 As Salaamu Alaykum, welcome to lecture number 32 of the course on statistics and probability. Students, you will recall that in the last lecture, I started the third and last segment of this course that of statistical inference and in this regard the very first concept that I dealt with in some detail was the concept of the sampling distribution of X bar. We discussed that the mean of the sampling distribution is equal to the population mean. The standard error of this sampling distribution is equal to the population standard deviation divided by the square root of the sample size in that situation where the finite population correction is not required and towards the end of the lecture, I discussed with you an extremely important theorem known as the central limit theorem. You will recall that according to this theorem, the sampling distribution of X bar approaches normality as the sample size small n tends to infinity. Students, let us begin today's lecture by discussing a real life example and a real life application of this particular concept. As you now see on the screen, suppose that a construction company has 310 employees who have an average annual salary of rupees 24000 and the standard deviation of the annual salaries is rupees 5000. Suppose that the employees of this company launch a demand that the government should institute a law by which their average salary should be at least rupees 24500 and suppose that the government decides to check the validity of this demand by drawing a random sample of 100 employees of this company and acquiring information regarding their present salaries. What is the probability that in a random sample of 100 employees, the average salary will exceed rupees 24500 so that the government decides that the demand of the employees of this company is unjustified and hence does not pay attention to the demand although in reality it was justified. Students, you have seen that this is an interesting problem and we are assuming that the actual figures are the ones that the employees are saying but when the government draws a sample of size 100, then if the average salary of that sample is more than 24500 then obviously the demand is unfounded. Let us see how we are going to solve this problem as you now see on the screen. The sample size 100 is large enough to assume that the sampling distribution of X bar is approximately normally distributed with mean and standard deviation given as follows, mu X bar is equal to mu and that we already know is rupees 24000 and sigma X bar is equal to sigma over square root of n multiplied by the square root of capital N minus small n over capital N minus 1 and putting the values of the population standard deviation capital N and small n sigma X bar comes out to be rupees 412.20. Students, you have seen that in this example we have applied the standard deviation of X bar which is valid in the case of sampling without replacement from a finite population. You have seen that the population size is only 310 or the sample size is 100. So it is obvious that the sample size is approximately one third of the population size and therefore it is much bigger than 5 percent of the population size or in the lecture I told you that if the sample size is less than 5 percent of the population size then we say that this FPC is not required because it is approximately equal to 1. Here definitely not because 310 minus 100 over 310 minus 1 is not going to be equal to 1. Now we have found the standard deviation and the mean of the sampling distribution of X bar. What is the probability that my X bar will be greater than rupees 24500 and students I will remind you of the lecture on the normal distribution. Since we are saying that according to the central limit theorem our distribution here is approximately normal therefore we are going to find this probability by computing the area under the normal distribution. As you will recall in the lecture on the normal distribution that is lecture number 30 I discussed with you the process of standardization and as you will remember the formula was Z is equal to X minus mu over sigma. This formula is called the variable minus its mean divided by the standard deviation. So students how are we going to apply this concept in this particular situation? As you know in this situation our variable is not X but X bar. So our formula becomes Z is equal to X bar minus mu X bar over sigma X bar. And as you now see on the screen when we substitute the values of mu X bar and sigma X bar in this formula we obtain Z is equal to X bar minus mu over sigma over square root of n over sigma over square root of n in that particular situation where we do not require the FPC and in the situation where we do require this quantity Z is equal to X bar minus mu over sigma over square root of n multiplied by the square root of capital N minus small n over capital N minus 1. Normal distribution here we have the distribution of X bar. The point is 24,500 because the probability that we wish to compute is the probability that in my sample of size 100 X bar will be greater than or equal to rupees 24,500. So what are we going to do? We are going to convert this particular value of X bar into Z by this formula and as you now see on the screen we obtain Z is equal to 24,500 