2022_Summer_MM2_Tutorial_2_Mon_2_Jan_Week_4_recording

Added: 2026-05-12

[00:00:01]hello everyone welcome we're doing the tutorial now normally i would do this in my group and post it afterwards but as it's um a public holiday there's a problem of getting it to you on the monday so not being able to expect anybody to come on a monday i have to record it so this is the recording all right so we'll go quickly through these um suppose that the diameter of steel rods for a particular machine have a mean of 20.02 and a standard deviation of 0.01 i'll find the z score for the following observations now the z-score if you recall was about having a mean a way to take any kind of data set okay and standardize it and that's the equation i gave at the back of one of the lectures i think the second lecture and instead of having data sets with a mean around you know 100 or what have you we go through this transformation process and we have a z score around zero and what it means is that lots of different distributions can be transformed in this way okay they can be moved on to a standardized system so the purpose of the z score is to do and then one standard deviation two standing there it's very easy to do okay it's very easy to pick it up in terms of just using minus one two three in terms of your z scores okay so um if your z score is close to that standardized mean of zero it'll be small the larger the z-score the further away you are okay and you can actually relate it directly to standard deviations one two three standard deviations so standard deviations will tell you the percentage of the data that you're taking in 68 98 so on and so forth okay so um the computation itself is very straightforward so what this is trying to do is to highlight for you that as you go through your as you go through your z scores your measurements what's the implication of the z it's called what's its meaning so if it's if it's a z score of 1 it will mean it's a standard deviation of 1.

[00:02:43]so to work this z score out what we would do we would see that we've got a standard deviation of 0.01 we've got a mean of 20.002 these are our random variables for a we've got xa and xb and then we're going to plug it in our z score uh we will get um our value so x is 20.002 minus mu which also happens to be 20.002 divided by that now seems superfluous because it's zero so the z score when the sample value is on top of the mean the z score is zero so that means uh zero okay if it's the same mean then it has a z-score of zero now look very carefully this is a 19.980 i wonder what the distance is between the two so if i put in 20.002 subtract 19.980 i get two twenty two thousandths the difference of these two is this very very small number twenty two thousands is that what's the statistical significance of that we know if it's the same value it's zero but if it's and this is what the z score is trying to get you to do the z score is trying to get you to realize that if it's one it's still a very long way even though the value or the distance of the machining size of the bar is small it's statistically significant so this is like an introduction to the concept of statistical significance okay so for such a small value of 19.8 if we put the z-score in again we'll get our 19 it'll actually be a negative because it's a little bit smaller 0.002 divided by 0.01 now that'll give us minus 0.023 on the top and uh 0.01 so it's still going to be quite a big number so if i just get that calculator back if i make that a negative 0.022 it was and i divide that by.01 oh minus 2.2 now what that tells us is that this b value of 19.980 is over two standard deviations away from the mean in other words it's outside 97.8 percent or 98.7 percent of all of the data around the mean in other words it's a long way away from the mean so this is a way of us translating if you like the know-how confidence idea such a small value in your reckoning 0.8 22 000 is hardly worth thinking about is a 22 thousandths of a millimeter 0.022 a millimeter is very small okay but statistically it's a very large value it's a long way from the mean in fact it's over 2.2 standard deviations so when we think about this idea of standard deviations this is where it gets a bit tricky okay so we don't i don't labor all of the fundamentals too much but if you've been reading up on distributions you can do this a number of ways but if you've been reading up in the text or perhaps even googling it thinking about the data covered by standard deviation so it's easy just to do a simple one like i could get the text out for you but it's just as easy for for us to do it this way uh google and if we go to google and we can or whatever it is that your preference is and i could do a standard deviation and if you look at a standard deviation we've got lots of things there there's our standard deviation okay and if we look at the uh if we look at a picture for standard deviation and we can see that 68.2 percent of the data is covered by one standard deviation we have to add a further 26 27.2 percent of the data when we go to two standard deviations okay so look at that once we get past two standard deviations in terms of where we are with regard to the mean we're a long now this is a standardized main it's got a mean of zero so it's exactly what matches our thinking but there's lots of examples here look i don't want to labor the idea but 68.2 at one standard deviation 95.44 at two standard deviations so when we're 2.2 we're actually beyond 95 percent so that's a big variation and 99.72 so see if you can remember those numbers 68 95 and 99.

