Lecture 30

[Music] hello friend welcome to the course multivariate procedure with r so you can recall that in the last lecture we have understood that what is the concept of test of hypothesis which is related to the testing the value of some parameter on the basis of given sample of data now from this lecture we are going to consider the case of normal dist distribution which has two parameters mu and sigma square and primarily we are more interested in developing a test of hypothesis for the mean now this can be done under two conditions when the variance Sigma square is known or unknown so in this lecture we are going to consider the case when Sigma square is known and in the next lecture we will discuss the case when Sigma square is unknown which is also estimated on the basis of given sample of data gradually we will move forward and I will try to develop the similar test of hypothesis for the two samples as well as for multivariate case well when I am saying here that we are going to develop the test of hypothesis for the mean in the case of normal distribution it does not mean at all that the test of hypothesis for other parameters from other type of probability density functions and probability Mass function cannot be done but as I have said several times our job here is not to learn the test of hypothesis but I want to give you here an exposure or some knowledge about those tools which I'm going to use in the future lectures so in this lecture we are going to consider the univariate random variable and we will try to construct the test of hypothesis when Sigma square is known also we had understood in the last lecture that there is a close connection between uh this confidence interval estimation and the decisions on guest of hypothesis and earlier we had talked about what are the confidence interval estimation so in this lecture we will also consider uh the construction of confidence interval estimation under the same setup that is confidence interval for the mean when Sigma square is known and then I will try to show you that how you can compute these things in the our software so let us begin our lecture so now we are going to consider here the test and confidence interval for mean in one sample with known variance in univariate data right so I already have explained you the concept of test of hypothesis right and we are going to Now work on it so as we discuss that we need uh some assumption for every test of hypothesis so when we are trying to test or or develop our test of hypothesis for the mean when Sigma is know then it is popularly called as Z test or gaus test right GS and in order to execute this test or use this test there are certain assumption uh which are known to us for example the null hypothesis is going to be specified by the statement H not mu is equal to Mu not where this mu not is some known value pre-sp specified value Sigma is known to us from some outside sample sources for example someone can tell me okay in this case the value of Sigma square is supposed 25 we assume that the population is normal that means the sample which we are dry from here belongs to a normal population or the second option is this if we don't know about the normality or in case if the sample size is here large which is more than 30 the reason for it is that when we try to consider the case when Sigma square is unknown and is estimated from the sample then in that case the decision rule for both the case becomes almost the same right so if you want to test this hypothesis s not mual to Mu right then the test statistics is like this z c is equal to xar minus mu upon Sigma by root n right so xar is the sample mean which is obtained from the given set of data and mu will be coming from here mu is equal to Mu not and Sigma Here is known and small n here is the sample size right you have to be careful in the further lectures that where the N is going to indicate the degrees of freedom but here in this case this n is the sample size and now we have to fix the level of significance that is the value of alpha say 0.05 or 0.10 and uh now the critical values that whether we are going to accept or reject the hypothesis on the computed value of zc that is what we have to decide on the basis of distribution of test statistics z c which is which has got a normal distribution with mean zero and variance one when s not is true so that is why these critical values for this test they are obtained from the probabilities of normal 01 right so I can say in general that z c follows normal 01 under h not right that is what I told you in the last lecture that we need to find out the probability distribution of the test statistics which is here zc right so now how are we going to conduct that test of hypothesis we are we have to State our null and alternative hypothesis H not and H1 then we need to compute the value of the test statistics zc on the basis of given set of data and then we have to obtain the critical value for a fixed value of alpha according to the hypothesis whether it is right tail left tail or two tail test then we have to compare the computed value of z c with the critical value and then we have to make the decision accordingly alternate alternatively if you want to use the P value also then you can use the P value to make the decision about the acceptance or rejection of the null hypothesis right so how to compute this P value we have discussed in the last lecture just for your quick revision that here in this case z c is the value of the test statistics which is computed on the basis of given sample of data and this follows a normal 01 so the P value is given by this following probabilities when we are trying to consider a two tail test then it is comped the twice the probability of Z greater than absolute value of z c for right tail test it is computed as probability of Z greater than z c and for left tail test it is commuted by here probability Z less than z c and uh in in case if you want to use it then the decision is like this that H not is rejected in favor of H1 at 100 Alpha per level of significance if P value is less than Alpha and this P value is the smallest level of significance at which s not would be rejected so if you try to see when you are trying to work for the test of hypothesis that becomes very simple right