 As Salaamu Alaikum, welcome to lecture number 39 of the course on statistics and probability. Students you will recall that in the last lecture I discussed with you hypothesis testing regarding the equality of two population means mu 1 and mu 2. After that we proceeded to hypothesis testing regarding P, the proportion of successes in a binomial population. In today's lecture I will begin where hypothesis testing regarding two population proportions. Let me explain this concept with the help of an example. As you now see on the screen a leading perfume company in a western country recently developed a new perfume which they plan to market under the name fragrance. A number of comparison tests indicate that fragrance has very good market potential. The sales department of the company wants to plan the strategy so as to reach and impress the largest possible segments of the buying public. One of the questions is whether the perfume is preferred by younger women or older women. A standard scent test will be used where each sampled woman is asked to sniff several perfumes one of which is fragrance and the woman will be asked to indicate which one she likes best. A total of 100 young women were selected at random and each was given the standard scent test. Twenty of the hundred women chose fragrance as the perfume that they liked best. Also two hundred older women were selected at random and each was given the same standard scent test of the two hundred older women one hundred preferred fragrance. Note the hypothesis that there is no difference between the proportions of younger and older women who prefer fragrance. Students, how do we approach this problem? The first thing is what is success and what is failure. It is obvious that we are interested in the fact that we found out that how many women like this particular perfume and how many older and younger women have any difference in this matter. This means that preferring fragrance is success and not preferring fragrance is failure. The second thing is that now we have two populations, a population of younger women and a population of older women, so who should be denoted by subscript one and who should be denoted by subscript two? As I talked to you in the last lecture with detail, this is on your own that who you want to keep one and who you want to keep two. The only difference is that whoever you keep one, then you adopt the mathematical procedure according to that. If we use one subscript for younger women in this problem, then what does P1 mean? The proportion of young women who prefer fragrance. Now what is the first step of our hypothesis testing procedure, the null hypothesis and the alternative? So null hypothesis, what will be the students? H naught P1 is equal to P2, that is the proportion of younger women who prefers this particular perfume is the same as the proportion of older women who prefers this perfume. What is the alternative? H alternative P1 is not equal to P2. The second step is the level of significance and if we want to set it as 5 percent, then of course our critical region will be Z is greater than or equal to 1.96 or Z is less than or equal to minus 1.96. You may say because of the alternative which I just mentioned, this is a two tail test and half of the 5 percent that is two and a half percent is going to be on the right tail and two and a half percent on the left tail or as per the Z values that we get, now you may remember that they are plus and minus 1.96. Students what is the test statistic in this particular problem? As you now see on the slide, Z is equal to P1 hat minus P2 hat minus 0 divided by the square root of Pc hat Qc hat multiplied by 1 over n1 plus 1 over n2. In this formula, Pc hat is equal to total number of successes in the two samples combined divided by the total number of observations in the two samples combined. In other words, Pc hat denotes the pooled proportion of successes in the two samples. Now you have said that we have got the chance to get confused again. Students, please come and pay attention to this. The thing is that our null hypothesis is that P1 is equal to P2. If this is true and you remember, we always begin by assuming that H0 is true and the entire process is based on the underlying mathematics that the null hypothesis is true. Then P1 is equal to P2 and I can say that both P1 and P2 they are equal to P where P is the unknown proportion of successes in either of the two populations. We are saying that