course Mth 152 xwЙyڜassignment #019
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14:57:04 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 --> x y x(y) x^2 62 120 7400 3844 62 140 8680 3844 63 130 8190 3969 65 150 9750 4225 66 142 9372 4356 67 130 8710 4489 68 135 9180 4624 68 175 11900 4624 70 149 10430 4900 72 168 12096 5184 663 1439 95708 44059 a = 2.94 b = 51 y' = 3(70) + 51 = y' = 261
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15:06:44 ** 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 did not get that answer, I have reworked it but cannot figure out what I did wrong.
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15:29:57 **** 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 --> x = 811 y = 1104 x(y) = 90437 x^2 = 66225 a = (10(90437) - (811)(1104)) / (10(66225) - (811)^2) a = (904370 - 895344) / (662250 - 657721) a = 9026 / 4529 a = 1.99 b = (1104 - 2(811)) / 10 b = (1104 - 1622) / 10 b = -518/10 = -51.80 y= 2(x) - (-51.80)
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15:30:53 ** 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 rounded the 1.99 to 2. Also, I forgot to put 51 as a positive in the equation instead of a neg.
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15:54:08 **** 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 --> x = 15 y = 418 x(y) = 2374 x^2 = 55 y^2 = 30266 r = (6(2374) - (15)(418)) / Sq.(6(55) - (15)^2) * Sq. (6(30266) - (418)^2) r = 9.6
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15:55:29 **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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RESPONSE --> I came up with 7974/830 = 9.6
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15:59:12 **** 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 --> reg line y = 1.44x - .39 y = 1.44 (18) - .39 y = 25.92 - .39 = 25.53 %
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15:59:46 ** 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 don't come up with the exact figures, I am always a couple of tenths off.
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