[00:08:35]just as a shortcut so this one here is over what do we say 68 95 and 99 this is between 95 and 99 of all the data points away from the zero mean so while it might look like a very small value statistically this is a big issue this is statistically significant this particular value so being able to determine where your observations are fitting within your group is is quite uh quite important all right so a team of environmental engineers were contracted to uh check out the distributions of quality in a water uh dumping an old oh well there's a movie john travolta was in a movie about lawyers and the lawyers were going around getting benefits for people who had been affected by the dumping of uh chemicals from uh from a factory that was dealing in leather and um and that was actually based loosely based on a true story now the true story uh one of our professors and a very distinguished uh human being she um actually did all the statistics for that court case and uh she's won a medal through harvard and all of this kind of thing so yeah you know you never know where you're going to end up and what you're going to do i don't think that's my lot i've got a different path but maybe that'll be you you'll be following in louise ryan's professor louise ryan's footsteps and maybe doing some wonderful help with your statistics so they measured the concentration of now prohibited probably thanks to louise they measured the concentration of the substance at the water surface at the riverbed and midway between the two okay so we do a cross section of the river there it is a cross section so that's the water surface so we'll call it ws and midway we'll call it midway and riverbed so what they're doing is they're telling us the values okay so then it's given us some information and this is good because i want you to be able to do this kind of thing determine whether the three samples have skewed or symmetric distributions if the data of a particular sample is skewed further classify them as left or right very good very good so what we've got if we look down at surface mid depth and we've got our table to interpret we have a mean of 4.06 we'll assume they're all normally distributed actually i'm going to leave them on one thing there a mean of 4.06 and then we go along and find the medium now medium is 4.3 oh so the median 4.3 what i might do is i might just take that off and leave that as the the sample test and then what i'll do is i'll put the mean 4.026 and the other one the middle the middle score is 4.33 so the mean has been pulled away from the middle score so what do we think is happening okay the mean has been pulled away and it's been pulled to the left so what we what we understand from our understanding of that is that it's left skewed or negative skewed okay left skewed or negative skewed now are they all doing that so 5.096 5.1 yes 6.259 6.250 okay so the bottom ones and the mid-depth they're a bit closer median they're a bit closer so they're they're i'd say fold the surface left skewed for the surface and mid way and riverbed they're pretty close 5.896 5.1 6.259 they're almost there so we'll call them symmetrical we'll put it approximately symmetrical because of this idea here okay that's all we're doing we're just having a look at the means and the medians pretty easy what would be the most appropriate measures of central tendency and variation for each of the samples well where we have a reasonably symmetric case so for midway and for the riverbed because they're symmetrical we could use the mean but for the skewing on the surface we can't use the mean so we would be better off to going to the water surface we would be better off going to other measures like interquartile range and median typical score is median get the middle 50 percent of the data and have a look at it there okay because the cement the symmetry of the um the symmetry of the system has been upset a little bit using appropriate measures of central tendency and variation compare the data from the riverbed and the mid depth data okay well the concentration of the main mean river bed and the mid depth so riverbed and mid depth so the oh well these these are these two these are our two relatively symmetrical cases and all we're doing is characterizing the data so we're not making any big statements okay so we're saying the mean concentration okay so we're going to use our measures of central tendency and we've already indicated mean so the mean is uh mid depth riverbed so 6.3 6.25 96.3 nanograms per liter versus 5.1 5.096 is 5.1 and that's all we're doing we're not being smart we're not being over the top we're just using the mean measure okay so the standard deviation is also an indicator of variation okay and they're very very close 1.16 also standard deviation uh 1.16 nanograms per liter for the riverbed versus 1.155 they're almost the same aren't they 1.55 nanograms per liter okay so there's slightly less variable less variable not only slight slightly less variable and that's all we're really saying we're just making very very strict observations we're not drawing any conclusions we're just saying the mean concentration is slightly lower at the root at the mid depth there it is slightly lower and also the variation at the uh riverbed are less variable at the bottom so you see those two contrasting ideas so that's all we're saying guys we're not making any other great sweeping statements we're characterizing the data using these particular measures of central tendency and measures of variation to characterize our data we're saying these two particular comparative ideas of riverbed and mid depth both symmetrically pretty symmetrical 6.3 nanos versus 5.1 nano is very close but slightly less at mid depth but the variation is less around that 6.3 on the riverbed the slightly bigger variation as you come up to the mid depth that's all we do now assign us we we make that statement and we just put that down and we say that's what we that's what we uh are observing a little bit of variance a variance in the mid depth slightly lower concentration um uh sorry a little bit less variation at the riverbed and slightly less concentration at mid depth and that's all you do you just make your statement and away you go that's all that's dated okay using the mean and standard deviation for mid-depth find the z-score for a reading of seven grams okay so what's the z-score how many standard deviations are we away from the mean at mid-depth so we've got x equals seven nano grams per liter the mean is uh 5.096 at mid depth there it is there 5.096 okay and the sigma is 1.155 very straightforward okay so our z score is going to be x minus divided by and so we'll get a z score of 1.648 wow so 1.648 well at 63 percent of the data 1 standard deviation this is heading up to two standard deviations which we know is 95 okay so it's not quite at 95 but it's on the y and so a a value of seven by way of comparison nanograms per liter um is getting quite a long way from the mean in terms of a normal distribution okay so uh that's all we do that's it that's all we've had to do okay so uh it's a very light introduction i hope that you can appreciate um what that was all about in terms of our ongoing development with our statistical intelligence but it's not too hard just keep it toned down in terms of expectation with regards to what you say the big errors come when you start making judgments let the data speak at this moment in time we haven't got a lot of other information we need to do some hypothesis testing before we can start making big statements so at this point all we're doing is characterizing it talking about slight changes in concentration slight changes in variation and a little bit about skewing because we had some knowledge about the way the mean and the median work together to explain skewing okay so we'll move on to the math part for this particular tutorial oh there's the uh box plots and you can clearly sorry i should have uh pulled these apart but um you can clearly see the concentrations at the various places and the amount of variance and the skewing this has got massive skewing on it okay massive skewing on it these are almost symmetrical slightly higher concentration at the surface and we talked a little bit about the variation so that's all we're doing characterizing the data okay don't get too carried away with the absolutes so initially we're just characterizing the data that's all we're doing all right so we've justified that use all right so let's have a look at our mathematics uh if it's possible to calculate a times b then do so so what this means is of course we're going to do a row down column row if it's possible if it's possible to do row down column then we'll do that can we do row down column we sure can so what this means is we've got uh two times one plus zero times one row down column row down next column two times minus one plus zero times zero row down column two times two plus zero times minus two and finally two times three and zero times zero okay so that's our our first set if you like okay now we go again the row down column then bro down column again and wrote down column and finally row down column again all right so what have we got 0 2 0 -2 0 4 6 and 0.

