Assignment 19

course Mth 152

xwЙyڜassignment #019

019. `query 19

Liberal Arts Mathematics II

04-19-2008

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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.

You appear to be using the right quantities; chances are it's a simple error. You're welcome to send me the details of your calculation, but if you're getting the right answers on similar problems it's probably not necessary.

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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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One of us miscalculated the appropriate value of x--it's either 16 or 18 and I don't have full information here, so it might have been me. However that's the only difference between the solutions.