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Twelve families have been referred to a counselor, and she has rated each of them on a scale that measures family cohesiveness. Also, she has information on family income and number of children currently living at home. Take family cohesion as the dependent variable. Family Cohesion Score Income Number of Children A 10 30,000 5 B 10 70,000 4 C 9 35,000 4 D 5 25,000 0 E 1 55,000 3 F 7 40,000 0 G 2 60,000 2 H 5 30,000 3 I 8 50,000 5 J 3 25,000 4 K 2 45,000 3 L 4 50,000 0 a. Find the multiple regression equations (unstandardized). b. What level of cohesion would be expected in a family with an income of $20,000 and six children? c. Compute beta-weights for each independent variable and compare their relative effect on cohesion. Which was the more important factor? d. Compute R 2 . e. Write a paragraph summarizing your findings.

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Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836

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Chapter
Section
BuyFindarrow_forward

Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836
Chapter 13, Problem 13.3P
Textbook Problem
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Twelve families have been referred to a counselor, and she has rated each of them on a scale that measures family cohesiveness. Also, she has information on family income and number of children currently living at home. Take family cohesion as the dependent variable.

Family Cohesion Score Income Number of Children
A 10 30,000 5
B 10 70,000 4
C 9 35,000 4
D 5 25,000 0
E 1 55,000 3
F 7 40,000 0
G 2 60,000 2
H 5 30,000 3
I 8 50,000 5
J 3 25,000 4
K 2 45,000 3
L 4 50,000 0

a. Find the multiple regression equations (unstandardized).

b. What level of cohesion would be expected in a family with an income of $20,000 and six children?

c. Compute beta-weights for each independent variable and compare their relative effect on cohesion. Which was the more important factor?

d. Compute R 2 .

e. Write a paragraph summarizing your findings.

To determine

(a)

To find:

The unstandardized multiple regression equation with family cohesion as the dependent variable.

Explanation of Solution

Given:

The table for a sample of twelve families is given below.

Family Cohesion Score Income Number of Children
A 10 30, 000 5
B 10 70, 000 4
C 9 35, 000 4
D 5 25, 000 0
E 1 55, 000 3
F 7 40, 000 0
G 2 60, 000 2
H 5 30, 000 3
I 8 50, 000 5
J 3 25, 000 4
K 2 45, 000 3
L 4 50, 000 0

Formula used:

Let the data values be Xi’s.

The formula to calculate mean is given by,

X¯=i=1NXiN

The formula to calculate standard deviation is given by,

s=(XiX¯)2N

Where, N is the population size.

The formula for calculating the correlation coefficient r is given as,

rxy=(XiX¯)(YiY¯)[(XiX¯)2][(YiY¯)2]

Where, X and Y are the two variables.

The formula to calculate the partial slopes for the independent variables is given by,

b1=(sys1)(ry1ry2r121r122)

And,

b2=(sys2)(ry2ry1r121r122)

Where, b1 is the partial slope of X1 and Y, b2 is the partial slope of X2 and Y, sy is the standard deviation of Y, s1 is the of standard deviation of the first independent variable X1, s2 is the of standard deviation of the second independent variable X2, ry1 is the bivariate correlation between Y and X1, ry2 is the bivariate correlation between Y and X2, r12 is the bivariate correlation between X1 and X2.

The least square multiple regression line is given by,

Y=a+b1X1+b2X2

The formula to calculate the Y intercept of the regression line is given by,

a=Y¯b1X¯1b2X¯2

Calculation:

From the given information,

The cohesion scores is the dependent variable. Thus, it is represented as Y.

The rest of the variables, i.e. income and number of children are the independent variables and are represented as X1 and X2 respectively.

The mean is given by,

X¯=i=1NXiN

For income, substitute 12 for N, 30, 000 for X1, 70, 000 for X2 and so on in the above mentioned formula,

X¯=30000+70000+...+5000012=51500012=42917

The mean of income is 42917.

For children, substitute 12 for N, 5 for X1, 4 for X2 and so on in the above mentioned formula,

X¯=5+4+...+3+012=3312=2.75

The mean of children is 2.75.

For cohesion score, substitute 12 for N, 10 for X1, 10 for X2 and so on in the above mentioned formula,

X¯=10+10+...+2+412=6612=5.5

The mean of cohesion score is 5.5.

Consider the following table of sum of squares of income and Cohesion Score.

Family Income (Xi) (XiX¯) (XiX¯)2 Cohesion Score (Yi) (YiY¯) (YiY¯)2 (XiX¯)(YiY¯)
A 30, 000 12,917 166848889 10 4.5 20.25 58126.5
B 70, 000 27, 083 733488889 10 4.5 20.25 121873.5
C 35, 000 7,917 62678889 9 3.5 12.25 27709.5
D 25, 000 17,917 321018889 5 0.5 0.25 8958.5
E 55, 000 12, 083 145998889 1 4.5 20.25 54373.5
F 40, 000 2,917 8508889 7 1.5 2.25 4375.5
G 60, 000 17, 083 291828889 2 3.5 12.25 59790.5
H 30, 000 12,917 166848889 5 0.5 0.25 6458.5
I 50, 000 7, 083 50168889 8 2.5 6.25 17707.5
J 25, 000 17,917 321018889 3 2.5 6.25 44792.5
K 45, 000 2, 083 4338889 2 3.5 12.25 7290.5
L 50, 000 7, 083 50168889 4 1.5 2.25 10624.5
Totals 515000 (XiX¯)2=2,322,916,668 66 (YiY¯)2=115 (XiX¯)(YiY¯)=22500

Substitute 30,000 for X1 and 42,917 for X¯ in (X1X¯).

(X1X¯)=(33,00042,917)(X1X¯)=12,917

Square the both sides of the equation.

(X1X¯)2=(12,917)2(X1X¯)2=166848889

Proceed in the same manner to calculate (XiX¯)2 for all the 1iN for the rest data and refer table for the rest of the (XiX¯)2 values calculated. Then the value of (XiX¯)2 is calculated as,

(XiX¯)2=166848889+733488889+...+50168889=2322916668

The formula for calculating the correlation coefficient r is given as,

rxy=(XiX¯)(YiY¯)[(XiX¯)2][(YiY¯)2]

Substitute 22500 for (XiX¯)(YiY¯), 2322916668 for (XiX¯)2 and 115 for (YiY¯)2 in the above mentioned formula.

rxy=22500(2322916668)(115)rxy=0.04

The correlation coefficient between income and cohesion score is 0.04.

The formula to calculate standard deviation is given by,

s=(XiX¯)2N

For income, substitute 2322916668 for (XiX¯)2 and 12 for N in the above mentioned formula.

s=232291666812=193576389=13913.17

The standard deviation of income is 13913.17.

For cohesion score, substitute 115 for (YiY¯)2 and 12 for N in the above mentioned formula.

s=11512=9.58=3.09

The standard deviation of cohesion score is 3.09.

Consider the following table of sum of squares of number of children and cohesion score.

Family Number of Children (Xi) (XiX¯) (XiX¯)2 Cohesion Score (Yi) (YiY¯) (YiY¯)2 (XiX¯)(YiY¯)
A 5 2
To determine

(b)

To find:

The expected cohesion in a family with an income of $20, 000 and six children.

To determine

(c)

To find:

The beta weights for each given independent variable and the effect of the variable on turnout.

To determine

(d)

To find:

The coefficient of multiple determinations.

To determine

(e)

To explain:

The relationship among the given three variables.

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