136

#$&*

course Mth 152

1158 pm 73113

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution.

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

019. ``q Query 19

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Question: `q 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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

62*120=7440 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 95748 44059

10(95748)-663(1439)/10(5184)-(663)^2

957480-954057/440590-439569

3423/1021

A=3.35

1439-3.35(663)/10

1439-2221.05/10

B=-78.21

Y=3.35x-78.21

=3.35*70-78.21

156.29

confidence rating #$&*:

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3

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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. **

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Self-critique (if necessary):

Ok!

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Self-critique Rating:

3

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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:

83 120 9960 6889

76 104 7904 5776

75 98 7350 5625

85 115 9775 7225

74 87 6438 5476

90 127 11430 8100

75 90 6750 5625

78 110 8580 6084

95 134 12730 9025

80 119 9520 6400

811 1104 90437 66225

10(90437)-811(1104)/10(66225)-811^2

904370-895344/662250-657721

9026/4529

a=1.99

1104-1.99*811/10

1104-1613.89/10

b=-50.989

y’=1.99x-50.989

confidence rating #$&*:

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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

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Self-critique (if necessary):

ok

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Self-critique Rating:

3

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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:

0 48 0 0 2304

1 59 59 1 3481

2 66 132 4 4356

3 75 225 9 5625

4 80 320 16 6400

5 90 450 25 8100

15 418 1186 55 30266

R=6*1186-6270/sqrt6*55-15^2*sqrt6*30266-418^2

846/sqrt] 330-225*sqrt181596-174724

846/sqrt721560

846/849

R=.996

confidence rating #$&*:

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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 **

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Self-critique (if necessary):

ok

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Self-critique Rating:3

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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:

0 0.8 0 0 0.64

2 2.6 5.2 4 6.76

4 5.0 20.0 16 25.0

6 7.7 46.2 36 59.29

8 10.0 80.0 64 100.0

10 13.3 133.0 100 176.89

12 17.1 205.2 144 292.41

14 21.2 296.8 196 449.44

56 77.7 786.4 560 1110.43

A=8(786.4)-56(77.7)/8(560)-56^2

6291.2-4351.2/4480-3136

1940/1344

A=1.44

77.7-1.44*56/8

77.7-80.64/8

b=-0.3675

Y’=1.44*16-0.3675

Y’=22.67

R=8(786.4)-56*77.7/sqrt8(560)-(56^2)*sqrt8*1110.43-(77.7^2)

6291.2-4351.2/4480-3136*8883.44-6037.29

1940/37*53

1940/1961

R=.989

confidence rating #$&*:

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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. **

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

"

Self-critique (if necessary):

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Self-critique rating:

"

Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Good work. Let me know if you have questions. &#