Question
Problem #5: Given the following five pairs of (rV) values_ 10 4 2 4Determine the least squares regression line_ (Be sure to save YOur unrounded values of bo and b1 for use in Problem #6 below:) (b) Draw the least squares regression line accurately On1 scatterplot Then look to see which (x pairs are above the regression line. Then add up the y-values for all of the (x pairs that fall above the regression line: For example. if you draw your least squares regression line accurately Onl scatterplot
Problem #5: Given the following five pairs of (rV) values_ 10 4 2 4 Determine the least squares regression line_ (Be sure to save YOur unrounded values of bo and b1 for use in Problem #6 below:) (b) Draw the least squares regression line accurately On1 scatterplot Then look to see which (x pairs are above the regression line. Then add up the y-values for all of the (x pairs that fall above the regression line: For example. if you draw your least squares regression line accurately Onl scatterplot, and you find that the first tWO pairs [i.e.. (2,8) and (3.6)] are above the regression line_ then since the SUI of the two corresponding y-values is 8 + 6 = 14, YOu would enter 14 into the answer box: enter the values of bo and b1 (in that order), separated by comma (numbers correct to decimals) Problem #S(a): sum of y-values for points above the regression line Problem #5(b):
Answers
Fitting a line to Data To find the least squares
regression line $y=a x+b$ for a set of points
$\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)$
you can solve the following system for $a$ and $b$
$$
\left\{\begin{array}{c}{n b+\left(\sum_{i=1}^{n} x_{i}\right) a=\left(\sum_{i=1}^{n} y_{i}\right)} \\ {\left(\sum_{i=1}^{n} x_{i}\right) b+\left(\sum_{i=1}^{n} x_{i}^{2}\right) a=\left(\sum_{i=1}^{n} x_{i} y_{i}\right)}\end{array}\right.
$$
In Exercises 55 and $56,$ the sums have been evaluated.
Solve the given system for $a$ and $b$ to find the least
squares regression line for the points. Use a graphing
utility to confirm the result.
$$
\left\{\begin{array}{c}{6 b+15 a=23.6} \\ {15 b+55 a=48.8}\end{array}\right.
$$
Okay. So I multiply my first equation by -15 x six. And so that gives me a -15 b. Which will then cancel out with the 15 B from the second equation those cancel out. And so I'm going to have 55 times six plus 15 square over six times a prepare. That's a minus equals minus 59 Us 48.8 which is -10.2. So I end up getting equals minus 10.2 times six, divided by 55 times 6 -15 square. And this gives me -0.58 approx. So from this second sulphur B I can use this equation here all the way down here, So it's gonna be 23.6 -1580, divided by six. Okay. And that will give me approximately 5.39. So those are the values of A. And B.
We need to solve the following system for A and B. So we can rewrite the equation of the regression in Y equals a times X plus B. So first we're going to multiply our top equation by negative two. So this becomes negative 10 B minus 20 A. And that becomes negative 40.4. So RBS cancel out we're left with 10 A. Is equal to 50.1 minus 40.4. Comes out to be 9.7. We divide everything by 10 and we get a is equal to 9.7, divided by 10 is 0.97. So now we take that 0.97 and plug it back into one of the original equations to solve for B. So we have five B plus 10 times 0.97 is equal to 20.2. So that's five B Plus 9.7 is equal to 20.2. We subtract 9.7 from both sides and we get 5B is equal to 20.2 -9.7. That equals 10.5. We divide both sides by five, and we get B is equal to 2.1. So we can now write the equation of the linear Equal Y is equal to 0.97 x plus 2.1.
This is probably number nine. We are given a set of data in X and Y and a test with performing a regression analysis. So to begin you will draw a scatter diagram, then select two points from that scattered diagram and find the equation of the line through those points. And finally you can plot this line on your scattered diagram. Next we will be determining the true least squares regression line to do this, we will use these equations for our equation coefficients and finding these five values from our data. So in doing so we get some of our X values is zero. The summer of the Y values is sex. Some of our X times Y is 22. Some of our the squares of X is 10 and the sum of the squares of why he's 58. So using these two equations we can find our regression line. You get y hat equal to first RB one which we find is 2.2 times X plus our slope which is plus our intercept, which is B zero. So plus 1.2. So we can now find our residuals squared. We do this by why my my hat minus Y squared or why hat is the predicted value from our regression equation. And why is our exact data points? So taking the sum of all of these for all of our data points, we find our residuals, some of these squares of our residuals to be equal to 2.4. This value should be significantly lower than your regression than your arbitrary regression line as the tourism regression line is meant to minimize this.
My system set up And I multiply the first equation by -2. That gives me -10 b. And that ends up canceling out with the second equation, so then I have 10 A equals 25.6 -23.4, which is 2.2, And so is just going to equal 2.2 divided by 10, which is 0.22. So from this I can find the value of B. I can just use this equation 11.7 -10 A, divided by five. So that gives me 1.9. So those are the values of the n B.