[00:24:33]one [Music] minus two minus one zero two plus four three and zero four three and one is four minus three and zero six and minus two nine and zero so we should be getting two minus two four six one minus two is negative one negative one two and four is six three uh three and one is four minus three four six minus two is four nine and that's our result throw down column okay there you go very very straightforward okay so um just that patience to work your way through it are you worth it i bet you are i bet you are it's important that you do okay all right so the next one we want to have a look at is our linear systems what's our linear systems like so if x is a solution of this particular question what we will find is that if we substitute x in we will get b so for part a all we need to do is do uh 4 two four negative two and yes we can use technology for all of these things but as i have suggested to you those that know tend to wait and just do the work so we'll do that now this time we'll do the x that's supposedly and this won't be as hard as you might suggest we're going to do a row down column and it's easy because the zeros so when we do four will be a match so i'll do it the first time so we would have four times one plus oh then it's two times zero plus four times one and then it's minus two times zero so the ones with the zeros are out but it's a repeat so all we're going to get is four and four so the first one's eight so i'm gonna i'm gonna skip the steps there and just go straight to doing that by hand okay so the first one is eight tick now this one's going down so immediately we can cross those off because they're out of it so three ones are three and three ones are three so three and three is six six tick this one's going down the column row down column the second ones are out so it's one plus one is two tick and the final one their second ones are out so we could have eliminated all of those so three ones are three three ones are three is six confirmed okay so we've done that uh quite uh quite well all right so how many uh how many solutions does the system have in total so how do we know if it's a unique cross your arms or parallel put your arms so that you can only see one of them how do we do that well we check the determinant in order for it to have a unique solution it must have a determinant oops a determinant of some number or if it's infinite we would expect the determinant to be zero we know there's a solution because we just verified a solution okay so what we need to do is to check the determinant and so far we've learned that we go down the determinant rows and columns so the debt of a uh we're just going to go down one and we'll go down uh we can go down this back one i think our initial one will be that so that'll be uh now where we've got zeros it it will they'll zero out so if we were to start at the zeros we would be multiplying zero by all of the sub matrix now what i mean by that is if we were to start with zeros here and start down this end we would get zero times the sub matrix well zero times the sub matrix is going to say gone so there's nothing there so forget the first one the next one is if we go this way but again the sub matrix is made up of these components and times zero so gone okay zero gone so there's no point us doing those two we only have actually two remaining that means it means that we have that row and that row which gives us -2 outside a sub-matrix of four two four one negative one one three one three that's that one and if i just change the color of the pen and do that one there so we've come down this column then we've got minus two sub matrix three three three one minus one one three one three so the answer becomes what are they now these two ones here i just wanna check something they are addresses row one column four now row 1 column 4 is a positive when they're added so we need to check 1 plus 4 which is minus 1 to the power 5 which is an odd so we do have a multiplier times minus one so minus one times four for that one oh that was the first one was that no that was right so minus 1 times -2 so that's going to make a plus all right the next thing we want to do is check this address which is the blue one so this is address two four so this is minus one to the two plus four which is minus one to the sixth now this is an even so we don't need to worry about changing the sign but we were affected by one of them but not the other okay so to fill out the determinant for this we will need to do a sub matrix across there okay so um it's very straightforward you just pick your you pick your next one and you go across your row it doesn't matter which way you go you can go across or down it doesn't matter the fact that you went down a column over here that's irrelevant you can go across the row only if you want and you'll make some sub matrices so -2 will be 4 and 4 sorry i'm just uh balancing my desk okay so it'll be four sub matrix minus one one three and we also have two one one three three and we'll also have four that's