 Fine. Let us continue. We left at this particular point probability of x bar-z alpha by 2 sigma by root n less than or equal to mu less than or equal to x bar-z alpha by 2 sigma by root n that is equal to 1-alpha. A few things to note. We are using capital X here and this is small z representing the upper alpha by 2 percentage points of the standard normal distribution. Sigma is assumed to be known. n is the sample size. This is the unknown population parameter mu and here we are having 1-alpha. That alpha divided by 2 is used in the subscript for z. Hence, if x bar is the sample mean small x bar is the sample mean of a random sample of size n obtained from a normal population with known variance sigma squared then the 100 into 1-alpha person confidence interval on mu is given by x bar-z alpha by 2 sigma by root n less than or equal to mu less than or equal to x bar plus z alpha by 2 sigma by root n. So this is based on the earlier definition. So the values are so identified on either side of mu such that the probability that mu lies between these 2 values is 1-alpha. So what are those values? And after identifying those values we can project them as the lower limit and the upper limit for mu and that is what we have exactly done here. We will now look at z alpha by 2. z alpha by 2 is the upper 100 into alpha by 2 percentage point of the standard normal distribution. You have the normal distribution sketched here. What is this? This is the standard normal distribution with mean 0 and standard deviation 1. And the upper 100 into alpha by 2 percentage points are shown here and also here. This is minus z alpha by 2. We are defining with respect to z alpha by 2 and if we choose alpha as 0.05 then 1-alpha will be 0.95. So that would represent the 95% confidence interval and alpha by 2 will then become 0.025. So that 0.025 is the area under the curve beyond the point z alpha by 2. So the probability of the standard normal variable taking a value greater than z alpha by 2 is 0.025 when alpha is 0.05. So that is what is done here and according to symmetry if you have located z alpha by 2 here for this particular case of alpha 0.05, z alpha by 2 takes the value of 1.96. So probability of the standard normal variable greater than 1.96 is 0.025. Similarly here you are having minus 1.96 and that represents minus z alpha by 2 and the probability of the random variable z taking a value less than minus 1.96 is 0.025. This is a random variable x bar the point estimator is also a random variable and the confidence interval we have constructed is bounded by 2 random variables. We can see here x bar minus z alpha by 2 sigma by root n less than or equal to mu less than or equal to x bar plus z alpha by 2 sigma by root n. So this came from the definition of the confidence interval and x bar is a random variable and x bar minus z alpha by 2 sigma by root n is also a random variable. x bar plus z alpha by 2 sigma by root n is also a random variable. So the confidence interval is based on 2 random variables. Hence the confidence interval is also a random variable and it can theoretically have different bounds. To demonstrate what is meant by a confidence interval we will take a small example. Probability of x less than or equal to x bar we will find after assuming that mu star is equal to 50. I am putting star because it is assumed the mu is assumed. We really do not know the exact value of the population mean but for the purpose of demonstration let us see what happens if mu is 50. Sigma is equal to 15 standard deviation of the normal distribution and n is equal to 16 which is the sample size. So the probabilities of various sample means x bar belonging to the population of mean 50 may be computed as shown in the table below. What we are doing is we are going to take different sample means and we are going to see what is the probability of choosing that sample mean value or lower from a distribution of sample means centered around 50. So given mu assumed mu star is equal to 50 what is the probability of a random variable x bar taking the specified value of x bar or lower. So we are going to find that probability using the assumed mu star and the known sigma by root n. So we are also going to fix the sample size and for example we have n is equal to 16. We know z alpha by 2 from the previous graph corresponding to the alpha value of 0.05 z alpha by 2 had a value of 1.96. So we take z alpha by 2 as 1.96 for the 95% confidence interval. This becomes 1.96 sigma we assume or take that value as 15. We have taken a sample of size 16. So if you plug in these numbers in that formula we will get z alpha by 2 sigma by root n is equal to 7.35. So this number you please remember right. This is a very interesting table. You have sample means here. You can have infinite number of sample means because it is a continuous distribution. You can have 40, 40.001, 40.001 etc. So there can be infinite sample means. For the purpose of demonstration I have chosen increments of 1 up to 42. Then you have 42.65 and then increments of 2 from 44, 44, 46 so on to 56. Then you have 57.35 and then 58. We calculate z. z is x bar-mu by sigma by root n and that comes as 40-50 divided by 15 by 4. So I compute these z values because I have to convert it into the standard normal variable form. So these are the values taken by the standard normal variable. So I have values from-2.67 to 2.13 and then I am finding the probability that z less than or equal to z. What is the probability that the sample mean of 40 or lower could have come from the sampling distribution with mean 50 and standard deviation 15 by 4. 