 As Salaamu Alaikum, welcome to lecture number 42 of the course on statistics and probability. Students, you will recall that in the last lecture, I discussed with you the chi-square distribution. I discussed the main properties of the distribution and then its role in statistical inference. You will recall that the chi-square distribution plays an important role in estimation and hypothesis testing regarding the population variance sigma square. In today's lecture, I will be talking about the F distribution and statistical inference about two population variances on the basis of the F distribution. But before I proceed with that, let us consolidate what we did last time with the help of an example. As you now see on the slide, the manager of a bottling plant is anxious to reduce the variability in the net weight of fruit bottled. Under a long period, the standard deviation has been 15.2 gram. A new machine is introduced and the net weights in grams in 10 randomly selected bottles all of the same nominal weight are 987, 966, 955 and so on grams. Would you report to the manager that the new machine has a better performance? Let us first try to understand this problem. This fruit that is being bottled and the desire is that the specifications are not more than that, not less than that, but whatever you try, there is some variability. But, of course, you would like that the variability should be minimal. So, in this, we are trying to test whether the new machine has been introduced or not the mean students, the variability, the variation from bottle to bottle has that variation reduced. So, how do we proceed with this test as you see on the screen? Step one is the null and the alternative hypothesis H naught sigma is equal to 15.2 gram that is the standard deviation is still the same as before and H 1 that sigma is less than 15.2 gram. In other words, the standard deviation has reduced. Step two is the level of significance and we may set it as 5 percent. Step three is the test statistic and students, you will recall from the last lecture that n s square over sigma square or in other words summation x minus x bar whole square over sigma square. This quantity follows the chi square distribution where n minus 1 degrees of freedom. Since we always begin by assuming that H naught is true, therefore we can write n s square over sigma naught square where sigma naught square is the value of sigma square that we hypothesize under H naught. The next step is the calculation and in this problem the sample size is 10. Sum x comes out to be 9713 and sigma x square is equal to 9 4 3 5 3 4 7. Now n s square is equal to sigma x square minus sigma x whole square over n according to the short cut formula and substituting the values n s square comes out to be 1110.1 hence n s square over sigma naught square which can be called chi square our test statistic. This comes out to be equal to 4.81. All right. 4 steps are complete and the fifth step is critical region and students in this particular problem what is the alternative that sigma square is less than what we hypothesized according to the null hypothesis. I hope you will all agree that the entire five percent area should be to the left side and therefore when we look at the chi square table we look under 0.95 because if left side pay five percent area chahiye to is ka matlab hai ke us point ke right side pay 95 percent area chahiye. Or jaisa ke maine last time kaha tha chi square table johana us me aapko jo areas milenge wo hain jo us point ke right pe ho. So as you now see on the screen the critical value is chi square 0.95 at 9 degrees of freedom and this comes out to be 3.32. To jaisa me kahiye ho ke ab hum left side pe jo tail hain a usko critical region maane hain kyunke hain our alternative hypothesis less than sign liye ho hain. But you will recall from the last lecture that the chi square distribution starts from 0 and goes up to infinity to is ka matlab yeh hoa ke yeh jo hain our critical value hai 3.32 uske left side pe jo area hai wo minus infinity tak nahi jaahara students it is going only up to 0. Agar hamari value computed value 0 aur 3.32 ke darmean liye kar jahegi then of course we will say that it is lying in the critical region. Lekin agar hamari value 3.32 se right side pe liye karegi then we say that it is in the acceptance region. So in this problem as you now see on the screen our computed value is equal to 4.81 as we calculated a short while ago and hence we say that it does not fall in the critical region and hence we would not report to the manager that the new machine has a better performance. After all hamara jo alternative tha wo yeh kahra