Okay. What do we have with us? We have X as a random variable. That represents the average daily temperature in Para Night in January. For the town off Hannah, Now X has a mean off approximately 68 F. Okay, so we can see that the mean is 68 all right? And the standard deviation sigma is approximately 4 F. All right, Now, there is a 20 year study that is one of 620 January days, and we have their observations with us in the table. All right, so it is Just look at the table. Okay, So the first column is reason under. Of course. Okay, after that, we have the temperature. Then we have expected person from the normal. Then we have the expected person from normal curve. Okay. After that, we have the observed values. After that, we have the observed values. We have the observed values. All right. Now, the region under the curve is the first category Is the reason between three standard deviation and to standard deviation towards the left. Okay, so this is going to be new, minus three sigma. And let me just write this anti. Call him. It'll just take a minute. Then we have new minus sigma. Then we have new. We just have milk. Then we have new plus sigma. And then we have mute lis. All right, X, then we have new minus two sigma. Mean minus two Sigma new minus Sigma men, New plus sigma. New plus two Sigma New plus three sigma. All right. Now, what are the exact boundaries this is given in this column? We have 56. We have 60 than 64. 68 72 76. All right, these are all less than equal to science. Okay, 80 76 70 to 68 64 60. All right. Now, if the normal distribution actually fix the, you know, actually fix the observations that we have, then what should be the expected percentage from the normal Come? Like if I draw this normal distribution? Let's say that this is the mean right? This is my one standard deviation of it. I'm to standard deviation of I'm 300 deviation of. Okay, so this one is mu minus three sigma. And this one is mu minus two. Signal what is actually the area what percent off values lie in this region. If this actually is a normal distribution, the expected present attribute 2.35%. Right? So these are the values over here in this column. So this is 2.35%. And since it is symmetrical, the last one will also be 2.35%. Then it is 13.5% 13.5%. And since this is symmetrical, this is also 13.5%. This is 64% and this is also 64%. Now, what is exactly the observed values that we have with us? Okay, The observed values are 14 86 207 Then we have 215 Then we have 83 then we have 15. All right, so now let's just look at the questions. What exactly are they asking us? They're saying impact one that mu and sigma this and this. Now what? Our columns 12 and three. In the context of this problem, well, call them one. What has actually happened happened over here is they have divided the entire distribution into various categories. So column one is going to tell us that this particular reason the first region is the reason between three Sigma and two Sigma towards the left like and this value is the difference between two sigma and one Sigma towards the left and so on. Okay, so this is column one column six Give Sorry. This column to gives us the exact boundaries and column three is nothing but the expected value. If this really were to follow normal distribution, what is the percentage of values that we expect to find between new minus three Sigma Nu minus two Sigma that is in this region, right? What is expected amount? It is 2.35 What is expected amount in this region, that is between 23 sig mind to sigma it is 13.5%. Right. And then over here we have the observed values, the values that have actually been observed. All right, now, what is this question to question two says we have to use a 1% level of significance, which means what is our Alfa R Alfa for this question is 0.1 all right to test the claim that the average daily January temperature follows a normal distribution with Mu 68 sigma for All right. So in orderto do this, in order to perform this analysis, we're going to use the chi Square statistic. And the first time of the Chi Square statistic is to find the expected values to find the expected values to find the expected values. And the formula for that is the total sample sizes. The total sample size multiplied by the probability multiplied by the probability multiplied by the probability off each category off each category. Okay, off each category. All right, so what exactly is going to be my sample size? If I add all of these up, this should be 6. 20 as it is given to us, right? 6, 20 days. All right, so let us just use that formula way. This is the expected values for all the categories expected values. All right, so this comes over here, so let's put the formula into action. Okay? So I'm going to use my calculator for this. I want to 0.35% off 6. 20. So this is 0.235 multiplied by 6. 20. This is 14.57 expected values. 14.57 Then I have 13.5% off. 6. 20 0.135 in 26 20. This is 83.7, 83.7. Then we have 64% or 0164 into 6. 20. This is 3 96 3 96 point eight. All right. Just a moment. This is 0.6. For how can this be 0.64 This has to be 34. My mistake. I apologize for this. This should be 34 right? So this is 34. 34%. All right, so this is 0.34 into 6. 20. This is 210.8. So if I just erase this, this thing is 210 pointed Now, since this is symmetrical, this is a normal distribution. So this will also be 210.8. This will be 83.7 and this will be 14.57 These will be the expected values. Now what now? In order to calculate the Chi Square statistic for all the categories we're going to apply the formula observed value minus the expected value. Whole square divided by the expected value. And in the end, I'm going to sum them Allah, and it will give me the overall chi square statistic for my question. All right, so let's just look at this inaction. These will be the individual chi square value, the individual chi square values for all the categories. We're going to send them all up in the end. So this is the difference between 14 point 57 and 14 and we square this and this. I'm getting us 0.32 and I divide this by the expected value, that is 14.57 So this is 0.2 0.2 Similarly, this is 86 minus 83.7. I square this and I divide this by 83.7, 83 0.7. This is 0.6 0.6 Then we have the difference between 210.8 on 207 I square this and I divide this by 210.8. So this is 0.685 Recognized this as zero point 0.7 All right, then we have difference between two and five and 210.8. We square this and divide this by 210.80 point 80.8 Okay, then The difference between these two is 20.7, and we square this divide this by 83.7. So this is 0.0 0.55 Texas. Right. This is six. And then the difference between 15 and 14.57 We square this and divided by 14.57 So this is 0.1 0101 Now we, some all of these up. So this is 0.22 plus 0.6 plus 0.7 plus 0.8 plus 0.6 plus 0.1 And this is 0.248 So I can say that my guys question to stick for this problem is 0.248 All right, now that I have Mike ice Question District, what else do I need in order to get my p values? I need the degrees of freedom DF I need d f. And how do I find d f? This is given by the formula Number of categories. Number off categories minus one. What is the number of categories that we have? We have six different categories, Right? 123456 So this is going to be six minus one six minus one. Or I can write this as five. So my degrees airfield I was five. Now I have my kaist question the steak. I have my degrees of freedom. So you can either use a chi square table and it will give you an approximate P value. Or what you can do is you can use an online software or statistical tool so that you get your exact P value. My guys question is thickest 0.248 This is 0.248 and my degrees of freedom is five, as we just saw are significant level is 0.1 and we can see that a P value is 0.9985 My P value is 0.9985 What was my Alfa? My Alfa was 0.1 hence I can say that as my P value is greater than Alfa I failed to reject. I fail to reject my null hypothesis. H not Okay now I think we also forgot to write the hypothesis initially. Okay, so let's just look at it briefly. What is going to be my null hypothesis for this question? What was my null hypothesis? My hypothesis waas that normal distribution normal distribution fits the distribution off days Just a moment for the distribution of the average daily January temperature fits the distribution fits the distribution off the average daily January temperature, January temperature and what would be my alternative hypothesis? My alternative hypothesis would be that the are not similar, right? That the normal distribution does not fit the average daily January temperature. So they are not similar. Let me just write it like this. They are not similar. All right? So I cannot reject my null hypothesis. So what is going to be my answer? This line is very important. I will say that I don't have I don't have enough statistical evidence. Enough statistical evidence to suggest I do not have enough statistical evidence to suggest that the normal distribution dozen doesn't fate The January temperatures. The distribution doesn't fit the distribution off average daily January temperatures, and this is how we go about doing this question.