15 by 4 is 3.75. So we are having a probability distribution centered around 50 and having a standard deviation of 3.75. So what is the probability of picking up a sample mean of 40 or lower from that distribution and that comes as 0.0038. So that is a very small chance of picking up a random sample with mean value of 40 from a sampling distribution of means centered around 50 and standard deviation of 3.75. Then you have 41 the probability increases slightly and when you come to 42.65 then the z value is-1.96 and the probability is 0.025 okay and then the probability values keep increasing and then we have 57.35 which is 50 plus 7.35 which is mu plus z alpha by 2 sigma by root 10. We saw that value z alpha by 2 sigma by root 10 as 7.35. So we have mu star plus z alpha by 2 sigma by root 10 as 57.35. Here you have 50-7.35 or mu star-z alpha by 2 sigma by root 10 and that comes as 42.65. So at 42.65 the probability of z less than or equal to small z is 0.025 and the probability of z less than or equal to z at a sample mean of 57.35 is 0.975. So you are having 0.025 here for 42.65 and 57.35 you are having 0.975. The probability difference is 0.975-0.025 which is 0.95. In other words we can say what is the probability of picking up a sample mean between 57.35 and 42.65 when the sampling distribution is centered around 50 with the standard deviation of 3.75. So that probability is 0.975-0.025 which is 0.95. So I have represented the information given in the table graphically. Here we have 42.65. Here we have 57.35. The probability of the random sample having a value less than or equal to 42.65 when the distribution is centered around 50 and standard deviation of 3.75 is 0.025 here and this entire probability for x bar less than 57.35 is 0.975. So the region between 42.65 and 57.35 will have an area under the curve of 0.95 or the probability of x bar lying between these 2 values will be 0.95. That is what I have done here. I have shaded that portion and that comes as 0.95. This plot and the previous one were generated using Minitab. So the plots generated with Minitab show that the sample means with values 42.65 and 57.35 or the 95% confidence lower and upper bounds on the population mean mu of 50. The probability is mu lying between 42.65 and 57.35 is 0.95. During the actual sampling process we get only one sample and hence we get only one sample mean x bar. So how do you know that your samples confidence interval bounds mu? We have assumed mu star equals 50 for illustration purposes only. Mu also may not be 50. It may be some other value. We are working on a confidence interval length around mu. What is that confidence interval length? That is what we are going to see. The confidence interval length is defined as the absolute value of x bar-mu. The distance between x bar and mu is termed as the length and we are taking the positive value only. So we put a modulus on x bar-mu. If x bar is greater than mu it is positive. If x bar is less than mu, x bar-mu is negative but modulus of x bar-mu is positive only. So the confidence interval length x bar-mu is less than or equal to z alpha by 2 sigma by root 10. So what we can do is we can construct an infinite number of confidence intervals. So we can generate a large number of confidence intervals. Okay, continuing. If the population mean is indeed 50 then among the different possible confidence intervals generated based on the randomly drawn sample means, those confidence intervals overlapping the region between 42.65 and 57.35 will constitute 95% of the selected confidence intervals, okay. So we can generate infinite number of 95% confidence intervals based on the selected or chosen random samples. So let us say we have a large number of confidence intervals. If the population mean was 50, I am putting a big if there. Then among the so many different confidence intervals we have, the confidence intervals that overlap the region between 42.65 and 57.35 will constitute 95% of them, 95% of the generated confidence intervals and of course the confidence intervals overlapping the region between 42.65 and 57.35 will encompass the population mean mu of 50. Let us see what this means. I will put a table here. So here we have different sample means. We can choose infinite number of them but for purpose of demonstration I have chosen only a few typical sample means starting from 40 going up to 58. Then x bar-mu is equal to z alpha by 2 sigma by root n which we know as 7.35. So we can generate our confidence intervals with the upper limits given by this column and lower limits given by this column, okay. So each represents a confidence interval and the upper limit for a sample mean or the confidence intervals upper limit based on a sample mean of 40 is 47.35 and the lower limit is 32.65 and whether the population mean of 50 is included, no. Similarly 41 will not include the population mean of 50 because its upper limit is 48.35 and lower limit is 33.65. 