we do not know the value of the proportion but both the populations are equal. If we are saying this, then this means that both P1 hat and P2 hat are estimating the same quantity, the common population proportion P. So, students, we think that we do not combine the information of these two samples, we pool them and we get an estimate of P from this combined information, because if we combine them, then we will have more information at our disposal from which we will do this estimation process. This is the concept of PC hat, C stands for combined and PC hat means the proportion that we compute from the combined sample information. Now, what is the formula? As I mentioned, it is the total number of successes in the two samples divided by the total number of observations in the two samples. As you now see on the screen, this formula can also be written as N1 P1 hat plus N2 P2 hat divided by N1 plus N2. And if we write the formula in this manner, then we realize that PC hat is the weighted mean of P1 hat and P2 hat. The formula for the weighted mean x bar w is w1 x1 plus w2 x2 over w1 plus w2. And the formula for PC hat is very similar in pattern to the formula of x bar w. It is N1 P1 hat plus N2 P2 hat over N1 plus N2. And we can see that the sample sizes N1 and N2 are acting as the weights. Now, students coming back to our test statistic, Z is equal to P1 hat minus P2 hat minus 0 over PC hat into QC hat multiplied by 1 over N1 plus 1 over N2. If you take PC hat into QC hat, then Z is equal to P1 hat minus P2 hat minus 0 over PC hat into QC hat over N1 plus PC hat into QC hat over N2. And you can see the similarity of this particular formula with the formula which we had in the last lecture, Z is equal to x1 bar minus x2 bar minus 0 over sigma 1 square over N1 plus sigma 2 square over N2. The formula that we had in that situation where we were testing H0 mu1 is equal to mu2. I would like to encourage you to work on this point on your own and to see that everything is very similar to what we did before. That we are dealing with the sampling distribution of P1 hat minus P2 hat. Or, when we standardize, the variable P1 hat minus P2 hat minus its mean which is equal to 0 if our analysis hypothesis is true. P1 minus P2 is equal to 0 divided by the standard deviation which is the square root of PC hat into QC hat over N1 plus PC hat into QC hat over N2. Yani, asal me to PQ hona jahiye thana, PQ over N1 plus PQ over N2. Lekin 2 ke P unknown hai. Isliye, we are going to replace it by its estimate PC hat. Now let us compute the value of Z so that we can proceed further in this problem. As you see on the screen, P1 hat comes out to be 20 over 100 because out of the 100 young women who were sampled, 20 preferred the perfume fragrance. Therefore, P1 hat is equal to 0.20. Similarly, P2 hat comes out to be 100 over 200 and that is equal to 0.50. Substituting these values in the formula for PC hat, we obtain PC hat is equal to 0.40. Therefore, QC hat is equal to 1 minus 0.40 and that is equal to 0.60. Substituting all these values in the formula for Z, we obtain Z is equal to minus 5.00. Z ki value to nikalahi minus 5.00 aur aapko yaad hoga ke kuch de pehle maine kaha kye because it is a two tail test. Therefore, the critical values are minus 1.96 and 1.96. Aap yeh ju value minus 5.00 i hai, obviously it is lying in the left tail of the distribution as you now see on the screen. And because of this, we can of course say that we will reject H naught. In other words, we conclude that the population proportions are not equal. Students zahe re ke agla sowal yeh ke agar yeh baraabar nahi hai, toh is it the younger women or is it the older women who prefer this brand more? Well, of course, we go back to our data and we find that the proportion is higher in the sample of older women than in the sample of younger women. Aur zahe re ke yehi waja hai ke humara jo answer aaya hai minus 5.00 that has the negative sign. Hum p1 hat minus p2 hat ki sampling distribution ki baat kar rahe thena and one denotes younger women. To agar answer negative hai to iska yeh matlab hai na, ke p1 hat is less than p2 hat. Yaani proportion of younger women is less than the proportion of older women who are preferring this particular perfume is less than the proportion of older women. All right, students, yeh pe aapko mai ek aur mathematical point aapke saath share karna chahati hoon. Yeh jo answer aaya it is very much in the tale of the sampling distribution. Deke, humari