just going across the top one minus one three one okay so that's fine one minus one three one we just need to check those addresses as well so this is a one one so that's okay the two is going to get a minus one because of the odd so that's a minus two we're going to convert that the 4 is ok we'll leave that alone okay so that's your that's your proposition so for that one there we have 4 outside of minus one times three so that's the product subtract the other product one one so just being patient and working your way down minus two big bracket 1 times 3 minus 3 times 1 so that's the diagonals and then 4 and then we've got one one minus three negative one so let's work through that and don't forget your minus two out there so minus three minus one is minus four minus four times four minus sixteen uh three minus three is zero so that's out one plus three is four four fours of plus sixteen oops plus sixteen minus sixteen remembering that this is a product now is zero so that whole thing is zero so the blue one is zero so i think i know where this is headed but i'll need to double check this because i need to be certain so i'll put my orange in and so now we're going to go across the plus two oops so i'm going to do plus two outside of which way you want to go just across the top it doesn't matter so we've got three sub matrix minus one one one three three sub matrix one one three three three again sub matrix one minus one three one just checking the relative standing so 1 minus 1 to the power 1 2 is an odd so that's going to get a minus so that 3 is becoming a minus that 3 is row two column two so that's an even that won't change row two column three is five so that's a minus so that's going to get a minus okay so we've managed to get that far all right so let's do this again minus three so we're going to do it across there minus one times three minus one times one just being patient that's all it is plus three if you can see what's happening i can see it already well that's a giveaway that one's repeated the problem three minus three is naught so that's out and then we've got minus three so minus one of the minus is a plus okay so we've got and there's a two out of the front so we've got minus three minus one minus one times three yep so we get minus three minus four plus twelve then we've got uh plus three minus three is nine i think i must have made a mistake there okay so i'm going to have to go back over this to correct it so what i'm going to do is i'm going to get some help and i'm going to be a bit pragmatic let's see what that looks like uh see if i've got that right four two four minus two three three three minus two one minus one one zero three one three zero so that looks good what's the determinant of that let's determine zero okay so the determinant of zero tells us uh that i've not only made it that one's okay but this one i've made a mistake somewhere uh and i'm not sure where it is so i don't want to be too so we've got this case because we had a solution and the determinant is zero that tells us that there's infinite solutions because we had a solution it tells us that it's either unique or there are many and by looking at the determinant we're able to do to say that it is in fact infinite solutions so i i'm not sure what i did i'm not sure what i did here if i want to pick that up uh it was the sub matrix wasn't it yep so one three four so when i went down the first one minus one one one three that was okay then i go down the middle one one one three three and then the end one is one minus one three one so that should have been okay unless i got these minuses but one of these is an error so um but we've already determined the determinant so all right so um a little bit of little bit of work there um just when you're doing that by hand just take your time be patient and double check all right let's have a look at the next one which is the solution we'll see if we can't do that i might come back and check that after if we depending on how much time there is okay so um usually time's a big issue here so we're going to augment this okay all right so if i divide this by 2 or 3 i'll get 1 one one three if i were also to divide this by two i'd get two negative one negative one zero negative two two two sorry negative three so negative 6 divided by negative 3 is positive isn't it so we get a negative and negative so that's all right we can do that all right so this is the one that we're looking for so we've got to do row 2 row three is going to become row three plus row two i think so two minus two minus two zero i think it's going to be minus see that's annoying when i do that so just picking that back up so we too minus two plus one and minus zero zero minus two plus one one minus two plus one which is minus one and zero so when we come to write this again we'll have one one one three zero minus one minus one zero all right so the next thing we could like to do is target this 2 here so we'll take row 2 and from row 2 we'll subtract row 1 so we'll get 2 negative 1 negative 1 0.