42 again will be having an upper limit close to 50 but not quite 50 yet. So 49.35 and 34.65 it is not having the population mean of 50 but if you take a sample mean of 42.65 the upper limit is just touching 50 and the lower limit is 35.3 and it is including the population mean. Similarly you go all the way up to 57.35 and the lower limit will just touch 50 and will include the population mean but any value greater than 57.35 will not include the population mean mu. So out of so many confidence intervals the confidence intervals which are present between 42.65 and 57.35 will encompass the population mean of 50 and the percentage of such confidence intervals falling between 42.65 and 57.35 will be 95% of all the chosen confidence intervals from the randomly chosen samples of the population. So we have to choose a large number of random samples from the population and calculate the random sample means using them we can construct the confidence intervals. So we have a large number of confidence intervals and if your confidence limit specification if our confidence interval specification is 95% then 95% of the generated confidence intervals will encompass or surround the population mean mu of 50. So after constructing the confidence interval of a given width based on a predetermined value of alpha we claim with 100 into 1-alpha percentage confidence that the interval contains the elusive population parameter mu. Alpha need not be fixed at 0.05 it may also take 0.025, 0.01 etc. If alpha is 0.01 we are constructing a 99% confidence interval. So you can see that if the value of alpha decreases our confidence level increases. When the alpha value was 0.05 we had a 95% confidence interval. If alpha is 0.01 we have a 99% confidence interval. So when alpha value decreases our confidence level increases. So it looks like we have to choose a low value of alpha. As alpha decreases 100 into 1-alpha increases. But what about precision? Precision means accuracy. What is the precision associated with the estimate for the chosen level of confidence? What is the relation between precision and the confidence interval? So that is what we are going to look at now. Broader or wider the interval we enjoy a higher confidence when stating that the interval does have the population parameter mu. But as the interval becomes wider the precision suffers. Our interval estimates become less precise when they become wider. Suppose you do experiments and the very frequently you are told to put the error bars. If the variability around the experimental points are quite high, the error bars are quite wide then we really do not know the precise value of the variable we are measuring. So it is important for us to have an estimate of the population mean mu in this case that can be claimed with both high confidence and high precision. The interval estimate we are proposing should be having a high level of confidence and also a high level of precision. But we saw that if we increase the level of confidence then the precision becomes less. So how to have both? It is important for us to have precise estimates of the population parameter so that our decisions can be made correctly. How do we plan the random sampling so that the desired confidence as well as desired precision are both achieved. So look at the equation the basic equation probability of x bar-z alpha by 2 sigma by root n less than or equal to mu less than or equal to x bar-z alpha by 2 sigma by root n and that is equal to 1-alpha. We want a high value of 1-alpha but the same time the interval bound on mu should be narrow. What it means is we want to have a high value of 1-alpha but the same time the interval should be narrow. In other words the lower limit should be approaching the upper limit so that the interval is narrow and our precision improves. On the other hand we also need to have a high value of 1-alpha so that the probability value on the confidence level increase. If you choose a lower alpha value it leads to a higher value of z alpha by 2 and hence the lower and upper bounds will be moving away from each other. If you reduce the value of alpha the z alpha by 2 value based on the standard normal curve will increase. Reduce alpha z alpha by 2 will increase. When z alpha by 2 increases-z alpha by 2 will decrease. So the interval between z alpha by 2 and 1-z alpha by 2 will widen. How to control the spreading of the interval when alpha value decreases? We have another handle here that is the sample size n which is very well in our control. We may require a bit more investment in having a slightly larger sample size but that is going to pay us back in terms of increased precision. So the sample size has to be so adjusted that we