critical value thi minus 1.96 left side pe. Lekin humara jo answer hai that is minus 5, which is far out in the tale. Agar humara level of significance, 5% ki bhejai, 1% hota, then because it is a two-tail test our critical value on the left hand side would have been minus 2.575. Ya, minus 2.5 8. Aur aap note ki je yeh jo answer humara aaya hai minus 5 that is even further to the left of minus 2.58. Asi situation mein we say that our result is not only significant but highly significant. To repeat what I have said, students, baat yeh ho rahe hai, ke agar aapka null hypothesis, 5% level of significance pe reject ho, to aap kahate hai, ki mera jo statistic hai that is significant. It is large enough, if it is on the right side will say large enough and if it is on the left side perhaps we will say it is small enough to signify that I should reject H naught. Agar humara hypothesis, 5% ki bhejai, 1% pe reject ho jai, then we say that our result is highly significant. Yaani siro significant, significant hi nahi, baat ke highly significant. Why are we using this term? Isliye ke ab the risk of committing type 1 error has been reduced to as low a level as 1%. Sirif 1% chance of being wrong ke saath hum ye conclude kah rahe hai that we should reject H naught. To aise situation mein we say highly significant. Students fars ki jai, ke iss problem mein, z came out to be minus 2.20. Aap kya situation hai? De ke minus 2.20 is lying on the left tail of the sampling distribution and it is to the left of minus 1.96, baat it is to the right of minus 2.58. Iska matlab ye hua ke agar hum 5% level of significance rakh rahe hai, tabto hum H naught ko reject kar deenge. Lekin agar hum level of significance ko reduce kar ke, sirif 1% kar deen, then we are not able to reject H naught. Kyuke 1% ke hi saab se to wo andar aage hai na in the acceptance region. To pere iss ka matlab kya bana? Ye toh kuch confusing baat lag rahe. Baat ye nikhli that if we are willing to tolerate 5% chance of committing type 1 error, then we can reject H naught. But if we are not willing to allow more than 1% chance of being wrong in rejecting H naught, then we cannot reject H naught with this sample information. Alright, let us consolidate some of these ideas with the help of another example. As you now see on the slide, a candidate for mayor in a large city believes that he yields to at least 10% more of the educated voters than the uneducated voters. He hires the services of a poll taking organization and they find that 62 of 100 educated voters interviewed support the candidate and 69 of 150 uneducated voters support him. At the 5% level of significance, should this hypothesis be accepted or rejected, students aaye iss ko usi tarah si step by step tackle karte hain, jistra humne pichle wale probleme kya. The first thing is what is success? The preference for this particular candidate is success agar wo usko pusand kar rahe, to that is success agar wo usko nahi pusand kar rahe to be the mayor of the city that is failure. The second step is the subscript kya hum one subscript kis ke liye rakhe or two kis ke liye rakhe. Agar hum one subscript educated voters ke liye rakhe, then what will be our null hypothesis and what will be the alternative? As you now see on the slide, in this case we will write H naught p1 minus p2 is greater than or equal to 0.10 and H1 p1 minus p2 is less than 0.10. Students aaye jo hypothesis form kya ispe gaur kiji. This is quite different from the hypothesis that we had in the last example or the one that we had in the case of testing mu1 equal to mu2. Yaha pe hum ye nahi kya rahe null hypothesis me that the proportion of educated voters is the same as the proportion of the uneducated voters, the ones who prefer this particular candidate. Hum ye kya rahe nahi hypothesis kim utabe ke ye jo dono proportions hain me jo difference hain that is at least 10 percent. So, if you look at the slide once again you find that what we have written is in accordance with what we are trying to hypothesize. H naught p1 minus p2 that is the proportion of educated voters favoring this candidate minus the proportion of uneducated voters favoring this candidate. This difference is greater than or equal to 0.10 that is 10 percent aur hum bilku ye hi kya nahi cha rahe theke bo jo difference hain wo kamas kam 10 percent hain. All right now that the null is clear obviously the alternative hypothesis will be that p1 minus p2 is less than 0.10. The second step is the