#$&* course MTH 152 4/10 11 019. ``q Query 19*********************************************
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Given Solution: `aThe equation is obtained by substituting the weights for y and the heights for x in the formula for the regression line. You get y = 3.35 x - 78.4. To predict weight when height is 70 you plug x = 70 into the equation: y = 3.35 * 70 - 78.4. You get y = 156, so the predicted weight for a man 70 in tall is 156 lbs. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query problem 13.6.12 reading 83,76, 75, 85, 74, 90, 75, 78, 95, 80; IQ 120, 104, 98, 115, 87, 127, 90, 110, 134, 119 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In order to find the equation of a regression line for this group of IQ scores we must find the total combination of both x and y values. The sum of all x values in the table is 811. When we square this total we get 66225. The sum of all y values in the table is 1104. When we square this total we get 124060. The sum of x and y, which we get by multiplying each corresponding x and y value then adding each result together, is 90437. We can use these numbers to create an equation of [10(90437)-(811)(1104)]/[10(66225)-(811^2)]. This equation is equal to 1.993. For the next step in the problem, we must fashion another equation. This one reads [1104-(1.993)(811)]/10. This equation is equal to -51.23. Therefore, the equation of the regression line is y'=1.993x-51.23. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query problem 13.6.15 years 0-5, sales 48, 59, 66, 75, 80, 89 What is the coefficient of correlation and how did you obtain it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In order to find the coefficient of correlation, we must first find the sums of the values in the table. The sum of all x values in the table equal 15. When we square this value we get 55. The sum of all y values in the table equal 418. The sum of x and y together equals 1186. In addition, we must find that Ey^2 is equal to 16. We need also remember that n is equal to 6, as 6 is the number of pairs in the data. We can use all these sums to construct an equation and discover that the coefficient of correlation is .996. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a **STUDENT SOLUTION: X Y XY X^2 Y^2 0 48 0 0 2304 1 59 59 1 3481 2 66 132 4 4356 3 75 225 9 5626 4 80 320 16 6400 5 90 450 25 8100 Sums= 15 418 1186 55 30266 The coefficient of the correlation: r = .996 I found the sums of the following: x = 15, y = 418, x*y = 1186, x^2 = 55 n = 6 because there are 6 pairs in the data I also had to find Ey^2 = 30266 I used the following formula: r = 6(1186) - 15(418) / sq.root of 6(55) - (15)^2 * sq. root of 6(30266) - (418)^2 = 846 / sq. root of 105 * 6872 = 846 / sq. root of 721560 = 846 / 849.4 = .996 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query problem 13.6.24 % in West, 1850-1990, .8% to 21.2% What population is predicted in the year 2010 based on the regression line? What is the equation of your regression line and how did you obtain it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In order to find the equation of a regression line for the population of the year 2010 we must first find the sums of x and y. The sum of x is 56. This value squared is 560. The sum of y is 77.7. This value squared is 1110.43. The sum of x and y together is 786.4. Using these sums we find that the value of a will be equal to 1.44, the value of b will be equal to -.39, and the value of r is equal to .99. If we look to the table and note the pattern of the regression line, we can conclude that the next number in the x sequence will be 16. This gives us an equation of y' = 1.44(16) - .39 = 22.65. When we solve the equation we discover that the population of 2010 will increase by 22.65%. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!