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Solutionsto Statistics Problems

P.408, #7

X

(Effects of Number of Siblings)

Y

(Happiness of the Child)

1

5

0

3

2

4

4

7

2

5

3

9

1

8

1

4

  1. Draw a scatter plot of the data.

  1. Calculate the regression slope and Y-intercept.

X

Y

XY

X^2

Y^2

(Effects of Number of Siblings)

(Happiness of the Child)

&nbsp

&nbsp

&nbsp

1

5

5

1

25

0

3

0

0

9

2

4

8

4

16

4

7

28

16

49

2

5

10

4

25

3

9

27

9

81

1

8

8

1

64

1

4

4

1

16

14

45

90

36

285

RegressionSlope

Slope=(NΣXY – (ΣX)(ΣY)) / (NΣX2&nbsp-(ΣX)2)

Slope= [8(90) – (14)(45)/8(36)-(14)^2]

Slope= [(720 – 630)/(288-196)]

Slope= [90/92]

Slope= 0.978

Y-intercept

  1. Draw the regression line on the scatter plot.

  1. Predict the happiness of an only child and of a child with two siblings.

Thehappiness of an only child and of a child with two siblings ismoderately correlated.

  1. Find the coefficients of determination and non-determination. What do they mean?

CorrelationCoefficient ( r ) = N x ∑ XY – ( ∑ X ) ( ∑ Y ) / √ N x ( ∑X2&nbsp-( ∑ X )2&nbsp√N x ( ∑ Y2&nbsp-( ∑ Y )2&nbsp

R= (8 x 90) – (14) (45) / √8 (36-14)^2 √ 8 (285-45)^2

R= 0.58760

Coefficientof Determination = r x r

Coefficientof Determination = 0.58760 x 0.58760

Coefficientof Determination = 0.345

• 34%of the variability in the Y scores can be predicted from the relationwith X

Coefficientof Non-determination = 1 – r^2

Coefficientof Non-determination = 1 – 0.345

Coefficientof Non-determination = 0.655

• 65%of the variability in the Y scores can be predicted from the relationwith X

  1. Construct an analysis of variance table and perform an F-test of the significance of the regression.

SUMMARY

Groups

Count

Sum

Average

Variance

X

8

14

1.75

1.642857

Y

8

45

5.625

4.553571

ANOVA

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

60.0625

1

60.0625

19.38617

0.000601

4.60011

Within Groups

43.375

14

3.098214

Total

103.4375

15

&nbsp

&nbsp

&nbsp

&nbsp

X

Y

X

1

Y

0.587596

1

F-VALUE= S12/S22

Where,S12&nbspisthe variance of first set of values and S22&nbspisthe variance of second set of values

F-VALUE= 1.642857^2/4.553571^2

F-VALUE= 2.699 / 20.735

F-VALUE= 0.13

p.408, #8

X

(Days)

Y

(Hours)

2

4

7

12

4

6

1

4

1

1

3

4

2

7

5

2

2

4

3

6

  1. Draw a scatter plot of the data.

  1. Calculate the regression slope and Y-intercept.

X

Y

XY

X^2

Y^2

(Days)

(Hours)

&nbsp

&nbsp

&nbsp

2

4

8

4

16

7

12

84

49

144

4

6

24

16

36

1

4

4

1

16

1

1

1

1

1

3

4

12

9

16

2

7

14

4

49

5

2

10

25

4

2

4

8

4

16

3

6

18

9

36

30

50

183

122

334

RegressionSlope

Slope=(NΣXY – (ΣX)(ΣY)) / (NΣX2&nbsp-(ΣX)2)

Slope= [10(183) – (30)(50)/10(122)-(30)^2]

Slope= [(1830 – 1500)/(1220-900)]

Slope= [330/320]

Slope= 1.031

Y-intercept

  1. Draw the regression line on the scatter plot.

  1. Predict the length of jury deliberation for a recently completed trial that lasted 6 days.

Thelength of jury deliberation for a recently completed trial thatlasted 6 days

Y(prediction) = 1.031 (6) + 1.907

Y(prediction) = 8.093 hours(Length of jury deliberation)

  1. Find the coefficients of determination and non-determination. What do they mean?

CorrelationCoefficient ( r ) = N x ∑ XY – ( ∑ X ) ( ∑ Y ) / √ N x ( ∑X2&nbsp-( ∑ X )2&nbsp√N x ( ∑ Y2&nbsp-( ∑ Y )2&nbsp