level of significance and as I have said earlier normally we take it to be equal to 0.05. The third step is the test statistic students isko gaur se dekhye this time we are writing z is equal to p1 hat minus p2 hat minus p1 minus p2 over the square root of p1 hat q1 hat over n1 plus p2 hat q2 hat over n2. All right ab sara gaur kiche ke jo pehle problem tha aur ab jo hain usme to kaafi fark hain. Students note kere ke pichle problem mein according to the null hypothesis p1 was equal to p2 yani p1 minus p2 was equal to 0. Isliye uske numerator mein humne p1 hat minus p2 hat minus p1 hat minus p2 hat wo pehle kya hoega. Z is equal to 0.10. Now we are saying according to the null agar hum uske andar jo equal sign hain uske khisaab se baat karein to hum karein ke p1 minus p2 is equal to 0.10. To phir humara jo formula hain the standardized version of p1 hat minus p2 hat wo pehle kya hoega. Z is equal to p1 hat minus p2 hat minus p1 minus p2 jiske value according to the null hypothesis 0.10 hain. So that is why we write 0.10 and not 0. Students numerator ki baat toh dae hoge aapne dekhha ke iske saath hi aapka denominator bhi change hoge hain. Yani last time to aap karein hain ke square root of pc hat qc hat over n1 plus pc hat qc hat over n2 aur agar usko aap bahar nikaal ne to phir wo bracket ke andar 1 over n1 plus 1 over n2 aajat hain. Is waakth hum pc hat compute hi nahi karna chaare. Isliye ke pc hat toh humne tab compute karna tha na if both p1 hat and p2 hat they were estimating the same quantity the common population proportion p. Is waakth we are not saying that the two populations have the same proportion of voters who favour this particular candidate. Aap toh humara null hypothesis yeh kaha hain ke in proportions mein kamas kam 10% ka farkh hain. To misaal ke toh ar pe agar educated voters messe 70% usko pasankarthe hain aur uneducated voters messe 50% usse pasankarthe hain. Toh yeh farkh jo hain 20% ka yeh 10% se zyada hi hain na toh aise situation mein yeh toh nahi kaisakthe na ke p1 is equal to p2 proportion of people who are favouring him in the first population is the same as the proportion of people who are favouring him in the second population. Is sari kahaani ka lubbe lubab yeh hai students ke we are not going to compute pc hat in this particular situation. p1 hat is estimating p1 and p2 hat is estimating p2. Le haza hum siddha siddha p1 aur p2 ki jaga pe p1 hat aur p2 hat hi substitute karenge in our formula of z. So, as you once again see on the screen our test statistic in this particular problem is given by z is equal to p1 hat minus p2 hat minus 0.10. This value we get according to the null hypothesis and this whole quantity divided by the square root of p1 hat q1 hat over n1 plus p2 hat q2 hat over n2. The next step is to compute the value of z. Now, p1 hat is equal to 62 over 100 which is 0.62 so that q1 hat is equal to 1 minus 0.62 and that is 0.38. Similarly, p2 hat is equal to 0.46 so that q2 hat is equal to 0.54. Substituting these values in the formula for z, z comes out to be 0.95. Students, what is the fifth step? Obviously, the critical region. Is this a two tail test? No. What was the null hypothesis? p1 minus p2 is greater than or equal to 0.10. But you remember that it is the alternative to determine whether it is a right tail test or a left tail test. So, in this problem what is the alternative? p1 minus p2 is less than 0.10. This means that we are going to talk about the left tail of the sampling distribution. Our area of level of significance that has to lie on the left tail of the sampling distribution or level of significance 0.05. So, what is the critical value? Minus 1.645. Now, what is the last step? The conclusion? What is our z value students? You have seen that our z value is 0.95. So, it is not even negative. It is very much in the acceptance region. Therefore, we accept H0 and what was H0? That p1 minus p2 is greater than or equal to 0.10. What does this mean? The mayor thought that he appeals to at least 10% more of the educated voters as compared with the uneducated ones that belief is supported by this particular data set. That means, no evidence has been found from this data that we contradict the case belief. Students, we have had a detailed discussion regarding testing p1 equal to p2 or p1 minus p2 is equal to some proportion. But in this discussion, why are we using the z statistic to conduct these tests? The reason is I hope obvious that if the sample size is large and