tha na ke the variability has reduced to agar ham us alternative ko accept nahi kar rahe, bal ke null ko accept kar rahe hain because our value has fallen in the acceptance region to humari yeh khahish to puri nahi hui na kyun ke hamara jo data hai wo iss baat ko support nahi kar rahe. Alright students let us now proceed to the F distribution another very important distribution in the theory of statistical inference. I will begin by the formal definition and then I will present to you some important properties of the F distribution. As you now see on the slide the mathematical equation of the F distribution is as follows F of x is equal to gamma nu 1 plus nu 2 over 2 into nu 1 over nu 2 whole raise to nu 1 over 2 into x raise to nu 1 over 2 minus 1 and this whole thing divided by gamma nu 1 by 2 gamma nu 2 by 2 into 1 plus nu 1 x over nu 2 and this whole expression raise to nu 1 plus nu 2 over 2 and this entire equation is valid for x lying between 0 and infinity. This distribution has two parameters nu 1 and nu 2 which are known as the degrees of freedom of the F distribution. Yes another very complicated equation just to make you realize that the F distribution is different from the chi-square distribution because its mathematical equation is different and although just like the chi-square distribution is positively skewed the F distribution is also positively skewed but they are not one and the same distribution. If you do not go into the intricacies of this equation just note one point that the equation that you just saw it has two quantities which can be called two parameters of the F distribution we denote them by nu 1 and nu 2 and students this may do order that is very important I mean if you are saying that our F distribution is with the nu 1 comma nu 2 degrees of freedom then its equation will be what you just saw but if you are saying that our distribution is F distribution with nu 2 comma nu 1 degrees of freedom then it will be completely interchanged in the equation. So, this is a point that you must keep in mind alright let me now present to you formally some basic properties of the F distribution as you now see on the screen the F distribution is a continuous distribution ranging from 0 to plus infinity as is evident from its equation. Number two the curve of the F distribution is positively skewed but it is important to note that as the degrees of freedom nu 1 and nu 2 become large the F distribution approaches the normal distribution. The third property students is that for nu 2 greater than 2 the mean of the F distribution is nu 2 divided by nu 2 minus 2 and this quantity is obviously greater than 1. Similarly, for nu 2 greater than 4 the variance of the F distribution is 2 times nu 2 square into nu 1 plus nu 2 minus 2 and this whole quantity divided by nu 1 into nu 2 minus 2 whole square into nu 2 minus 4. Obviously the square root of this quantity will give us the standard deviation of the F distribution. The fifth property is that for nu 1 and nu 2 both greater than 2 the F distribution is unimodal and the mode of the distribution is at nu 2 into nu 1 minus 2 over nu 1 into nu 2 plus 2 and this quantity is always less than 1. Students, you have just seen that the mean is greater than 1 and the mode is less than 1. So, we were saying that the F distribution is positively skewed. So, it should not be skewed positively. Obviously, if it is positively skewed the mode will be to the left and the mean will be slightly towards the right. So, these properties are according to what we would expect. What is the next property? As you now see on the screen, if X has an F distribution with nu 1 and nu 2 degrees of freedom then the reciprocal 1 over X has an F distribution with nu 2 and nu 1 degrees of freedom. This is a very interesting property students and it is also very useful in certain ways. We are saying that our original variable X that is following the F distribution with nu 1 comma nu 2 degrees of freedom. Now, if we transform or introduce this new variable as we have said that let Y be equal to 1 over X. So, this new variable Y or 1 over X, this new variable distribution that is also an F distribution, but it is that one which has nu 2 comma nu 1 degrees of freedom which I said earlier that nu 1 nu 2 is different for example, if nu 1 is equal to 7 and nu 2 is equal to 9. We are saying that our original X variable that is following the F distribution with 7 comma 9 degrees of freedom. Now, this new variable is 1 over X that will follow the F distribution, but another F distribution that one which has 9 comma 7 degrees of freedom. Having discussed the basic properties of the F