollars and 3.62 dollars. In this situation, students note that we have said 1 middle income shoppers and 2 low income shoppers. So, if we are saying x 1 bar minus x 2 bar and after that our interval is coming like you just saw. So, this means that on the average, middle income shoppers are saving a lot of money as compared with the low income shoppers. If our interval minus something to minus something comes, then we say that low income shoppers are saving a lot of money as compared with the middle income shoppers. See, this is a very simple thing that you subtract from which and after that your answer is positive or negative. According to that, you will interpret that situation and that answer. Students, in today's lecture, we have discussed in considerable detail the concept and the method of constructing confidence intervals. I discussed with you the confidence intervals for mu and mu 1 minus mu 2. In the next lecture, we will be talking about confidence intervals for p and p 1 minus p 2. Until next time, best of luck and Allah Hafiz.