[00:44:50]to 2 row 1 2 times 1 minus 2 times negative 1 minus two times one and minus two times zero so that's a zero minus three minus three zero so we get one one three four row two now we've got zero negative three negative three zero zero negative one y zero okay so the next thing we wanna do is uh consider this guy here so row three is going to become row three minus minus is a plus so minus one third row two a little bit annoying because i've made that did that division before but you just pick it up as you go so we've got zero negative one negative one zero minus a third minus a third and in row two we've got zero minus three minus three zero so that's out they'll cancel minus times minus is a plus so minus one plus one is a zero same thing is going to happen there and minus one third minus one third is zero now let's just double check this again um i think it's good for you to see it i'm going to i'm going to do it again i'm going to do it again i could put it into the technology but i'm not going to do that all right um like i hope everybody had a merry christmas and a happy new year by the way i hope you all are staying safe and all of that stuff so let's try this again please so we've got four minus six zero zero i don't know that swapping is going to help so row three will become row three minus six lots of row one at six rots of lower run okay so in this case here so we've got minus six six and six and zero so that's zero that's twelve 12 and 18.

[00:48:51]so we get 1 1 3 4 2 minus 2 0 0 which is what we're after 12 12 18.

[00:49:05]i was being too smart by half i think before all right so we're going to target this guy here so row r2 row 2 will become row 2 minus 4 times row 1 so we'll have 4 2 minus 2 0 minus 4 lots of one minus four lots of one minus four lots of one minus four lots of three so we get a zero two minus four is minus two uh minus two minus four is minus six and minus twelve so our new list will be one one one three zero minus two minus six minus twelve zero twelve twelve eighteen and we're identifying this one so row three will become row three adding uh six lots of row two so we've got zero twelve twelve eighteen plus six spots of naught plus six watts of negative two plus uh six lots of negative six i thought that's a mistake and six lots of 18 uh minus 12.

[00:51:17]so that's zero that becomes zero now that's problematic because it's -36 24 and this one is 72 64.

[00:51:55]so we get one one one three zero minus two minus six minus twelve zero zero uh sixty four is it now that's got an error in it i'm sure that's got an airiness so let me check this one minus two minus six so i'm i'm getting 0 12 18.

[00:53:05]you know what when it happens this way um don't shy off don't shy off have another go now it could be merry christmas i'm not going to lie but uh i'm getting a little bit worried okay so if we go for the four the minus six take tick tick okay so i'm happy with that now that's got that's that was the one we're after let's go for this one that so that's going to give us a zero now two minus four is minus two minus two minus four is minus six look at that that's where i made an error okay so that's minus uh six minus six and minus twelve okay so it's worthwhile going back over it when i transposed the columns i made a silly transpositional error in that i didn't include that negative with that 2 and you saw me have 3 goes at it because i did that so that was an initial error i was looking for something later but i hadn't made it later i'd made it earlier so anyway it's it's okay we got we got there in the end so if i now go for this guy row three will want to become row three now we're going to add two lots of row two so we're going to have 0 12 12 18 plus two lots and two lots of zero two lots of minus six again and two lots of minus twelve so they're minus uh 12 4 plus 12 is zero minus 12 12 plus 12 is zero uh 24 minus 24 plus 18 is uh negative six so what we've got here is we've got a one one one three zero minus six minus six minus twelve and then we've got a row of zeros so who can tell me what that is okay this does not compute because we can't have a variable being composed of uh adding up to zero so this line here tells us that this system is inconsistent so thank you for bearing with me it's important that we get things right in cheats because the best place to learn about what's happening in the final exam and uh who who knew so easy just to put one thing across okay so well done and um uh oh you can put up with me now again for one second all right thanks very much everyone i will see you next time bye now