get a predefined error x bar-mu. So if we decide how far x bar should be away from mu okay. It does not mean that we should know mu here. We are only telling how far x bar should be away from mu and that is the number we are going to project. Then based on that number we have to adjust the sample size n such that we get both the desired level of confidence as well as the desired precision. So the error which is defined as x bar-mu should be of a certain limit which should not be exceeded and that error x bar-mu is equal to z alpha by 2 sigma by root n. Again look at the fundamental expression. So you are having this term. Let us take x bar-mu here and we get z alpha by 2 okay. x bar-mu will then become less than or equal to z alpha by 2 sigma by root n. x bar-mu the absolute value of which is termed as the error e. So once you have stuplated the value of e and since you already have defined error as z alpha by 2 sigma by root n, you square this expression then you get e squared. Then we do not have to use the modulus e squared is always positive for real values of e and so we have e squared is equal to z alpha by 2 squared sigma squared by n. So n is equal to z alpha by 2 squared sigma squared by e squared and so we take n is equal to z alpha by 2 sigma by e whole squared okay. Is it alpha by 2 is the upper alpha by 2 percentage point of the standard normal sigma is the known standard deviation of the population and e is the stuplated error. So if x bar is used as an estimate for mu then we may be 100 into 1-alpha percent confident that the error given by x bar-mu will not exceed a specified amount e and the sample sizes n is equal to z alpha by 2 sigma by e to the power of 2. So I will make a small correction here. If x bar is used as an estimate of mu then we may be 100 into 1-alpha percentage confident that the error given by x bar-mu the absolute value of x bar-mu mind it will not exceed a specified amount e when the sample sizes n is equal to z alpha by 2 sigma by e whole squared. If the value of n we compute turns out to be a non-integer it must be rounded off to the next highest integer. After deciding upon the sample size n we get an interval of length such that it is twice the stuplated error that is 2e. Here it is assumed that the parent population distribution is known and is specified to be normal. The variance sigma squared of this population distribution is also known. The sample size required will increase when the desired interval 2e decreases for given values of sigma and the specified confidence. When the standard deviation of the population increases that means there is more spread in the population then the sample size will increase for a specified value of error and a specified level of confidence. So let us say that we say that the error is so much and the confidence interval is 95% then if the standard deviation of the population increases there is more uncertainty there is more variability around the population mu then we need to of course invest in a larger sample. The sample size also increases when you also want a high level of confidence for a fixed desired length 2e and a fixed standard deviation sigma. So when you want to have a higher value of confidence level for a given precision e and a given standard deviation sigma then you have to go for a larger sample size. In certain cases we do not need the upper and lower bounds it may be enough if you specify the lower bound on mu or the upper bound on mu. Lower bound on mu probability of x bar-z alpha not z alpha by 2 sigma by root n less than or equal to mu is equal to 1-alpha. Please note that I am using z alpha and not z alpha by 2 as I was using earlier that is because I am now constructing a lower bound on mu. So this is termed as 100 into 1-alpha percentage lower confidence bound for mu. Similarly the upper bound on mu may be defined as probability of mu less than or equal to x bar-z alpha sigma by root n equals 1-alpha this is termed as 100 into 1-alpha percent upper confidence bound for mu. So we are chopping off one side of mu and using the other side only and when we do that we use alpha and not alpha by 2 when finding out the upper percentage points. So the lower bound on mu may be written as x bar-z alpha sigma by root n less than or equal to mu upper bound on mu may be written as mu less than or equal to x bar-z alpha into sigma by root n. So let us now come to the actual situation where the population variance is not known okay but let us say that we have chosen a large sample how large is large in our particular case let us say that the sample size is greater than 40. So the problem of unknown sample okay so the problem of unknown population variance sigma squared is mitigated or attenuated if you take a large sample. What are the advantages of taking a large sample? Then we can relax the assumption that the parent population is normal the parent population can be anything it can be normal it need not be normal but our sample size is quite large then the central limit theorem helps us by saying that the sampling distribution of the means