so now we see how to conduct this test of hypothesis in the art software so this test is available in the our software in a package composition c o m p o s i t i o NS which means compositional data analysis for this package is to be installed first using the command install. packages and then you have to upload the library compositions now this test is available by the command here gus. test gs. T but you have to be careful that g here is uppercase right so from this test we can have one sample test two sample test with equal mean of normal variates and known variant of what is this mean initially I'm going to consider here only the one sample test but later on I will extend it to two sample test right so in that case you will see the same package same command is going to work there only small modification will be needed so the command to conduct this test here is like this gaus do test and then you have to see here X which is the data on the on which we want to conduct that test of hypothesis X Y here is actually null because as I said this the same command can be used to test the equality of the means from the two population which is a two sample test but here we are using it only in the case of one sample test so that is why we are writing here Y is equal to capital n u l so if you remember once in the lecture we had discussed the concept of n and n that null is something which does not exist now we have to give here the value of mean me a an then we have to give here the value of a standard deviation which is the value of Sigma right and then now this alternative hypothesis can be of three type two-sided less than greater than so that has to be indicated here by the option alternative is equal to two-sided less or greater and you have to choose only one whatever you are going all these values are given here so that you can understand that that there are three possible values and then you have to choose one value right so you have to choose one option depending on your requirement right so as I said that X is a numeric Vector providing the first data set Y is the optional second data set which will be used when we try to conduct the two sample test of hypothesis mean is the mean and to compare with LD is the standard deviation which is known and alternative is that the nature of the alternative hypothesis so now let me try to show you this with an example uh suppose if we have a random sample of size 20 on the day temperature in a city and we have collected 20 observation and suppose the temperature in the population follows a normal distribution whose mean mu is unknown but Sigma square is 36 it is known so the sample values of temperature in degrees celsius they are reported here like this and based on that we estimate the MU had as sample mean that is esate of me right so on an average I can say that the temperature in the city at this time is [Music] you will see that I will be using this example and the same data set in different lectures so that you can compare that how the things are changing right now if I want to test that the null hypothesis is mual to 30 versus alternative hypothesis mu is not equal to 30 what does this mean I am trying to make a statement here that the value of the mean in the population is 30° C and my alternative statement against which we want to T the null hypothesis is that the MU is not equal to 30 that means the population temperature is not equal to 30° C so I try to give here the data using the command gs. test xal to Temp now y equal to here n mean is equal to here 30 this is coming from here mu0 30 and SD you can see here this variance was given to be here 36 you can see here so this Sigma becomes here 6 Sigma square is here 36 now alternative here is two-sided how if you try to see here the structure of mu not equal to 30 from there I'm trying to say that it is two-sided just of hypothesis and this is here the outcome so you can see here this is here T so this is the value of here zc mean here is this equal to 30 this is the value of here mu0 SD here equal to 6 this is the value of your Sigma and now you can see here P value is coming out to be like this 1 into 10 power of your - 6 which is very small very close to zero and the alternative hypothesis is here two-sided so now you can see here your rule was H not is rejected if P value is less than Alpha so if you try to choose here Alpha is equal to 0.05 which is 5% level of significance and your P value here is 1 into 10^ of minus 6 what do You observe that P value is much lower than the value of alpha so I can say that here H not is rejected what does this mean h not is rejected mean h not is mual to 30 is rejected that means mu is not equal to 30 and if you try to see this also makes sense if you try to see here that you had obtained the sample mean to be that means the data is saying that the average temperature is close to 42° and we are saying that the temperature here is Mu equal to 30 so definitely there is a huge difference between the two value and that is the same thing it is is implied by saying that s not is rejected so s not is rejected means H1 is accepted H1 is accepted mean mean is not equal to 30 right and that makes sense now it is not telling you that what is the correct value of mean remember one thing test of hypothesis cannot tell you that what is the correct value of mean it can only tell you whether the value you have chosen is correct or not so now we try to change the alternative hypothesis and other things remain the same s not will remain as Mu equal to 30 and H1 is Mu less than 30 so the same data set i'm going to use here so this the command here will remain the same x value y value mean SD they will remain the same only alternative is going to be changed as less so yeah I'm just trying to show you that if your null hypothesis remain the same and alternative hypothesis is changed then how are you going to control it right so now if you try to conduct this test the our software outcome has this type of values so you can see here this is data here is the same temp t value will remain the same as earlier mean SD that is given now P value here is one so you can see here as you try to change the alternative hypothesis this P value is changing right so if you try to see here p is equal to here one and Alpha suppose