distribution, I think I should now share with you the table of areas for the F distribution. Students, in this case may it is a slightly more complicated situation not really, but it is very interesting. Now, in the top row we will be writing the various values of nu 1 and in the first column we will be writing the various values of nu 2. We have different tables, separate tables for 1 percent area on the right tail and for 2 and a half percent area on the right tail and for 5 percent area on the right tail. So, as you now see on the screen the table which is pertaining to 5 percent area on the right tail is as you see in the top row the values of nu 1 are 1, 2, 3, 4, 5, 6, 8, 12, 24 and infinity and in the first column we have various values of nu 2 in a similar manner. If you look at the very first cell we have the number 161.4 and students this is that value on the x axis under the F distribution to the right of wage the area is 5 percent and which distribution are we talking about that F distribution which has 1 comma 1 degrees of freedom. Similarly, if you look at the fourth entry in the row corresponding to nu 2 equal to 1 you find that the number is 224.6 and this is that value of x under the F distribution having 4 comma 1 degrees of freedom to the right of wage the area under the F curve is 5 percent. In a very similar way we have the table for 2 and half percent area on the right tail and also a table where 1 percent area on the right tail. Having discussed the basic definition the basic properties and also the table of the F distribution students I think it is about time that we begin the discussion of the role of this particular distribution in statistical inference. As I mentioned earlier this distribution enables us to talk about 2 population variances and the condition is that the 2 populations whose variances we are wanting to compare the 2 populations should be normally distributed. As you now see on the slide let 2 independent random samples of size n 1 and n 2 be taken from 2 normal populations with variances sigma 1 square and sigma 2 square and let small s 1 square and small s 2 square be the unbiased estimators of sigma 1 square and sigma 2 square respectively. Then it can be mathematically proved that the quantity s 1 square over sigma 1 square divided by s 2 square over sigma 2 square has an F distribution where n 1 minus 1 comma n 2 minus 1 degrees of freedom. Students I A is point to understand kane ki koshish karte hain dekhye aapke pass 2 normal populations. For example the heights of the men and the heights of the women in a particular country are normally distributed. So it is quite an interesting problem isn't it? So now what should we do? We have drawn 1 sample from the first population of size n 1 and we have drawn 1 sample from the other population of size n 2 and now the sample is here. So it is obvious that we can compute the sample variance of s 1 square unbiased. From here we have drawn sample from the women population and we can compute the variance of s 2 square unbiased. Now for a short while you assume that the first population's population variance is known and the other one is also known. So sigma 1 square and sigma 2 square are known. Just for a short while you assume that. Now we have 4 quantities s 1 square sigma 1 square s 2 square and sigma 2 square. Then of course I can compute this quantity that we just short while ago saw on the screen as you see once again s 1 square over sigma 1 square. This whole thing divided by s 2 square over sigma 2 square. The quantity you have computed for these 2 samples that you had drawn one from the first population one from the other. I want to say that again you assume that it is not just one sample that you are drawing from here and one from here. Think of all possible samples that you could have drawn from the first population of size n 1 and all possible samples that you could have drawn of size n 2 from the second population. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . N1 minus 1 comma N2 minus 1 degrees of freedom. Once again, I would like to say as I said in the earlier lecture, this is the non-mathematical way of doing it, but my point is that you should have some basic idea in your mind about what is going on. Now that we are clear about this, we have this knowledge that this particular statistic follows the F-distribution. After all, sampling distribution, the probability distribution of our statistic, the sampling distribution of this particular statistic is the F-distribution having N1 minus 1 comma N2 minus 1 degrees of freedom. You can also construct the acceptance region and the critical region if you are wanting to do hypothesis testing. Let us begin with interval estimation and as you now see on the screen, the confidence interval for the variance ratio sigma 1 square over sigma 2 square is given by s 1 square over s 2 square multiplied by 1 over F alpha by 2 N1 minus 1 comma N2 minus 1 degrees of freedom. This is the lower limit and the upper limit is given by s 1 square over s 2 square multiplied by F alpha by 2 N2 minus 1 comma N1 minus 1 degrees of freedom. Students, you have just seen the formula. Note two things from this question. The first thing is that it is the ratio of the two population variances that we are talking about. When we were talking about means, I did not construct confidence interval for mu 1 over mu 2. Similarly, when we were talking about proportions, we did not talk about p 1 over p 2. You remember that we constructed the confidence interval for mu 1 minus mu 2 and p 1 minus p 2. Now we are interested in the ratio. Students, instead of that, the underlying mathematics here, which I have said a little while ago, that s 1 square over sigma 1 square over s 2 square over sigma 2 square follows the F distribution. Since we do not want to go into mathematical derivations in this course, you can just see the pattern of the statistic, the F statistic. That tells us that it will be the confidence interval for sigma 1 square over sigma 2 square and not sigma 1 square minus sigma 2 square that we can construct on the basis of the F distribution. The second point to understand is that in the formula that I just presented, you saw that F alpha by 2 N 1 minus 1 comma N 2 minus 1 was written in the denominator and F alpha by 2 N 2 minus 1 comma N 1 minus 1 was written for the upper limit in the numerator. Obviously, there is no reason to be confused. That is the amount of area that we want to the right of the F value that we want to compute. Let me explain this point with the help of an example. A random sample of 12 salt water fish was taken and the girth of the fish was measured. The standard deviation small s 1 came out to be 2.3 inches. Similarly, another random sample of 10 fresh water fish was taken and their girth was measured and the standard deviation small s 2 came out to be 1.5 inches. Find a 90 percent confidence interval for the ratio between the two population variances sigma 1 square over sigma 2 square and in doing so assume that the populations of girth are normal. In order to solve this question of course, we will resort to the formula that is mathematically proved and that was presented a short while ago. The formula is lower limit s 1 square over s 2 square into 1 over F 0.05 N 1 minus 1 comma N 2 minus 1 and the upper limit is s 1 square over s 2 square into F 0.05 N 2 minus 1 comma N 1 minus 1. Students, you note that I said 0.05 in the place of alpha by 2 and why is that? Because we want 90 percent confidence. So, the derivation that we did earlier when we tried to construct the confidence interval for mu, you remember that the basic thing was that if you want 90 percent confidence then you will keep 90 percent area in the middle and on the right tail you will have 5 percent and on the left tail also you will have 5 percent. That is why in this problem alpha by 2 will be equal to 5 percent. Now, the rest that is fairly simple as you now see on the slide. We have s 1 square is equal to 2.3 square and that is 5.29. Similarly, s 2 square is 1.5 square and that is 2.25 N 1 minus 1 is 12 minus 1 and that is 11 and N 2 minus 1 is 10 minus 1 and that is 9. Now, in order to find the value of F 0.05 11 comma 9 we will need to consult the F distribution that one which has been constructed for 5 percent area in the right tail and consulting that distribution we find that F 0.05 11 comma 9 is equal to 3.1. Similarly, we also need to find F 0.05 9 comma 11 and looking at the same table we find that this value is equal to 2.9. Abhijo table manai present ki thi that is the abridged version of the larger table aur usme aapne shahid note kiya ho that in the top row we had 1, 2, 3, 4, 5, 6 likin badme the maam values nahi thi kuch values ki spacing ke baad we have some values. To aesi suratme you will be interpolating as you now see on the slide. In the top row after the number 8 we have the number 12 in this abridged table, but in the first column we do have numbers going continuously in the beginning of the table. So, that we have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 and so on. To aap agar hum 11 comma 9 dekhna chahathe. To iska matlab ke top row se 11 ke niche ja aana chahiye aur first column me 9 ke agaist hume jaana chahiye. Now agar hum 9 ke