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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)       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

 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)       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

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)       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

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)       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

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.

   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   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   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)