ID Number
IDNumber
CourseName, Course Number
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 

Draw a scatter plot of the data.

Calculate the regression slope and Yintercept.
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)) / (NΣX^{2} (ΣX)^{2})
Slope= [8(90) – (14)(45)/8(36)(14)^2]
Slope= [(720 – 630)/(288196)]
Slope= [90/92]
Slope= 0.978
Yintercept

Draw the regression line on the scatter plot.

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.

Find the coefficients of determination and nondetermination. What do they mean?
CorrelationCoefficient ( r ) = N x ∑ XY – ( ∑ X ) ( ∑ Y ) / √ N x ( ∑X^{2} ( ∑ X )^{2} √N x ( ∑ Y^{2} ( ∑ Y )^{2} 
R= (8 x 90) – (14) (45) / √8 (3614)^2 √ 8 (28545)^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 Nondetermination = 1 – r^2
Coefficientof Nondetermination = 1 – 0.345
Coefficientof Nondetermination = 0.655
• 65%of the variability in the Y scores can be predicted from the relationwith X

Construct an analysis of variance table and perform an Ftest 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 
Pvalue 
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
FVALUE= S1^{2}/S2^{2}
^{ }Where,S1^{2} isthe variance of first set of values and S2^{2} isthe variance of second set of values
FVALUE= 1.642857^2/4.553571^2
FVALUE= 2.699 / 20.735
FVALUE= 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 

Draw a scatter plot of the data.

Calculate the regression slope and Yintercept.
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)) / (NΣX^{2} (ΣX)^{2})
Slope= [10(183) – (30)(50)/10(122)(30)^2]
Slope= [(1830 – 1500)/(1220900)]
Slope= [330/320]
Slope= 1.031
Yintercept

Draw the regression line on the scatter plot.

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)

Find the coefficients of determination and nondetermination. What do they mean?
CorrelationCoefficient ( r ) = N x ∑ XY – ( ∑ X ) ( ∑ Y ) / √ N x ( ∑X^{2} ( ∑ X )^{2} √N x ( ∑ Y^{2} ( ∑ Y )^{2} 
R= (10 x 183) – (30) (50) / √10 (12230)^2 √ 10 (33450)^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 Nondetermination = 1 – r^2
Coefficientof Nondetermination = 1 – 0.405
Coefficientof Nondetermination = 0.595
• 59%of the variability in the Y scores can be predicted from the relationwith X

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 
Pvalue 
F crit 

Between Groups 
18 
1 
18 
2.53125 
0.131174 
4.493998 

Within Groups 
113.7778 
16 
7.111111 

Total 
131.7778 
17 
  
  
  
  
FVALUE= S1/S2^{2}
^{ }Where,S1^{2} isthe variance of first set of values and S2^{2} isthe variance of second set of values
FVALUE= 3.555556^{2}/9.333333^{2}
FVALUE= 12.64/87.11
FVALUE= 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 

Draw a scatter plot of the data.

Calculate the regression slope and Yintercept.
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)) / (NΣX^{2} (ΣX)^{2})
Slope= [14(90) – (30)(24)/14(128)(30)^2]
Slope= [(1260 – 720)/(1792900)]
Slope= [540/892]
Slope= 0.605
Yintercept

Draw the regression line on the scatter plot.

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

Find the coefficients of determination and nondetermination. What do they mean?
CorrelationCoefficient ( r ) = N x ∑ XY – ( ∑ X ) ( ∑ Y ) / √ N x ( ∑X^{2} ( ∑ X )^{2} √N x ( ∑ Y^{2} ( ∑ Y )^{2} 
R= (14 x 90) – (30) (24) / √14 (12830)^2 √ 14 (7424)^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 Nondetermination = 1 – r^2
Coefficientof Nondetermination = 1 – 0.591
Coefficientof Nondetermination = 0.409
• 40%of the variability in the Y scores can be predicted from the relationwith X

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 
Pvalue 
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 
  
  
  
  
FVALUE= S1^{2}/S2^{2}
^{ }Where,S1^{2} isthe variance of first set of values and S2^{2} isthe variance of second set of values
FVALUE= 4.22^{2}/1.82^{2}
FVALUE= 17.81/3.31
FVALUE= 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 

Draw a scatter plot of the data.

Calculate the regression slope and Yintercept.
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)) / (NΣX^{2} (ΣX)^{2})
Slope= [12(94.09) – (32.1)(28.4)/12(106.77)(32.1)^2]
Slope= [(1129.08 – 911.64)/(1281.241030.41)]
Slope= 0.786
Yintercept

Draw the regression line on the scatter plot.

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

Find the coefficients of determination and nondetermination. What do they mean?
CorrelationCoefficient ( r ) = N x ∑ XY – ( ∑ X ) ( ∑ Y ) / √ N x ( ∑X^{2} ( ∑ X )^{2} √N x ( ∑ Y^{2} ( ∑ Y )^{2} 
R= (12 x 94.09) – (32.1) (28.4) / √12 (106.7732.1)^2 √ 10 (84.528.4)^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 Nondetermination = 1 – r^2
Coefficientof Nondetermination = 1 – 0.597
Coefficientof Nondetermination = 0.403
• 40%of the variability in the Y scores can be predicted from the relationwith X

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 
Pvalue 
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 
  
  
  
  
FVALUE= S1^{2}/S2^{2}
^{ }Where,S1^{2} isthe variance of first set of values and S2^{2} isthe variance of second set of values
FVALUE= 0.414^{2}/0.427^{2}
FVALUE= 0.17/0.18
FVALUE= 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 ( ∑X^{2} ( ∑ X )^{2} √N x ( ∑ Y^{2} ( ∑ Y )^{2} 
R= (10 * 299) – (57) (41) /√10 x (401 (57)^{2} √10 x (247 (41)^{2} 
R= (2990) – 2337) /√10 x (401 (3249) √10 x (247 (1681)
R= (2990) – 2337) /√10 x (401 (3249) √10 x (247 (1681)
R= 653 /√10 x (401 (3249) √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 ( ∑X^{2} ( ∑ X )^{2} √N x ( ∑ Y^{2} ( ∑ Y )^{2} 
R= (12 * 201) – (65) (48) /√12 x (477 (65)^{2} √12 x (250 (48)^{2} 
R= (2412) – 3120) /√12 x (477 (4225) √12 x (250 (2304)
R= (708) /√12 x (3748) √12 x (1434)
R= 0.69315 (not significant)