neither p nor q is very close to 0, then the binomial distribution tends to the normal distribution. I will encourage you to study these examples again and see for yourself whether you would like to decide whether or not n is large in these problems. If there are two populations, then you are talking about both n1 and n2. Students, it is now time to start another very interesting topic and that is tests based on the t distribution. We have conducted a lot of tests and estimated a lot of intervals but all of them were based on the standard normal distribution. There are situations where you will be testing about mu or about mu1 minus mu2 but the z statistic is no longer applicable. And that is the situation when we will be talking about the t distribution, a very important distribution in statistical theory. So, before we go to those tests, of course, first of all I should introduce the t distribution to you in a formal manner. As you now see on the screen, the t distribution is well known as the student's t distribution. Student is the pseudonym of the person who derived this particular distribution. The mathematical equation of the t distribution is f of x is equal to 1 over square root of nu multiplied by the beta function beta half comma nu by 2 and this whole expression multiplied by 1 plus x square over nu whole raised to minus nu plus 1 over 2 and this function is valid from minus infinity to plus infinity. Students, this is a very complicated equation and you will surely be able to solve this problem if you try to get confused. The reason why I gave you this expression is to consolidate in your mind the fact that for any curve that you can draw on the graph paper, for any mathematical curve that you may draw there is of course a mathematical equation. But students, as far as you are concerned, all that I would like you to note is that in this equation there is only one quantity that can be called the parameter of the t distribution and that is nu. Nu is a Greek letter and we use it for the one lone parameter of the t distribution. What I would like to concentrate on are the properties of the t distribution and as you now see on the screen the first property is that the t distribution is bell shaped and symmetric about the value t equal to 0 and it ranges from minus infinity to plus infinity. Students, you have noted that I have said that it is bell shaped, it is symmetric about 0 and it ranges from minus infinity to plus infinity. Student, the difference is that generally speaking the t distribution is wider than the standard normal distribution. And as you now see on the screen the parameter nu is that quantity that determines the shape of the t distribution. Students, nu is called the degrees of freedom of the t distribution and we find that as the degrees of freedom increase the t distribution becomes narrower and narrower. The parameter of the t distribution is called degrees of freedom. Now, a very interesting term is degrees of freedom up chi square distribution or f distribution. So, I will be explaining the reason why these parameters are called degrees of freedom. I will be explaining this later. At the moment I just want you to concentrate on this one point that it is that one quantity that occurs in the equation of the t distribution or jis ke change hone par hamari distribution change hoti hai, iss tara se that as the degrees of freedom increase the distribution gets narrower and narrower and narrower. But students, there is a limit to how narrow it can become. It is not possible. What happens is that as nu tends to infinity the t distribution tends to coincide with the standard normal distribution. Standard normal distribution, your t distribution tight hote hote hote, us ke saath ko inside kar sakti hai usse aur zyada tight nahi hosakti. Let us look at some other properties of the t distribution. As you now see on the screen the t distribution has a mean of 0 when nu is greater than or equal to 2. The mean does not exist when nu is equal to 1. Students, ye chu baat maini kahi that the mean does not exist when nu is equal to 1. Ye zyada mathematical point hai agar aap uski equation ko dekhain aur phir uska mean compute karein. To aap dekhainge ke agar nu ki value 1 ho to aap ka jo integral hai that will not converge. I would not want you to get worried about this point. Aap isska jo pehla hissa hai uspe gaur ki je aur iss ko yu samizli je ke generally speaking the mean of the t distribution is 0. To