R= (10 x 183) – (30) (50) / √10 (122-30)^2 √ 10 (334-50)^2

R= 0.63650

Coefficientof Determination = r x r

Coefficientof Determination = 0.63650 x 0.63650

Coefficientof Determination = 0.405

• 40%of the variability in the Y scores can be predicted from the relationwith X

Coefficientof Non-determination = 1 – r^2

Coefficientof Non-determination = 1 – 0.405

Coefficientof Non-determination = 0.595

• 59%of the variability in the Y scores can be predicted from the relationwith X

  1. Construct an analysis of variance table and perform an F test of the significance of the regression.

SUMMARY

Groups

Count

Sum

Average

Variance

2

9

28

3.111111

3.861111

4

9

46

5.111111

10.36111

ANOVA

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

18

1

18

2.53125

0.131174

4.493998

Within Groups

113.7778

16

7.111111

Total

131.7778

17

&nbsp

&nbsp

&nbsp

&nbsp

F-VALUE= S1/S22

Where,S12&nbspisthe variance of first set of values and S22&nbspisthe variance of second set of values

F-VALUE= 3.5555562/9.3333332

F-VALUE= 12.64/87.11

F-VALUE= 0.15

p.409, #10

X

(Number of hours spent watching TV)

Y

(Number of schoolmates attacked on the playground)

0

0

6

3

2

2

4

3

4

4

1

1

1

0

2

3

5

3

5

2

4

3

0

1

2

3

6

4

  1. Draw a scatter plot of the data.

  1. Calculate the regression slope and Y-intercept.

X

Y

XY

X^2

Y^2

(Number of hours spent watching TV)

(Number of schoolmates attacked on the playground)

&nbsp

&nbsp

&nbsp

0

0

0

0

0

6

3

18

36

9

2

2

4

4

4

4

3

12

16

9

4

4

16

16

16

1

1

1

1

1

1

0

0

1

0

2

3

6

4

9

5

3

15

25

9

5

2

10

25

4

4

3

12

16

9

0

1

0

0

1

2

3

6

4

9

6

4

24

36

16

30

24

90

128

74

RegressionSlope

Slope=(NΣXY – (ΣX)(ΣY)) / (NΣX2&nbsp-(ΣX)2)

Slope= [14(90) – (30)(24)/14(128)-(30)^2]

Slope= [(1260 – 720)/(1792-900)]

Slope= [540/892]

Slope= 0.605

Y-intercept

  1. Draw the regression line on the scatter plot.

  1. Predict the number of schoolmates attacked by a child who watches television in 3 hours daily.

Y(prediction) = 0.482(3) + 0.837

Y(prediction) = 2.28 or 2 schoolmates

  1. Find the coefficients of determination and non-determination. What do they mean?

CorrelationCoefficient ( r ) = N x ∑ XY – ( ∑ X ) ( ∑ Y ) / √ N x ( ∑X2&nbsp-( ∑ X )2&nbsp√N x ( ∑ Y2&nbsp-( ∑ Y )2&nbsp

R= (14 x 90) – (30) (24) / √14 (128-30)^2 √ 14 (74-24)^2

R= 0.76901

Coefficientof Determination = r x r

Coefficientof Determination = 0.76901x 0.76901

Coefficientof Determination = 0.591

• 59%of the variability in the Y scores can be predicted from the relationwith X

Coefficientof Non-determination = 1 – r^2

Coefficientof Non-determination = 1 – 0.591

Coefficientof Non-determination = 0.409

• 40%of the variability in the Y scores can be predicted from the relationwith X

  1. Construct an analysis of variance table and perform an F test of the significance of the regression.

SUMMARY

Groups

Count

Sum

Average

Variance

0

13

42

3.230769

4.025641

0

13

32

2.461538

1.435897

ANOVA

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

3.846153846

1

3.846154

1.408451

0.246929

4.259677

Within Groups

65.53846154

24

2.730769

Total

69.38461538

25

&nbsp

&nbsp

&nbsp

&nbsp

F-VALUE= S12/S22

Where,S12&nbspisthe variance of first set of values and S22&nbspisthe variance of second set of values

F-VALUE= 4.222/1.822

F-VALUE= 17.81/3.31

F-VALUE= 5.38

p.410, #16

X

(HS cumulative grade average)

Y

(Cumulative grade average in college)

3.3

2.7

2.9

2.5

2.5

1.9

4.0

3.3

2.8

2.7

2.5

2.2

3.7

3.1

3.8

4.0

3.5

2.9

2.7

2.0

2.6

3.1

4.0

3.2

  1. Draw a scatter plot of the data.

  1. Calculate the regression slope and Y-intercept.

X

Y

XY

X^2

Y^2

(HS cumulative grade average)

(Cumulative grade average in college)

&nbsp

&nbsp

&nbsp

3.3

2.7

8.91

10.89

7.29

2.9

2.5

7.25

8.41

6.25

2.5

1.9

4.75

6.25

3.61

4

3.3

13.2

16

10.89

2.8

2.7

7.56

7.84

7.29

2.5

2.2

5.5

6.25

4.84

3.7

3.1

11.47

13.69

9.61

3.8

4

15.2

14.44

16

3.5

2.9

10.15

12.25

8.41

2.7

2

5.4

7.29

4

2.6

3.1

8.06

6.76

9.61

4

3.2

12.8

16

10.24

32.1

28.4

94.09

106.77

84.5

RegressionSlope

Slope=(NΣXY – (ΣX)(ΣY)) / (NΣX2&nbsp-(ΣX)2)