if you try to take here 0.05 so your rule was reject H if p is less than Alpha right so in this case you can see here P value here is one and Alpha is here 0.05 right so you can see here now the decision is changed right so and this is here the screenshot of the same outcome right so let me try to show you these things on the our software so that you get here more confidence right so let me try to just uh copy this data over here and upload the Library command so Library composition command so you can see here now here we have the temperature is here like this and now if I try to use here this command here G do test on the temperature data it is going go to here like this you can see here this is the same value which is changed right and if you try to use here the same data set but if you try to say here use here can see here in this slide there were two options I will try to show you here both these options less and greater right you can see here this is here less and this is here greater so if I try to use here the command here greater you can see here this P value is here like this and if I try to change here this here less so then you can see here the P value here is now changed so depending on your alternative hypothesis this P value is going to change and uh yeah in the first case when I'm trying to take two-sided alternative that means H is mual to Mu 0 and H1 is Mu is not equal to Mu whereas when I'm trying to take it here greater than so it is alternative here is Mu greater than mu and when I try to say here less then the alternative is Mu less than mu right so this is how we can conduct such hypothesis or we can test such a hypothesis right now I come to the confidence interval estimation under the same setup so now we have got here a random sample X1 X2 X1 from a normal population with unknown mean mu and known variance Sigma not a square so we have already used the point estimator say this sample mean Capital xar to estimate the MU and now we want to construct the interval estimate right this xar equal to 1 upon n summ x I was the maximum likel estimate of mu so now I'm not going into that much detail but there is a central limit theorem which C us that xar follows distribution normal which is the mean mu and variance Sigma squ upon n so here the variance is Sigma not a squ so xar will follow a normal distribution with mean mu and variance Sigma Square upon n well this is the job of the statistician who are working in theory to derive such distribution and based on that if I try to create here a statistic square root of n xar minus mu upon Sigma then it will follow a normal distribution with mean zero and variance one right well you need not to worry those who are not from the statistics background that is the job of the statis to find out the probability distribution of such statistics so but now I know that because it is following a normal distribution so what I can do here that I would like to see for example if this is my here normal so I know that there will be some values on the left hand side and some value on the right hand side of this mean zero so this value on the left hand side can be minus Alpha by 2 then the value on the right hand side at the same distance that will be plus z alpha by 2 right so now I want to find out the probability that my here this statistic Zed s otk of n xar minus mu upon Sigma what uh is the probability that it will lie between minus z alpha by 2 and plus Z alha by 2 and I fix this probability as 1 minus Alpha what is this probability here I'm trying to take here this area as a alpha by 2 this area as say here Alpha by 2 now this area will become here 1 minus Alpha so now I'm trying to write down Qui equation that uh what will be the value of mu such that the probability that s root of n xarus mu upon Sigma is lying betweenus z alpha by 2 and plus by 2 whose probability is 1 minus Alpha right and if you simply try to solve this equation this will come out to be like this that mu should lie between xar minus z alpha by 2 Sigma upon root n and xar + z alpha by 2 Sigma by root n right so yeah as I told told you that square Alpha by 2 is the 100 Alpha by 2% points on the normal 01 distribution or in terms of quantiles this is the alpha by 2th quantile on the normal 01 so now I can see here that this is the lower limit of the confidence interval and this xar plus Alpha by2 Sig 0 by root n this is the upper limit of confidence interval and if you try to recall that when we introduce the concept of confidence interval estimation while introducing the estimation right or statal inference then we had indicated this lower limit to be head head Theta head L and this upper limit was indicated here as the Theta h u so now using the same terminology I can write down here that 1001 minus Alpha per confidence interval for Mu is is this interval Theta hat L Theta hat u based on the sample X which is here like this xar minus Alpha by 2 Sigma by root n and xar + Alpha by 2 Sigma by root n so you can see here if you get here some sample values so if you try to see here this is depending on xar this is depending on z alpha by 2 this is depending upon Sigma 0 and this is depending on here n so X this can be computed on the basis of given sample of data this alha by2 is known from the table Sigma 0 is already known and N is also known which is the sample size so you can compute this interval and then you can see that okay that with the for example if I say take Alpha is equal to 0.05 then I can say that there is a 95% confidence interval for new is that mu is lying between these two values so that is the same if you try to recall once I take an example that when uh you want to compute the time taken from your home to your college then there are two option that you try to collect some data that for cutter number of days you record the values of time taken and then you try to take the sample mean and then sample mean will come out to be some value that will be the point estimator and if you um suppose sample mean comes out to be suppose you 20 minutes then it is a point estimate but if you want to know the other alternative is that okay it will that you will take between 15 minutes to 25 minutes this was the cont confidence interval so now I have shown you here how are you going to construct this confidence