minus 24,000 and this expression divided by 412.20 and solving this expression Z comes out to be 1.21. Hence we need to compute the area between Z is equal to 1.21 and plus infinity. Now according to the area table we are able to find the area between Z is equal to 0 and 1.21 and we find that this area comes out to be 0.3869. Since the total area under the normal curve from Z is equal to 0 to Z is equal to plus infinity is 0.5. Hence subtracting 0.3869 from 0.5 we obtain the area under the curve from Z is equal to 1.21 to plus infinity and this area comes out to be 0.1131. Students, this is the answer we have obtained. So, we are going to interpret this as 0.1131. Obviously 0.1131 means 11 percent that the probability is only 11 percent that in a random sample of size 100 the mean of the sample X bar will exceed rupees 24,500. This means that there is no chance that this will not be possible and we can say that the probability is high that the government will consider the demand of the employees of this particular company. The next concept that I will discuss with you is the sampling distribution of P hat where P hat denotes the proportion of successes in the sample. What we are dealing with here is a binomial population in which every element can be classified as either success or failure and if we draw a sample out of this population then of course in the sample also every element can be classified as success or failure. So, in this situation we obtain a sampling distribution which also has very important properties just as we had in the case of the sampling distribution of X bar. Let me explain this concept with the help of an example as you now see on the screen. Suppose that a population consists of 6 values 1, 3, 6, 8, 9 and 12. If we regard the occurrence of an even number as success then we find that the proportion of successes in this population is 3 by 6 and that is 1 by 2 because there are 3 even numbers in this population and they are 6, 8 and 12. Now if we draw all possible samples of size 3 without replacement from this population then what will be the salient features of the sampling distribution of p-hat students. In order to solve this question we have to proceed step by step exactly the same way as we did when constructing the sampling distribution of X bar. Now since we are sampling without replacement the total number of samples of size 3 that can be drawn from a population of size 6 is 6 C 3 and that is equal to 6 factorial over 3 factorial into 3 factorial and students if we solve this expression the total number of samples as you now see on the screen comes out to be 20. Now what are these 20 samples as you see on the slide the samples are 1, 3, 6, 1, 3, 8, 1, 3, 9, 1, 3, 12, 168, 169, 1612, 189, 1812, 1912 and so on. Students when I have say last lecture may be abad kahi thi that it is very important that when you want to construct the list of all possible samples you go in a very systematic manner. Abhi main aapke saamne jo samples prezent kie aap dekh kie uske andar ek pattern tha. Hame pehla number 1 consider kia aur uske saath sabse pehle 3, 4, 5, 6, 7, 8, 9, 10, 10, 10, 10, 11, 12, 12, 13, 14, 15, 16, 16, 17, 18, 19, 19, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 21, 21, 21, 27, 27, 28, 28, 28, 29, 29, 29, 29, 33, 31, 33, 33, 34, 4, 35, 37, 38, 34, 34, 34, 34. Abhi main aapke saath sabse pehle 3 jo number hai usko milla笑 uske saath 3rd number was possible. And after that we went to 1-8. And as you now see on the slide we have after the first 10 which have 1 in the very beginning, the 11th one, 368, 369, 3612 and then 389, 3812 and then 3912. Now, 3 K Tamam groups mukambal hone par, we start with the number 6 and we have 689, 6812 and then 6912. Finally, the last triplet is 8912, agar aap ki confusion abhi dur nahi hui to aap istrasi dekhye, we can write the numbers 1, 3, 6, 8, 9 and 12 in a row and then you can start attaching an arrow from starting from 1 with the other ones. So, you can have 1, 3, 6, 1, 3, 8, 1, 3, 9 and 1, 3, 12 or istrasi you can then have this other ones 1, 6, 8, 1, 6, 9 and so on. What we are interested in students is the sampling distribution of p hat, where p hat represents the proportion of even numbers in the sample is liye case example mein humne even number ko success kaha hai. So, as you now see on the screen for the very first sample 1, 3, 6 there is 1 even number and that is 6 and there are 2 odd numbers and hence the proportion of even numbers in this particular sample is 1 by 3. Similarly, for the second sample p hat is equal to 1 by 3, but for the third one we find that there is not even a single even number in the sample and hence the proportion of even numbers is 0 by 3 and that is equal to 0. Similarly, you can compute the proportion of even numbers for every one of those 20 samples and we find that the proportion 1 by 3 as well as the proportion 2 by 3 is repeated many times. Hence, we are interested in constructing the frequency distribution of p hat and doing so we find that the proportion 1 by 3 occurs 9 times and the proportion 2 by 3 also occurs 9 times, but the proportions 0 and 1 each of them occurs only once and thus we find that the sum of the column of frequencies is 20 exactly the same as what it should have been because the total number of