agaist dekhne to 8 ke under the value is 3.23 jab ke 12 ke under the value is 3.07. Iska matlab yeh hua ke 11 ke under jo value hogi agaist 9 that will be somewhere between 3.23 and 3.07. Agar hume 11 ke bajaye 10 ke under dekhna hota tap to hum simple average leh sakte thein dono number ka jo 12 aur 8 ke under hain. So, that we would have got 3.07 plus 3.23 over 2 and that would have been equal to 3.15. Lekin agar hume 11 ke under chahiye to iska matlab hai ke 11 to 10 aur 12 ke darmean lai kar rahe. To hum chahiye to 3.07 aur 3.15 ka arithmetic mean leh sakte hain and that will give us the value against 11. So, if we do that 3.07 plus 3.15 divided by 2 will give us 3.11. Students, this is the way we interpolate between two values. Aap yeh assume karte hain ke us range mein jo missing values hain. Wo equi spaced hain aur iss tara se hum average leh ke hum kar sakte hain. Mane aasan tari ke se karne ki koshish ki ke pehle aap 8 aur 12 ke middle mein aajaye 10 ke agaist value mil gayi aur phir aap 10 aur 12 ke middle mein aajaye to aapko 11 ke agaist value mil gayi. So, jais aapne dekhah iss tara karne se humari value aah rahi hai 3.11. Lekin jo value mein e abhi tori deh pehle present ki thi wo kya thi? As you see on the slide, I said that by consulting the F table, we obtain F is equal to 3.1. To aapne note kia students ke manage pehle jo present ki thi value, that was correct to 1 decimal. Agar aap interpolation value ko dekhain, 3.11 to agar aap usko bhi round karlein correct to 1 decimal, you get 3.1. Iska matlab yeh hoa ke agar aap bohot zyada rigorous interpolation value method mein nahi bhi jaana chaate, to aap andaaza lagaasakte hain ke what would be your F value. Correct to 1 decimal jo 12 ke under thi, 3.07 even that is equal to 3.1 aur chuke 11, 12 ke bohot nas dikh hain. Isliye hum kaisakte hain ke 3.1 hi hogi for 11 because it is close to 12 and correct to 1 decimal, it is correct. Substituting these values in the formula of the confidence interval, we obtain 5.29 over 2.25 into 1 over 3.1 as the lower limit and 5.29 over 2.25 into 2.9 as the upper limit of our 90 percent confidence interval and solving these quantities, the interval is 0.76 to 6.81. Students yeh jo answer yeh jo result hain mila iss silsle mein aap do baate note karein. Pehli baat hi hai ke this one is the interval for sigma 1 square over sigma 2 square, yaani the variance ratio. Agar hum interested ho sigma 1 over sigma 2 ke liye interval construct karne mein, yaani an interval for the ratio of the standard deviations of the two populations to hum kaisakte hain. Zahir hai ke we will be taking the square root of the lower limit as well as the upper limit and if we do that then as we now see on the slide, we obtain the 90 percent confidence interval for sigma 1 over sigma 2 as 0.87 to 2.61. Yeh to hui pehli baat ki agar aap standard deviations ke ratios mein interested hain to aap square root liye sath hai. Dhusri jo baat hai that is more important and interesting in a way. Dhe khe yeh jo abhi humne variance ratio ke liye confidence interval construct kiya uski lower limit ai 0.76, upper limit 6.81. Zah la khaur ki jaye ke 0.76 ka kya matlab hai? Iska matlab yehi hai na that sigma 1 square is less than sigma 2 square. Isi liye to yeh ratio 1 se less aaya hai. Yeh thi lower limit. Upper limit ke mutabek the variance of the first population sigma 1 square is 6.81 times the variance of the second population. Isli to unka ratio jo aaya hai that is 6.81. To students yeh kuch confusing nahi hai. Ke aapka jo confidence interval hai agar aap uski lower limit ke taraf chale jaye to aap yeh interpretation aapko mil rahi hai ke sigma 1 square is less than sigma 2 square. Aur agar aap upper limit ke taraf chale jaye to sigma 1 square seems to be much much greater than sigma 2 square. Yes, this is a kind of a problem. Of course, if you lower the level of confidence so that it is not 90 percent, maybe if you bring it down to 80 percent or maybe even less to fir bo narrow ho jaye gana chaya aapko discussion yad ho that the higher the level of confidence the wider your interval is. Lekin iss waak, me uski opposite baat kar rahin ho. Ke agar aap low karein level of confidence ko to aapka yeh narrow down ho jaye gana. It will become narrower to fir ho sakta hai ki us case me lower limit might come out to be 1 point something and upper limit might come out to be 5 point something. To kumse kumse yeh to fayda hoa na ke kha aap lower limit ko dekhin kha aap upper limit ko dekhin. Your number is greater than 1 implying that sigma 1 square is greater than sigma 2 square. Shaaar aap karee honge ki kuch boh jyada complicate hota jaar