#$&* course Mth 152 Dec 1 - 9:55pm 019. ``q Query 19
.............................................
Given Solution: `aThe equation is obtained by substituting the weights for y and the heights for x in the formula for the regression line. You get y = 3.35 x - 78.4. To predict weight when height is 70 you plug x = 70 into the equation: y = 3.35 * 70 - 78.4. You get y = 156, so the predicted weight for a man 70 in tall is 156 lbs. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): “I’m looking at pg 848 and I don’t see where you are getting 3.35? I see al the numbers from above and I’m looking at the height of 70 (x) which corresponds with 149 pounds (y)?” I understand how you worked the problem, I just don’t see where the information came from? ------------------------------------------------ Self-critique Rating: 1 ********************************************* Question: `q Query problem 13.6.12 reading 83,76, 75, 85, 74, 90, 75, 78, 95, 80; IQ 120, 104, 98, 115, 87, 127, 90, 110, 134, 119 Find the equation of the regression line. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Pg 848 10 individuals = n Reading (x) = 811 Sum x^2 = 66225 IQ (y) = 1104 Sum y^2 = 124060 Sum xy = 90437 10 (90437) - (811) (1104) / 10(66225) - (811^2) = 1.993 904370 - 895344 / 662250 - 657721 = 1.993 9026 / 4529 = 1.993 1.99 1104 - (1.993)(811)/10 = -51.23 The regression line is 1.993x-51.23 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a n = 10 sum x = 811 sum x ^2 = 66225 sum y = 1104 sum y^2 = 124060 sum xy = 90437 a = [10(90437) - (811)(1104)] / [10(66225) - (811^2)] = 1.993 a = 1.99 b = [1104 - (1.993)(811) / 10 = -51.23 y' = 1.993x - 51.23 is the eqation of the regression line. ** STUDENT QUESTION How did you get sum x ^2 = 66225??? Is it not 811 * 811 = 657721?How did you come up with sum y^2 = 124060??? Is it not 1104 * 1104 = 1218816?I worked it out, can you tell me where I went wrong??? And I will try to rework the problem. INSTRUCTOR RESPONSE You didn't distinguish between sum x^2 and (sum x)^2. Sum x^2 means you figure out x^2 for every value of x, then add them. Remember that exponentiation precedes addition. (sum x)^2 means you add all the x values then square them. The same comment applies to sum y^2 vs. (sum y)^2. You didn't ask, but sum xy can also be confusing: • Sum xy means multiply each x value by the corresponding y value, then add the products. This is order of operations: multiplication before addition &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I wondered the same thing as the student question but I get it. ------------------------------------------------ Self-critique Rating: 1 ********************************************* Question: `q Query problem 13.6.15 years 0-5, sales 48, 59, 66, 75, 80, 89 Find the correlation coefficient. What is the coefficient of correlation and how did you obtain it? Year (x) 0 1 2 3 4 5 Sales (y) 48 59 66 75 80 90 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: X - 0 1 2 3 4 5 = 15 Y - 48 59 66 75 80 90 = 418 X*Y 0 59 132 225 320 450 = 1186 X^ 0 1 4 9 16 25 = 55 Y^ 2304 3481 4356 5626 6400 8100 = 30266 The coefficient of the correlation (6 numbers) = 0.995 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a **STUDENT SOLUTION: X Y XY X^2 Y^2 0 48 0 0 2304 1 59 59 1 3481 2 66 132 4 4356 3 75 225 9 5626 4 80 320 16 6400 5 90 450 25 8100 Sums= 15 418 1186 55 30266 The coefficient of the correlation: r = .996 I found the sums of the following: x = 15, y = 418, x*y = 1186, x^2 = 55 n = 6 because there are 6 pairs in the data I also had to find Ey^2 = 30266 I used the following formula: r = 6(1186) - 15(418) / sq.root of 6(55) - (15)^2 * sq. root of 6(30266) - (418)^2 = 846 / sq. root of 105 * 6872 = 846 / sq. root of 721560 = 846 / 849.4 = .996 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* Question: `q Query problem 13.6.24 % in West, 1850-1990, .8% to 21.2% What population is predicted in the year 2010 based on the regression line? What is the equation of your regression line and how did you obtain it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I can’t find this question in the book, I don’t understand where you are getting the stats from? confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aSTUDENT SOLUTION: Calculating sums and regression line: n = 8 sum x = 56 sum x^2 = 560 sum = 77.7 sum y^2 = 1110.43 sum xy = 786.4 a = 1.44 b = -.39 r = .99 In the year 2010 the x value will be 16. y' = 1.44(16) - .39 = 22.65. There is an expected 22.65% increase in population by the year 2010. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 0 ------------------------------------------------ Self-critique Rating: 0 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!