Query 19

#$&*

course Mth 152

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: y = 3.35 * 70 - 78.4

y = 156

confidence rating #$&*: 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:

y' = 1.993x - 51.23 regression line

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

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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: 846 / 849.4 = .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 **

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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?

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Your solution: X value for population at 16

y' = 1.44(16) - .39 = 22.65 = 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. **

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Self-critique (if necessary): ok

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Self-critique Rating: 3

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&#Very good work. Let me know if you have questions. &#