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10.10 marks)Briefly describe what is principal component analysis (PCA) algorithm. (2 marks) Briefly describe how to obtain the Ist principal component: 2 marks) Gi...

Question

10.10 marks)Briefly describe what is principal component analysis (PCA) algorithm. (2 marks) Briefly describe how to obtain the Ist principal component: 2 marks) Given a dataset consisting of 1000 samples and 30 variables, what are the dimen- sions of the transformed dataset using the Ist 2nd and 3rd principal component values? (2 marks) Given a zero-mean dataset (as shown below) consisting of four variables A B, C, and D, calculate the transformed dataset using the 1st principal component value

10. 10 marks) Briefly describe what is principal component analysis (PCA) algorithm. (2 marks) Briefly describe how to obtain the Ist principal component: 2 marks) Given a dataset consisting of 1000 samples and 30 variables, what are the dimen- sions of the transformed dataset using the Ist 2nd and 3rd principal component values? (2 marks) Given a zero-mean dataset (as shown below) consisting of four variables A B, C, and D, calculate the transformed dataset using the 1st principal component values, ie. PC_1. marks) Instance Instance Instance 0.367 0.500 0.000 0.033 ~0.134 ~0.700 00o 0.033 -0.233 0.200 0.000 -0.066 PC_1 = (-0.331,-0.943 , 0.000, 0.013)



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[M] A Landsat image with three spectral components was made of Homestead Air Force Base in Florida (after the base was hit by Hurricane Andrew in 1992 ). The covariance matrix of the data is shown below. Find the first principal component of the data, and compute the percentage of the total variance that is contained in this component.
$$
S=\left[\begin{array}{rrr}{164.12} & {32.73} & {81.04} \\ {32.73} & {539.44} & {249.13} \\ {81.04} & {249.13} & {189.11}\end{array}\right]
$$

Broken on 36 since the standard deviation off the city's is signal over squirrels of end and the smaller cities will have a greater variation. Therefore, it would make sense that and the small cities would both be the most safe and least and let's see.

Were given a data table of X and Y coordinates, and our first task was to use technology to create a scatter plot. This was done create created using excel to make our scatter plot and excel also through the correlation function calculated the are value for us negative 0.98336 That's an awfully good fit. Also calculated for us in the equation for the line of best fit in the slope. Here is negative 0.5 88 0.5844 Yeah, it's a pretty good fit now. Our second task was to repeat this process but eliminate this data point here the 20 negative 16 and we can kind of predict what's gonna happen when we look at this. The 20 negative 16 is very much in line with the rest of the data. So when we remove that data point we could is a pretty good guess. It's gonna slightly, uh, decreased the our value. But let's go ahead and move this over and take a look. So here we have our are five data points instead of just the it. So the whole six we've removed this data 60.20 negative 16. And what we see is the our value indeed went down just a little bit. Right Are is slightly less. Then the original will call this one our one and this one are two. It went down just a little bit. The slope also went down a little tiny bit, shallower, just the tiniest bit. So we can say there's not really an appreciable difference there still approximately the same.

This is problem number 11, We are given a set of data points in X and Y. Enter test with completing a regression analysis. To begin, you will draw a scatter diagram, select two points from that scatter diagram and find the line, Find the equation of the line that passes through those two points. You can then plot this on your scatter diagram. Next we will find the actual least squares regression line. Since we have you have found an arbitrary one. Using those two points to do this, we will use these two equations and these values, which will be substituted in for these equations. These are derived from your textbook in the case that we are not given a mean and standard deviation. So from our data we get the following, we have the sum of X is 200. The sum of why is 439 be some of our products X times y is 16,000, 840. The sum of the squares of our X values is 900 9000, sorry. And the sum of the squares of ry values is 39,000 and 95. So using these values substituting them into these equations, we get the following for our regression equation. We have a white hat equal to B. One which is our slope. So negative 0.72 times X plus B zero which is our intercept 116 6. So based on this we can find our residuals recall that. The equation for a residual is why hat minus. Why in this case it tasked us with finding the sum of the squares of the residuals. So if we take this square it and find the sum of all of them remembering that y hat is the predicted value for a certain X value. And why is the real data points? We get a value of 32.4. If you compare this to the residuals that you acquire from your own line that you've drawn, you should find that this value is much lower. And that is because the regression line, the least squares regression line is meant to minimize the squares of our residuals.

Alright, here we are given a scatter plot and asked to determine whether or not there's linear correlation that exists here. So we're going to be using some subjective knowledge to start. So by taking a look at this visually, what might we conclude from this? I personally think that there probably is no linear correlation, but we also have this point up here at 10 10. So we may end up running into some issues there because a computer may see a line that exists somewhere through here connecting that one, making it appear linear, although the majority of the data sits within this lower part of the quadrant. So it's my guess that 10 10, the fact that it's an outlier may skew our data a little bit. But all in all, I don't think that there is any linear correlation that exists Now. Let's go ahead and allow our graphing calculators to tell us what it thinks by inputting each of these coordinate points into our stat edit function and then running our linear, refreshing T test. In doing that, we will find that we have a correlation coefficient, which is equal to 0.9056 and that suggests to us that there's actually really high and positive linear correlation going on. We also have a P value. That's output it here of 0.0 308 What that does is tells us that this correlation coefficient is actually significant at the 1% level, which is outstanding. But considering the fact that we really don't think that there is any linear correlation, so let's go ahead and repeat the previous steps. But let's remove this outlier the 10.10 10 from our data on or graphing calculator doing the same thing again again, we're probably going to conclude that no, there is no linear correlation. But then running your linear regression T test as well. Let's see what we get there. Doing that. We end up getting a correlation coefficient equal to zero, and we have a P value equal to one. So what this tells us is that there is no correlation at all between these values, and the P value tells us that that's significant because if we were to have a null hypothesis where our correlation is equal to zero and our alternative where our correlation coefficient is not equal to zero. The fact that our P value is equal to one means that we actually failed to reject the null. So we cannot reject that in favor of the alternative, which means we must assume that there is, in fact, no correlation in this data set. And given all of this, let's start. Let's try to examine why this is meaningful. Looking at the effect of the single pair of values is outlier of 10 10. We can see that our results were incredibly skewed to suggest to us that we first had our, um, in our value of 9056 So 90% correlation right there, when in fact we actually had none. So it looks as though the effect of a single pair of values can actually be quite extreme. So this is where we need to start looking at our data in terms of where might the computer make the mistake? We need to be a little bit smarter than it and think ahead so that we can remove those data points that may give us the wrong conclusions.


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