samples that we drew was 6 C 3 and that is equal to 20. You saw that this is exactly the same procedure that we adopted during sampling distribution of X bar and you will recall that after finding all these frequencies we can divide each of them by the total frequency and those numbers represent the probabilities of those particular values of p hat. As you now see on the slide, the probability that p hat will be equal to 0 is 1 by 20. The probability that the proportion of even numbers in our sample is 1 by 3, this probability is 9 by 20. The probability of p hat equal to 2 by 3 is also 9 by 20 and that for p hat equal to 1 is 1 by 20. If we draw the graph of this particular sampling distribution of p hat students, we find that it is absolutely symmetrical as you now see on the slide. Next, we are interested in finding the mean and the standard deviation of this particular sampling distribution. The formula for mu p hat that is the mean of the sampling distribution of p hat is sigma p hat into f of p hat very similar to the formula that we had in the case of the sampling distribution of X bar. Also the formula for the variance of the sampling distribution of p hat is sigma p hat square into f of p hat minus sigma p hat into f of p hat whole square. In the case of the sampling distribution of X bar, we had a very similar formula sigma square X bar at that time was sigma X bar square into f of X bar minus sigma X bar into f of X bar whole square. Students, I hope that you have been able to see the similarity between the two situations. And now if we apply these quantities in our example and we compute the relevant quantities we find as you now see on the slide. The mean of the sampling distribution of p hat comes out to be 10 over 20 and that is 0.5 and the variance of the sampling distribution is 1 over 20 and that is 0.05. Of course, if I take the square root of 0.05, I will obtain the standard deviation of the sampling distribution of p hat which is also called the standard error of p hat. I have taken the mean 0.5. If we locate it on the graph it is absolutely symmetric. So, the mean has to be in the exact center of this particular distribution. So, our distribution may p hat values 0, 1 by 3, 2 by 3 and 3 by 3 that is 1. So, there may be two values students 1 by 3 or 2 by 3. What will that be? 1 by 3 plus 2 by 3 divided by 2 and is it or is it not exactly equal to half? Just decide for yourself. And the other point that I would now like to convey to you is that there are some very important relationships between the mean of the distribution of p hat and p the proportion of successes in the population. Also, relationships between sigma p hat that is the standard error of p hat and p and q the proportion of successes and failures in the population. As you now see on the slide, the relationships are number 1, mu p hat is equal to p and number 2, sigma p hat is equal to square root of p q over small n and this whole quantity multiplied by the square root of capital N minus small n over capital N minus 1. Of course, this particular formula is valid if we are sampling without replacement from a finite population, but if we sample with replacement then our equation is sigma p hat is equal to square root of p q over small n because in this situation the finite population correction factor is not required. Using the values of mu p hat p q capital N and small n in these equations, students as you see on the slide we find that both the equations valid in this situation are verified. Students you have seen that the mean or standard error is very similar to the sampling distribution of x bar. Now, what was the extremely important third property in case of the sampling distribution of x bar? I will remind you of the central limit theorem according to which the sampling distribution of x bar approaches normality as the sample size tends to infinity. In the case of the sampling distribution of p hat also students, we find that this distribution tends to normality as n tends to infinity and the rule of thumb in this connection is that if both n p and n q are greater than or equal to 5 then we can approximate the binomial sampling distribution of p hat by the normal distribution. Let me now explain this to you with the help of an interesting example. Suppose that 10 percent of the 1 kilogram boxes of sugar in a large warehouse are underweight. Suppose a retailer buys a random sample of 144 of these boxes. What is the probability that at least 5 percent of the sample boxes will be underweight? In order to solve this problem the first point to be realized is that the sample size 144 is large enough to assume that the sample proportion p hat is approximately normally distributed. According to the relationships that I mentioned earlier, the mean of this sampling distribution is given by mu p hat is equal to p and that is equal to 0.10 because we have the information that 10 percent of the boxes are underweight. Similarly, the standard error of p hat is equal to the square root of p q over small n and this is equal to 0.10 into 0.90 divided by 144 the square root of this entire quantity and this comes out to be 0.025. Hence, here you note that for the standard error the formula that I applied is square root of p hat q hat over small n and along with