hai. Students, mera makhse diye hai ke aap iss baat ko realize karein that as I said once earlier mukhtilif mathematical formulae develop kiye gayin with a lot of logic, rationale and very useful formulae. Lekin, har method or har formulae ki koi na koi limitation bhi to ho sakta hai na. So iss traki thodi si problem kisi kisi wak aaj aati hai. Yani interpret karne me shaaar aapko thodi si dekhat ho jaye. But believe me, the more you practice and the more experience you have with this kind of work, it will be become quite easy for you after some time. Now that we have discussed the confidence interval for sigma 1 square over sigma 2 square, which enables us to compare the variability of one normally distributed population with that of another students, let us proceed to hypothesis testing regarding the population variances and I would like to explain this to you with the help of the example that we now see on the screen. In two series of holes to determine the number of plankton organisms inhabiting the waters of a lake, the following results were found series 1, 80, 96, 102, 77 and so on and series 2, 74, 122, 92, 81 and so on. In series 1, the holes were made in succession at the same place. In series 2, they were made in different parts scattered over the lake. Does there appear to be a greater variability between different places than between different times at the same place? In order to solve this problem, the first thing to note is that if x denotes the number of plankton organisms per hole, then for each of the two series, x can be assumed to be normally distributed. Now the hypothesis testing procedure is exactly the same as before, H naught sigma 1 square is greater than or equal to sigma 2 square and H alternative sigma 1 square is less than sigma 2 square. But students please note that H naught can also be written as sigma 2 square is less than or equal to sigma 1 square and H alternative that sigma 2 square is greater than sigma 1 square. Is there greater variability between different places as compared with different time periods at the same place? Now the way I have stated it, it may be more compatible with the statement. In series 1, that is for the same place, different times, but in series 2, that is for different places. We want to know that the population variance of series 2, sigma 2 square is that greater than sigma 1 square. Now note that we have placed this in alternative hypothesis, not in null, because in this statement, equal sign is not occurring. We always have to put that one in the null, which carries the equal sign. So, what we want to know, what we want to test is actually in this problem, it is in alternative. And the equal sign is the complement that is placed in the null. What is the next step? Of course, the level of significance and as you now see on the slide, we may set it at 0.05 as usual. But of course, if we were interested in lower risk of committing type 1 error, then we could also have set it at 0.01. Students, aap jo test statistic hai uspe khas tawar pe gaur ki jie. Maini kushte k pehle aap se kaha tha that if we are drawing a sample of size n 1 from the first normal population and a sample of size n 2 from the second one, then the statistic s 1 square over sigma 1 square over s 2 square over sigma 2 square follows the f distribution with n 1 minus 1 comma n 2 minus 1 degrees of freedom. Agar aap ye 1 or 2 kaapis mein interchange karne, can I not say that the statistic s 2 square over sigma 2 square whole divided by s 1 square over sigma 1 square follows the f distribution having n 2 minus 1 comma n 1 minus 1 degrees of freedom. If you accept that students, the next point is that according to the null hypothesis, sigma 2 square is less than or equal to sigma 1 square. Agar aap equal signa 2 square is equal to sigma 1 square agar aasah hai, toh phir aamaara statistic kya hoga. If sigma 1 square is equal to sigma 2 square, then in the statistic s 2 square over sigma 2 square whole divided by s 1 square over sigma 1 square, sigma 2 square will cancel out with sigma 1 square and we are left with s 2 square over s 1 square. And students, we can say that under the null hypothesis, this statistic follows the f distribution having n 2 minus 1 comma n 1 minus 1 degrees of freedom. Of course, I should not have to remind you that we always begin by assuming that h naught is true. Toh agar aasah aap aasah hai or unko equal money, then they cancel out and we are left with s 2 square over s 1 square. Students, the fourth step is the computation of my statistic and as you now see on the screen, for the first data set, we have 10 values 80, 96, 102 and so on. And therefore, upon