course Mth 152 kέfpwx{ܖѬassignment #019
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14:36:58 query problem 13.6.9 wt vs ht 62,120; 62,140; 63,130; 65,150; 66,142; 67,13068,175; 68,135; 70,149; 72,168. Give the regression equation and the predicted weight when height is 70.
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RESPONSE --> 149 lbs (as given in the table)
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14:54:35 ** The 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. **
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RESPONSE --> I understand that I should have used the eqation to find the answer to this problem
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15:40:59 **** 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
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RESPONSE --> a=10(90437)-(811)(1104)/10(136225)-(811)2= 9026/704529=.01281 b= 1104-(.01281)(811)/10=109.36 y'=.01281x+109.36
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15:41:45 ** 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. **
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RESPONSE --> I got off on the math a little, but I understand how it is set up
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16:08:16 **** 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?
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RESPONSE --> r=6(1156)-(15)(418)/sqrt [6(55)-(15)^2]*sqrt[6(30266)-(418)^2] =666/849.36 =.784
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16:08:36 **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 **
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16:22:50 **** 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?
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RESPONSE --> r=8(782.4)-(56)*(84.9)/sqrt[8(547)-(56)^2]*[8(1173.76)-(84.9)^2] =1504.8/1644.368=.915
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16:24:03 ** STUDENT 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. **
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RESPONSE --> I understand that I should have found a and b to use the y' eqaution to get the solution
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