this the finite population correction factor square root of capital N minus small n over capital N minus 1. I did not use it. What is the reason for this? If you pay attention to the first statement in this problem then you will see that it will be alright. We had said that in a large warehouse it has been found that 10 percent of the boxes are underweight. When large warehouse is being talked about then this means that thousands upon thousands of boxes are being prepared. Therefore, the population size is very large as compared with the sample size which is 144 and we can say that as if we are sampling from an infinite population or you will remember that in such a situation we do not require the finite population correction. In this problem what are we trying to determine? Question what is the probability that in a sample of size 144 the proportion of boxes that are underweight will be at least 5 percent and it comes from 5 percent underweight. So, mathematically we express as you now see on the slide what we want to find is the probability that p hat is greater than or equal to 0.05. Students there is a very important concept that is to be discussed here and that is the concept of continuity correction. Whenever we approximate the discrete binomial distribution by the continuous normal distribution we need to apply what is called continuity correction. K discrete variable x is 0, 1, 2 and so on. We are converting it into a continuous variable by applying this correction and for example, the number 1 is replaced by the interval 0.5 to 1.5 and the value 2 of the variable x is replaced by the interval 1.5 to 2.5. And in this particular scenario that we are discussing now since I am not talking about the random variable x which represents the number of successes in my sample rather I am talking about p hat which is x over small n the proportion of successes in my sample x over small n number of successes divided by the total sample size. This is the proportion of successes in my sample. Because right now I am talking about this, that is why the continuity correction is not only plus or minus half but we will either subtract or add 1 over 2 n, i.e. below half, n is also attached because it is x over n that we are trying to apply this correction to. Therefore, in this particular example as you now see on the screen the probability that p hat is greater than or equal to 0.05 will be found by computing the probability that p hat is greater than or equal to 0.05 minus 1 over 2 times 144 because in this example the sample size is 144. Here note that I have minused and the reason is that I have to compute the area to the right of this value that I am interested in. When we expand 0.05, it will expand a little back and a little forward. I have to remove the area forward and I have to cover the minimum edge from which I should remove the area forward. So, I will subtract. But if there is any problem where I say that I want to compute the probability that p hat is less than or equal to a certain value, vahappi may students add karthi 1 over 2 n. Now, going back to this example, we have the probability that p hat is greater than or equal to 0.05 minus 1 over 2 n. This is equal to the probability that p hat is greater than or equal to 0.05 minus 1 over 288. Now, we have to find the area beyond this particular value and students, aapko patato hai ke jab kabhi bhi aapko normal curve ke under area compute karna hai. The first step is to standardize your variable and to convert it to z. And what is the formula for standardization? The variable minus its mean divided by its standard deviation. So, in this particular example as you now see on the screen, we have z is equal to p hat minus 0.10 divided by 0.025 and substituting the value 0.05 minus 1 over 288, we obtain z is equal to minus 2.14. So, this is the z value beyond which we have to find the area under the normal distribution. So, we are required to compute the area under the standard normal curve to the right of z is equal to minus 2.14. As explained in lecture number 30, we will first find the area between z is equal to 0 and z is equal to plus 2.14 and that comes out to be 0.4838. We know that because of the absolute symmetry of the normal distribution, the area from 0 to minus 2.14 will also be 0.4838 and adding this area to 0.5 which is the area between z is equal to 0 and plus infinity, the total area comes out to be 0.9838. What is the interpretation of this particular result that we have obtained students? We can say that the probability is as high as 98 percent that at least 5 percent of the boxes in our sample of size 144 will be underweight. So, agar is mulk mein as equal law hai ki jiske the heth not more than 5 percent of the boxes should be underweight to fir is warehouse koto kaafi zada problem agai na. The probability is as high as 98 percent that at least 5 percent of the boxes will be underweight whereas, in reality in that particular warehouse as many as 10 percent of the boxes are underweight. So, they cannot get away with it students. The sampling distributions that we have considered up till now pertained to that situation where we have only one population and we are drawing samples of a particular size from that one population. Let us now consider the situation where