interval based on the given sample of data so this is the confidence interval for mean whose confidence coefficient is 100 1 minus Alpha perc right so this 100 1us Alpha perc this is called as confidence coefficient right so and if you try to understand this thing through this graphic so as I said suppose if you try to consider here the distribution of normal 01 so the mean is here zero so this area here in the green color which I am shown you this is the area Alpha by two so corresponding value at on the xaxis there will be z alpha by 2 on the right hand side and minus Alpha by 2 on the left hand side and on these two side there will be somewhere minus infinity and plus infinity but when you are trying to compute the confidence interval now this will also follow on the normal contribution but now mean here is Mu and this is lying between these two limits so the limit on the left hand side is xarus z alpha by 2 Sigma 0 upon root and this area here is Alpha by by 2 and on the right hand side the the area is the same Alpha by 2 but this limit is now here xar + z alpha by 2 Sigma 0 by root n so this is the basic idea and this area in say red this is 1 minus Alpha right so this is the interpretation and the same interpretation will continue in all the cases whenever we want to do it now I would like to show you that how are you going to construct the confidence interval well in the case of confidence interval for Mu uh when Sigma square is known this is not available in the outcome of the art software but uh this will be available when in the software outcome when Sigma square is known so in this case I'm going to take the same example of the temperature where we have collected the data on say 20 days n equal to 20 and we have the data on the temperature and uh we are going to construct this confidence interval by writing our own program which is is not very difficult right so if you try to see here now the temperature data is here now you can compute here the if you try to see this confidence interval is depending on here mean and standard deviation right so so simply have to write down here xar which is obtained here by mean of Temp and then you have to obtain the value of z alpha by 2 which is here Norm 0.975 how if you try to see here uh this normal distribution this area here is 95% and so this area is here 5x 2% and this area on the right hand side is 5 by 2% so this is here uh means if you try to see here it is here I will use here a different color pen it is here 0 point means if you try to see here this is here 1 minus Alpha by two and this area here is Alpha by 2 like this so this comes out to be here 1us Alpha by which is here 0.975 and now Sigma 0 is given to us and then square root of length of temperature which is the number of observation and if you and similarly you can compute here the upper limit by just by replacing minus by here plus sign that is xar + Alpha by 2 Sigma by root 10 so it is not difficult and if you try to compute it you can see here that um it will here come like this so now you you can see here Sigma 0 is here six as given and uh this qor can be obtained directly from the r software and square root of length of temperature that can be obtained directly from the uh data Vector Temp and then Sigma 0 is given so this value comes out to be 39.3 and similarly if you try to compute the upper uh confidence limit this will be come out to be like this just by replacing minus by plus sign right this will come out to be 44. here 59 so I can say here that uh you can see here now if you try to consider this thing the point estimate was here 41.979747 n so yeah close to the point estimate was close to 42 so now the interpretation is that on the basis of given sample of data I can say that the point estimator is 42° and the confidence interval is a 39.33621 [Music] and now I try to write down here the lower limit which is here like this and and upper limit if you want you have to Simply make it here uh see here plus sign as you can see here it is here like this okay so now we come to an end to this here uh lecture and you can see here that we have considered here a very simple test of hypothesis my objective was to illustrate you how are we going to conduct such test of hypothesis so we have considered the case when we want to test a hypothesis about the mean from a normal population and we want to test s not me is equal to mu0 in a univariate case in a single Sample When Sigma square is known now similarly what will happen in the case when Sigma square is unknown that is the actually more realistic case so that we are are going to consider in the next lecture but the process is going to be the same as we have seen that the statistics square root of n xar minus mu upon Sigma that is following normal distribution similarly when Sigma square is unknown and that is estimated from the sample then we would try to replace Sigma by this estimated value and then we will try to see that the distribution of the statistics will become T distribution what was T if you try to recall sometime back you have done the sampling distribution Kai sare T and F so then we will see that how are we going to conduct the test of hypothesis but this steps and the way we have taken the decision for conducting the test of hypothesis and construction of the confidence interval that will remain the same so it is very important important for you that if you want to understand the further couple of lectures all the steps whatever I have explained you here they should be crystal clear in your mind what are we do uh doing why are we doing what are the different steps yes some steps will be varying in comparison to what I told you in the last lecture because now since we are using the P value so that the test of hypothesis through simple uh in the case of this software becomes easier for us and it is pretty simple in comparison to you try to look into the Z value from the table or similarly the mean other type of critical values from the table then then you try to compare it manually right so this is how we are going to proceed further so my request is once again that you please try to look into this lecture try to revise it so that you are well prepared for the next lecture and I will see you in the next lecture till then [Music] [Music] [Music] goodbye