is approximately normal irrespective of the shape of the parent population distribution this is very good. So for a large sample size the sampling distribution of the means is normal and previously we had used is that is equal to x bar-mu by sigma by root n but now we do not know the value of sigma what is to be done we are having a large sample and the sampling distribution of the means is normal. So we can normalize it and convert it into a standard normal variable z but we do not know sigma what is to be done what do we know we have the sample with us with the sample we have the sample mean we also have the sample standard deviation s so instead of sigma we can put s here and the resulting distribution will still be approximately normal. So we can put s instead of sigma and life will be nearly as usual as before. So I am replacing s instead of sigma here and so the distribution importantly is still approximately normal. So I am able to use a standard normal variable is z and that is equal to x bar-mu by s by root n. So we are taking a large sample from a population that may or may not be normal since the sample size is large the condition of a normal original population is not required. According to the central limit theorem the sampling distribution of the means is nearly normal or approximately normal with mean mu and variance sigma squared by n provided n is large. Since the variance sigma squared is unknown we assume that the sample distribution is nearly normally distributed with mean mu and variance s squared by n. We were able to make this assumption because of the large sample size. So we can now appreciate the merits in investing more resources for taking a large sample. So we need not take the entire population into consideration during the sampling exercise that is not going to be realistic but we can invest on a large sample size. So a large sample size helps us to increase the precision of our confidence interval and it also helps us to handle situations where the parent population is not normal. It helps us to handle situations when the population variance is not known. So when you have a large sample size and the parent population is not normal or the parent population distribution is unknown the sampling distribution of the means become normally distributed. If sigma squared is not known then we have to use s squared. If the population distribution shape is not known if the variance sigma squared is not known and we have a large sample size we can do the following. We can still assume approximately that the sampling distribution of the means is normal. Number 2, instead of sigma squared which is not known we can use s squared. Here s squared is the sample variance. So the large sample confidence interval is now defined for mu as probability of x bar – z alpha – s by root n. This is less than or equal to mu less than or equal to x bar – z alpha – s by root n that is equal to 1 – alpha. And the 100 into 1-alpha percentage confidence interval on mu is given as x bar – z alpha – s by root n less than or equal to mu less than or equal to x bar – z, less than or equal to signal Listino x bar plus z alpha by 2 s by root n. Similarly, the upper and lower bounds may be defined taking care to replace sigma with s, small s, the sample standard deviation. The sample size should be preferably 40 or more. We earlier saw that for the central limit theorem to apply, we need a sample size greater than 30. But since we do not know sigma, there is additional variability and hence we have to go for a larger sample size. So this completes our discussion on confidence intervals. This is very, very important because when you look at any statistical software output, you usually find the 95% confidence interval limits presented. The confidence intervals also have an important and interesting property. If the lower limit of the confidence interval is negative and the upper limit of the confidence interval is positive, then it is giving us some important information. For example, it is like a person when queried at what time the train is going to reach the station, he says, oh, the train has left 5 minutes back or it is expected in 5 minutes time. The person who is listening to this will get completely confused. Is the train going to come or has it already gone? So if the lower limit is negative and the upper limit is positive, then the 0 value is bounded between the upper and lower limit. So the population parameter may be 0. It has certain implications in this design of experiments and in linear regression analysis. We may construct the confidence intervals on the model parameters and we may see that the confidence bounds may be passing through negative as well as positive values. They have special significance attached to them and these are also very helpful to interpret the statistical estimates of these parameters. So I will wind up the discussion now. So I request you to think about what I have said and please remember that the confidence interval is probably bounding the value of mu with the stated level of confidence. Thank you.