we are sampling from two populations. In this regard I will be discussing with you the sampling distribution of the differences between sample means that is the sampling distribution of x 1 bar minus x 2 bar. Also we will be talking about the sampling distribution of p 1 hat minus p 2 hat. Let us begin with the first one and I will explain it formally as you now see on the screen. Suppose we have two distinct populations with means mu 1 and mu 2 and variances sigma 1 square and sigma 2 square respectively. Let independent random samples of sizes n 1 and n 2 be selected from the respective populations and let us compute the differences x 1 bar minus x 2 bar between the means of all possible pairs of samples that we can have. Then a probability distribution of x 1 bar minus x 2 bar can be obtained and such a distribution is called the sampling distribution of the differences between sample means. Then we will find the differences between x 1 bar and x 2 bar for all possible combinations of x 1 bar and x 2 bar. So, let me explain this to you with the help of a very simple example. As you now see on the slide suppose that we draw all possible random samples of size 2 with replacement from a finite population consisting of the values 4, 6 and 8. Similarly let us draw all possible samples of size 2 with replacement from another finite population consisting of the values 1, 2 and 3. Find the possible differences between the sample means of the two populations and construct the sampling distribution of x 1 bar minus x 2 bar. We have three small size populations and we have kept them small so that this example is small otherwise it will be quite lengthy. The first population is 4, 6, 8. But remember if we are sampling with replacement the total number of possible samples is n raised to n, 3 raised to 2 and that is 9. Similarly from the other population again population size is 3, sample size is 2 and the total number of samples that we can draw is 3 raised to 2 and that is 9. Now what are these 18, 9 and 9 samples? Let us see them on the screen. From the first population the samples are 4, 4, 4, 6, 4, 8, 6, 4, 6, 6, 6, 6, 6, 8 and 8, 4, 8, 6, 8, 8. On the other hand from the second population the samples are 1, 1, 1, 2, 1, 3, 2, 1, 2, 2, 2, 3 and 3, 1, 3, 2, 3, 3. Students you saw that we have a number repeat. We are not sampling without replacement, we are sampling with replacement. Now we have 9 samples from the first population, 9 from the second population, 9 from the second and therefore we have 9 values of x bar from the first population and we denote these sample means by x 1 bar. Similarly we denote the 9 sample means of the 9 samples that we drew from the second population by x 2 bar. Hence as you now see on this slide we have x 1 bar as 4, 5, 6, 5, 6, 7 and 6, 7, 8 whereas x 2 bar is 1.0, 1.5, 2.0, 1.5, 2.0, 2.5 and 2.0, 2.5 and 3.0. Alright 9 sample means plus 9 sample means. So, we are actually talking about 9 into 981 possible differences between x 1 bar and x 2 bar and as you now see on the screen these 81 possibilities are as follows. If we write the values of x 1 bar in the top row and the values of x 2 bar in the first column of a bivariate table of the form that you now see we will have 4, 5, 6, 5, 6, 7 and 6, 7, 8 on the top and we will have 1, 2, 1.0, 1.5 and so on in the first column of the table. Now I will be subtracting every value of x 2 bar from every value of x 1 bar and by doing that I will be obtaining 81 differences which I will be writing in the body of this table. In the very first cell I will write 4 minus 1.0 and that is equal to 3.0. Let us take another cell, let us take the cell corresponding to the third column and the fourth row of this table and students you note that in this situation I should be subtracting 1.5 from 6 by doing that I obtain 4.5 and this is exactly the value that we have written in that particular cell. All right, dekhah apne 9 from 1 and 9 from the other have given us 81 possible differences between x 1 bar and x 2 bar. Ab zahir hai ke this is a big pile of numbers and we would be very interested in forming a frequency distribution so that these numbers are arranged in a compact form and students if we do that we obtain the picture that you now see on the screen. We have a column of x 1 bar minus x 2 bar which can be denoted by small d. Of course, d means difference and the values in this column are 1.0, 1.5, 2.0 and so on. Ye wohi values hai jo hamari bivariate table ke andar bai jatiye. Now, after we do the tally method and tally every one of the 81 values in this table the frequencies come out to be 1, 2, 5, 6, 10, 10, 13, 10, 10, 6, 5, 2 and 1. So, what we have done is students I hope that you have realized that even if we do not draw the line chart of this particular distribution it is absolutely symmetric and if we are able to say this because if we place a mirror horizontally against the value d equal to 4.0 we see that the frequency is above this mirror are the mirror image of the ones below dividing every one of these frequencies by the total frequency 81 we obtain 1 by 81, 2 by 81, 5 by 81 and so on and students these numbers represent the probabilities of the various values of d. This particular distribution is called the sampling distribution of d or more appropriately