Variation and Prediction Intervals
In Exercises 17–20, f...

Question

Variation and Prediction Intervals

In Exercises 17–20, find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. In each case, there is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions.

Weighing Seals with a Camera The table below lists overhead widths (cm) of seals measured from photographs and the weights (kg) of the seals (based on “Mass Estimation of Weddell Seals Using Techniques of Photogrammetry,” by R. Garrott of Montana State University). For the prediction interval, use a 99% confidence level with an overhead width of 9.0 cm.

Accepted Answer

Welcome back, everyone. In this problem, a researcher requires the heights in centimeters and corresponding arm spans in centimeters of 5 individuals. We want to calculate the explained and unexplained variations. Now, before we find these variations, we'll need to compute the linear regression line. Recall OK. That the the linear regression line, Y, or let's say Y hat is equal to A plus B X, OK? Where Where or sorry B. is equal to the sum of x y values minus the sum of x values multiplied by the sum of y values divided by n multiplied by the sum of Xred values minus the sum of X values squared. And, OK. Uses B in that A is equal to the sum of Y values minus B multiplied by the sum of X values all divided by n. So in this case, if we can find our values of A and B, then we'll be able to compute our regression line that we can use to calculate the explained and unexplained variations. Now, let's set up a table to help us figure out these A and B values. So, now we have our X and Y values, OK? And then we have our X squared values and our XY values, OK? Now I'm intentionally making this table short because we'll be extending it in a few to help us find our variations, OK? But for now, let's just keep it pretty smart. So if we go back to our table to get our X and Y values, uh, we'll have X 160 and Y 162. Likewise, we'll have 165 and 168, uh, 170 and 171, 175 and 176, and 180 and 181, OK? Bearing in mind that we can, we would let X be equal to the heights, OK? And why be equal to the corresponding arm spans? Now, in this case, if we go ahead, let me just uh bring, make a little space here. OK. So, for X2d, we're gonna have 160 squared for the first one, that's 25,600 and XY would be 25,920. Next, X squared for 165 is 27225 and XY is 27, 720. For 170 and 171, we have 2, 28,900 and 290,070. 175 and 176 would be 30,625 and 30,800. And then 180 and 181. And 32,400 and 32,580 respectively. So now, if I were to put an extra line here. Then the sum of, and we sum our values, OK. Then the sum of X values would be 850, sum of Y values 858. The sum of Xgra values 144,750, and the sum of Xy values, 146,090. So now we can go ahead and compute our regression line. Because this means then that be. Is going to be equal to n, which is 5 or sample size, multiplied by the sum of Xy values 146,090. OK, minus the sum of X value is 850 multiplied by the sum of y values 858 all divided. By the sum of N multiplied by the sum of Xsquared values, 5 multiplied by 144,750, OK, minus the sum of X values 850 squared. And now when we go ahead and solve here, we should get B to be equal to 11, 1150 divided by 1250, which equals 0.92. And if B is equal to 0.92, that means A then will be equal to the sum of Y values 858 minus B multiplied by the sum of X values. All divided by N, OK, which again, when we work that out, that's 76 divided by 5, which equals 15.2. So if A equals 15.2 and B equals 0.92, OK, then that means for a regression line, Y hat will be equal to 15.2 plus 0.92 X. Now this is very important because this is what we're going to use to figure out our variations. Now, let's start with our explained variations, OK? Now for the explained variation. And It's going to be equal to the sum of each y value minus the mean Y value, OK, squared, OK? Uh, in other words, it's the sum of the squares of the differences between each Y value and their means. No, we need the meanwhile value, OK. And we know that the meanwhile value will be equal to the sum of wire values divided by the number of values. That's 858 divided by 5, which is equal to 171.6. So now we can compute each each Y value, OK, and just use that to help us figure out the difference for each and then square them to help us find the explained variation. Because now if we add that as a column to our table here, let me add that in red. And let me extend our table a little bit, OK. Then no, we're going to find our Yhat value using our formula for our regression line, OK? Find the difference between that value and the mean, OK. And then we're going to square our result, OK? So it's gonna be Y hat minus the mean squared. Now, for example, uh, when, when X is 160, that means we would substitute 160 into our formula for Y hat, that's 15.2 plus 0.92 multiplied by 160, and thus we would get 162.4. That means the difference between that and our. Um, mean, which is, uh, like we said earlier, 171.6, would be 9.2, and when we square that, it's 84.64. Likewise, when Y is 168, sorry, when X is 165, then Y would be 167. The difference between Y hat and the mean would be -4.6, and when we square that, it's 21.16. When X equals 170. Then Y equals 1 equals 171.6. The difference between that and the mean would be 0, since the mean is also 171.6 and thus when we square that we get 0. When it's 1 X is 175, Y hat is 176.2. Our difference is 4.6, and our square is 21.16. And finally, when it's 180, Y hat is 180.8. The the uh difference is 9.2, and the, the square is 84.64. So now when we go ahead and sum all our squares for our differences, that's what would give us our uh explained variation. You should find that is summing all the values in this column. You should find that it equals 211.6, OK? Therefore, the explained variation. OK. Explained variation. Equals 211.6. Now finally, what about our unexplained variation? Let's put that in blue. Well, the unexplained variation. Yeah. Unexplained variation is equal to the sum. Of y values minus Y have values squared, OK? In other words, it's the sum of the squared residuals. So let's go back to our table and now again I'm going to add another column for each Y value minus the corresponding Yha value which we've already calculated. And then a final column for the square of those residuals Y minus Y hath squared. No, starting with our Y minus Y values. When Y is 162, then our residual would be equal to negative 0.4, and the square of that is 0.16. Next, when it's 168, the residual is 1, and the square of that is 1. When it's 171, the residual is -0.6, whose square is 0.36. When it's 176, that's negative 0.2 and 0.04, and then when it's 181, that's 0.2 and 0.04. Therefore, this tells us then that the sum of the squared residuals when we add all of these values up will be equal to 1.60. So our unexplained variation is equal to 1.60. So, in summary, our explained variation is 211.6 and our unexplained variation is 1.60 approximately. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Explained Variation

Unexplained Variation

Prediction Interval

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