ye to bahati simple si baat hai jab hum karein hai that it is bell shaped and it is symmetric about the value 0 to phir zahir hai. Ye to hum bahot pehle pachu kahin ki for any symmetric curve the mean is in the middle aur iss case mein agar 0 ke around symmetry hume mil rahe hai to phir saaf zahir hai that the mean will be 0. Next let us talk about the median of the t distribution. Isn't it obvious students that the median will also be 0? Isli ye hum bahot pehle ye baat bhi pachu kahin that for any hump shaped symmetric distribution the mean median and also the mode they are exactly equal. This brings me to the next property of the t distribution and that is that the mode is also equal to 0. Let us talk about the spread of the distribution as you now see on the slide. The variance of the t distribution is given by sigma square is equal to nu over nu minus 2 and this formula is valid for all values of nu greater than 2. Now if you pay attention to this formula students it is obvious that the numerator is greater than the denominator implying that sigma square is greater than 1. In other words it is greater than the variance of the standard normal distribution. After all do you not remember that the standard deviation and the variance of the standard normal distribution is exactly equal to 1? Students aapne dekhha ke variance ka jo formula abhi main aapke saabne present kia. That is confirming what I said earlier agar t distribution ka variance once ye zyada hai. Tis ka matlab aake standard deviation once ye zyada hai. Yani standard normal distribution ke jo standard deviation hai usse zyada. And that means that the t distribution is more spread out than the standard normal distribution generally speaking. Now that we have discussed some of the basic properties of the t distribution students let us concentrate on its application in statistical inference. Ye sirf testing meini balke interval estimation mein bhi humare kaam aati hai in one important situation. Aur jo situation aap me aapke saamne present karne lagi hoon aap se mai request karungi. Kyus pe gaur ki jhe. There are three important points aur agar ye tino baate hoon then we should not apply the z statistic. Rather we need to apply the t statistic because it has been mathematically proved that in this particular situation t is the distribution and t is the statistic that we should use. As you now see on the screen if the population from which the sample is being drawn is normally distributed the population variance sigma square is unknown and the sample size is small that is less than 30. Then the statistic x bar minus mu over s over square root of n where s is the square root of sigma x minus x bar whole square divided by n minus 1. This statistic follows the t distribution having n minus 1 degrees of freedom. Students aaye is point ko understand karne ki koshish karte. Sabse pehle ye tasavur ki jaye that you have a normally distributed population. Ye baat to bohat tafa ho chukhi hai ki bohat se phenomena approximately normally distributed hote hain muslin, height, weight and so many other phenomena. To aap aap pehle baat to ye hoge hi ki hain maare pahas ek normally distributed population hain. Aap aap isme se small size ka sample randomly draw karne. May be the size is 15, 20, it's quite a small size. Students ye sample me ne draw kar liya to zahir hai ki me iska mean aur iska variance compute kar sakti hoon. I will get x bar and s square for this sample. Agar me small s square nikaal na chaati hoon jo ke me is case me nikaal na hi chaati hoon. Of course, the denominator of the formula will have n minus 1 rather than n. To ye jo ek sample draw kiya tha na uske liya x bar bhi mil kya or s square bhi mil kya. Aap tasavur ki jaye that you are not drawing just one sample from this population. But you are drawing all possible samples of this particular size from this normal population. To students ye to karodha samples aajainge na. Aap ne dekhah hi tha pehle lectures me ke jab hum without replacement ki baate bhi karte thi. To hamara jo result aata tha ho itna bada number hota tha aksa rakaat. Aur agar vidh replacement ki baate ho to usse bhi kahin zyada. To aap tasavur kar sakte hain ki agar ek large population hai aur usme se ek small size ka sample aap lein. To agar ek ki bhajaye all possible leine hoon to karodha samples aajainge. And then for every one of those samples you will have nx bar and ns square. Aap hi ek karodha x bars aur s squares aapko available hain. To