Slope= [12(94.09) – (32.1)(28.4)/12(106.77)-(32.1)^2]

Slope= [(1129.08 – 911.64)/(1281.24-1030.41)]

Slope= 0.786

Y-intercept

  1. Draw the regression line on the scatter plot.

  1. Predict the college grade average of a student who attains a 3.0 grade average in high school.

Y(prediction) = 0.786 (3) + 0.291

Y(prediction) = 2.649 or 3 students

  1. Find the coefficients of determination and non-determination. What do they mean?

CorrelationCoefficient ( r ) = N x ∑ XY – ( ∑ X ) ( ∑ Y ) / √ N x ( ∑X2&nbsp-( ∑ X )2&nbsp√N x ( ∑ Y2&nbsp-( ∑ Y )2&nbsp

R= (12 x 94.09) – (32.1) (28.4) / √12 (106.77-32.1)^2 √ 10 (84.5-28.4)^2

R= 0.77298

Coefficientof Determination = r x r

Coefficientof Determination = 0. 77298 x 0. 77298

Coefficientof Determination = 0.597

• 59%of the variability in the Y scores can be predicted from the relationwith X

Coefficientof Non-determination = 1 – r^2

Coefficientof Non-determination = 1 – 0.597

Coefficientof Non-determination = 0.403

• 40%of the variability in the Y scores can be predicted from the relationwith X

  1. Construct an analysis of variance table and perform an F test of the significance of the regression.

SUMMARY

Groups

Count

Sum

Average

Variance

3.3

13

67.51433

5.19341

66.261

2.7

13

59.72711

4.594393

51.92355

ANOVA

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

2.332339613

1

2.33234

0.039469

0.844195

4.259677

Within Groups

1418.214634

24

59.09228

Total

1420.546974

25

&nbsp

&nbsp

&nbsp

&nbsp

F-VALUE= S12/S22

Where,S12&nbspisthe variance of first set of values and S22&nbspisthe variance of second set of values

F-VALUE= 0.4142/0.4272

F-VALUE= 0.17/0.18

F-VALUE= 0.03

p.367, #8

Computea Pearson’s correlation coefficient and indicate whether thecorrelation is significant.

&nbsp

Distance to School (Miles)

Number of Clubs Joined

XY

X^2

Y^2

Lee

4

3

12

16

9

Ronda

2

1

2

4

1

Jess

7

5

35

49

25

Evelyn

1

2

2

1

4

Mohammed

4

1

4

16

1

Steve

6

1

6

36

1

George

9

9

81

81

81

Juan

7

6

42

49

36

Chi

7

5

35

49

25

David

10

8

80

100

64

&nbsp

57

41

299

401

247

CorrelationCoefficient ( r ) = N x ∑ XY – ( ∑ X ) ( ∑ Y ) / √ N x ( ∑X2&nbsp-( ∑ X )2&nbsp√N x ( ∑ Y2&nbsp-( ∑ Y )2&nbsp

R= (10 * 299) – (57) (41) /√10 x (401- (57)2&nbsp√10 x (247- (41)2&nbsp

R= (2990) – 2337) /√10 x (401- (3249)&nbsp√10 x (247- (1681)

R= (2990) – 2337) /√10 x (401- (3249)&nbsp√10 x (247- (1681)

R= 653 /√10 x (401- (3249)&nbsp√10 x (247- (1681)

R= 0.84272 (significant)

p.368, #10

Computea Pearson’s correlation coefficient and determine whether thecorrelation is significant.

Accountant

X

Y

XY

X^2

Y^2

A

2

8

16

4

64

B

7

3

21

49

9

Chi

5

4

20

25

16

David

12

2

24

144

4

E

1

5

5

1

25

F

10

2

20

100

4

G

8

1

8

64

1

H

6

5

30

36

25

I

5

4

20

25

16

J

2

6

12

4

36

K

3

7

21

9

49

L

4

1

4

16

1

&nbsp

65

48

201

477

250

CorrelationCoefficient ( r ) = N x ∑ XY – ( ∑ X ) ( ∑ Y ) / √ N x ( ∑X2&nbsp-( ∑ X )2&nbsp√N x ( ∑ Y2&nbsp-( ∑ Y )2&nbsp

R= (12 * 201) – (65) (48) /√12 x (477- (65)2&nbsp√12 x (250- (48)2&nbsp

R= (2412) – 3120) /√12 x (477- (4225)&nbsp√12 x (250- (2304)

R= (-708) /√12 x (-3748)&nbsp√12 x (-1434)

R= -0.69315 (not significant)