taking the square of each value and adding the squares, sigma x 1 square comes out to be 9, 3, 2, 7, 6, whereas sigma x 1 itself is 960. Similarly, for the second series, the sum of the x 2 column is 657, whereas the sum of x 2 square is 63129. Now, small s 1 square is equal to 1 over n 1 minus 1 into sigma x 1 square minus sigma x 1 whole square over n 1 and small s 2 square is defined in a similar fashion. Hence, substituting all the relevant values, small s 1 square comes out to be 124, whereas small s 2 square is equal to 244.14. Hence, the computed value of our sum of our statistic which we can call f is equal to 1.97. Students, the next step is the critical region. Our alternative hypothesis was that sigma 2 square is greater than sigma 1 square or since it is greater than cosine. Therefore, we are going to look at the right tail area. We want the entire 5 percent area on the right of our critical value. Now, in this problem, the number of observations for the second series was 7 and the number of observations for the first series was 10. Therefore, n 2 minus 1 is equal to 7 minus 1 and n 1 minus 1 is equal to 10 minus 1. Therefore, students looking at the area table of the f distribution, the one that is for 5 percent area to the right of our value, we obtain as you now see on the slide, f 0.056 comma 9 is equal to 3.37. The last step, of course, is the conclusion and since our computed value 1.97 is less than 3.37, therefore, we do not reject H naught. Our data does not provide sufficient evidence to indicate that there is greater variability in the number of plankton organisms per whole between different places than between different times at the same place. Students, this is the method of testing sigma 1 square equal to sigma 2 square. Sometimes we might be conducting two-tailed test and sometimes a one-tailed test as you noticed in this example. Now, let us consider another example which is all the more interesting. As you now see on the slide, two methods of determining the moisture content of samples of canned corn have been proposed and both have been used to make determinations on proportions taken from each of 21 cans. Method 1 is easier to apply, but appears to be more variable than method 2. If the variability of method 1 were not more than 25 percent greater than that of method 2, then we would prefer method 1. The sample results are as follows. N1 is equal to N2 is equal to 21, X1 bar is 50, X2 bar is 53, sigma X1 minus X1 bar whole square is 720 and sigma X2 minus X2 bar whole square is equal to 340. Based on the above sample results, which method would you recommend? Students, it is quite an interesting process to have. Now, let us consider the first method which is easier but has more variability. But if the variability is less than 25 percent greater than that of method 1, then we will prefer method 1. So, how we will formulate this mathematically? As you now see on the slide, H0 will be that sigma 1 square is less than or equal to 1.25 times sigma 2 square whereas, H1 will be that sigma 1 square is greater than 1.25 times sigma 2 square. Now, concentrate on the number 1.25. 1.25 means 125 percent or iskai ahima is equal to 1.25. So, it means that 25 percent more as compared with the variability of method 2. Students, I would like to encourage you to work on this particular problem on your own. The second point is the test statistic. In this case, our test statistic is not going to be S1 square over S2 square or S2 square over S1 square. As you now see on the slide, according to what I stated earlier, it can be shown that if sigma 1 square over sigma 2 square is equal to k, then f equal to S1 square over S2 square multiplied by 1 over k follows the f distribution with n 1 minus 1 comma n 2 minus 1 degrees of freedom. Hence, in this particular example, our statistic will be f equal to S1 square over 1.25 S2 square. Students, apne dekha k according to the formulation of the null hypothesis, our statistic changes slightly and once again I would like to encourage you to work on the earlier formula that I presented and the one that I have presented now and see that the two will tally with each other. After that, carry out all the steps and conclude for yourselves what is the conclusion in this particular problem. In today's lecture, I discussed with you the f distribution and its role in statistical inference as far as the comparison of two population variances is concerned. In the next lecture, students, we will be discussing the role of the f distribution with reference to analysis of variance and experimental design. Until next time, my best wishes to you and Allah Hafiz.