aap ye quantity jo abhi aapne screen pe dekhhi ye quantity bhi to karodha marthaba compute kar sakte hain na. Which quantity? X bar minus mu over s over square root of n. Aap dekhi is sab kuch hai aapke paas hai. X bar s n the sample size of course you know. Lekin mu is not there to phil vaakth aap iss hawale se sochein ki agar ham testing kar rahe hain. And we always begin by assuming that H naught is true. To phir mu is equal to something. Wo hamne H naught ke tahit kaha hoogan hain. That mu is equal to something. To wo jo number hain wo aap usme daal ne. To phir ye quantity compute ho gaye na. For one sample and for those millions of samples that you have drawn. Millions of such quantities are now available to you. Ham jo kah rahe hain wo ye hai ki ye jo ambar hain hain hain hain hain hain. Is khusam ki quantities ka. Agar aap inki frequency distribution banaal hain. Probability distribution construct kar lein. And then you graph this distribution students. You will get a curve which is the t distribution having n minus 1 degrees of freedom. Ye jistara se main aap ke saati saari baat ki. Ye ek non-mathematical tariqa hain isko explain karneka. But the point in my mind is that I would like to have you have a basic idea of what we are talking about. Of course agar aap isko mathematical tariqa se hum baat karein. So that may be a bit too advanced. So the gist of the whole discussion is that we are now dealing with a sampling distribution. Which is the t distribution with n minus 1 degrees of freedom. Let us apply this to interval estimation. And for this I would like you to have a look at the example that we now have on the screen. The masses in grams of 13 ball bearings selected at random from a batch are 21.4, 23.1 and so on. Calculate a 95% confidence interval for the mean mass of the population of ball bearings supposed to be normal. From which these 13 ball bearings were drawn. Students aapne dekha ke wo tin sharait din ka zikar maine pehle kia. The population of the masses of the ball bearings is normally distributed. The variance of this population is unknown and the sample size is small. It is only 13. Hence our sampling distribution is not z but t. And according to the same logic as we had in the case of z our confidence interval this time is as you now see on the screen. X bar plus minus t alpha by 2 at n minus 1 degrees of freedom multiplied by s over square root of n. Aapko yaad hoga ke jab z statistic tha uswak thamara formula tha. X bar plus minus z alpha by 2 sigma over square root of n. Aur agar sigma available nahi tha to s hum raktetethe uski jaga pe. So you can see the similarity. Aab sabaal yeh ke t alpha by 2 jo hai at n minus 1 degrees of freedom uski value kaisi milegi? Students just as we have a table of areas under the standard normal distribution. We also have a table of areas under the t distribution. And as you now see on the screen the values in the top row are the areas to the right of the t values that we would like to determine. And the values in the first column are denoting the degrees of freedom. In this example because n is equal to 13, therefore n minus 1 is equal to 12. And if we want 95 percent confidence then we will have to look under 0.025. And when we do so and look under 0.025 against 12 degrees of freedom our value comes out to be 2.17. Now that we have found the value of t of course we have to find x bar and small s. And from the values that we have x bar comes out to be 24.21 and s is equal to 1.77. Substituting these values in the formula of the confidence interval. We obtain 24.21 plus minus 2.179 multiplied by 1.77 divided by the square root of 13. Solving this expression the 95 percent confidence interval for mu comes out to be 23.14 to 25.28. Students, you saw that on the basis of 95 percent confidence we can say that the mean mass of those ball bearings lies somewhere between 23.14 grams and 25.28 grams. And you see how interesting it is that you drew such a small sample only size 13 and you are able to make an estimation for the mean of the entire population which consists of thousands upon thousands of ball bearings. In today's lecture students we discussed hypothesis testing regarding p1 minus p2 and after that we began the discussion of statistical inference based on the t distribution. I would like to encourage you to attempt a few questions on these concepts and next time we will be discussing the t distribution in further detail. Best of luck and Allah Hafiz.