the sampling distribution of x 1 bar minus x 2 bar. What are the basic properties of this particular sampling distribution students we find that the mean of the sampling distribution of the differences between sample means is equal to the difference between the population means. As you now see on the slide this can be expressed mathematically as mu x 1 bar minus x 2 bar is equal to mu 1 minus mu 2 and this formula is valid regardless of whether we are sampling with replacement or without replacement. As far as the standard deviation of the sampling distribution of x 1 bar minus x 2 bar is concerned in case of sampling with replacement we have the formula sigma x 1 bar minus x 2 bar is equal to the square root of sigma 1 square over n 1 plus sigma 2 square over n 2 and in case of sampling without replacement this formula will be modified. Students when I have said earlier that in this course we are not doing the mathematical derivations of the various formulae but if you are interested you are most welcome to look them up and you will find many of them in your own textbook. I will be concentrating here on applying them to various practical problems and going back to the example that we were considering just now what do we have as far as these two formulae are concerned. As you now see on the slide the mean of the sampling distribution of x 1 bar minus x 2 bar which is the same thing as expected value of x 1 bar minus x 2 bar it is given by summation x 1 bar minus x 2 bar multiplied by f of x 1 bar minus x 2 bar. Since we are denoting x 1 bar minus x 2 bar by small d therefore we can say that mu x 1 bar minus x 2 bar is equal to sigma d into f of d and multiplying the column of d by the column of f of d and adding these products the answer is 324 over 81 and that is equal to 4. Also sigma square x 1 bar minus x 2 bar which is the same thing as the variance of d it is given by the expected value of d square minus the expected value of d whole square and that is equal to summation d square into f of d minus summation d into f of d whole square and constructing the column of d square into f of d and adding this column and also substituting the other required values we find that the variance of the sampling distribution of x 1 bar minus x 2 bar comes out to be 1.67. All right, we will need to find the mean and variance of the first population as well as the mean and the variance of the second population. As you see on the screen applying the ordinary formulae of the mean and the variance the mean of the first population comes out to be 6 and the variance comes out to be 8 over 3. Similarly, the mean of the second population is 2 and the variance is equal to 2 by 3. What we have to show is that mu x 1 bar minus x 2 bar is equal to mu 1 minus mu 2 and if we compute mu 1 minus mu 2 we find that it is equal to 6 minus 2 and that is equal to 4 and students this is exactly what we obtained when we computed mu x 1 bar minus x 2 bar. Hence, this first property is verified and as far as the second one is concerned we now compute sigma 1 square over n 1 plus sigma 2 square over n 2 and substituting the values we obtain 5 over 3 and we note that this is exactly the result that we obtained when we computed the variance of x 1 bar minus x 1 bar minus x 2 bar. Students, you have seen that these two properties were verified in this particular example and you note that here this formula is that the variance of x 1 bar minus x 2 bar is equal to sigma 1 square over n 1 plus sigma 2 square over n 2. This is why it is valid because we have done sampling with replacement. If sampling without replacement, then finite population correction factor should apply. The third point is what is the shape of our sampling distribution as you now see on the slide. The first point is that if the populations are normally distributed then the sampling distribution of x 1 bar minus x 2 bar will be normal regardless of the sizes of the two samples. It will be normal with mean mu 1 minus mu 2 and variance sigma 1 square over n 1 plus sigma 2 square over n 2. But if we have a situation where the populations are non normal, but the sample sizes are large so that as a rule of thumb they are greater than or equal to 30, then by way of the central limit theorem students, we can say that the sampling distribution of the differences between sample means will be approximately normal in this particular situation. In other words, we are saying that if sample sizes are large, then we will come in this position that we can say that our sampling distribution will be approximately normal. And as a rule of thumb the number n equal to 30, this is regarded as the minimum size at which we can say that it is quite large, large enough for our sampling distribution to be approximately normal. In today's lecture I discussed with you in detail the sampling distribution of p hat and towards the end of the lecture we discussed that situation where we are sampling from two distinct populations and we discussed the sampling distribution of the differences between means. I would like to encourage you to practice with this concept as much as you can and students in the next lecture I will take up the sampling distribution of p 1 hat minus p 2 hat. My best wishes to you and until next time Allah Hafiz.