Alyssa Mikole O. Rodiño
BS Psychology 1-4
MATHEMATICS IN THE MODERN WORLD
Assignment(Variation_Correlation_Regression)
A. Construct an FDT of the following Data below using Class interval of ‘12’ and start
with the lowest
value as your lowest-lower limit:
48
73
57
57
69
88
11
80
82
47
Compute:
46
70
49
45
75
81
33
65
38
59
1. Range
94
59
62
36
58
69
45
55
58
65
2. QD
30
49
73
29
41
53
37
35
61
48
3. MAD
22
51
56
55
63
37
56
59
57
36
4. SD
12
36
50
63
68
30
56
70
53
28
5. Variance
6. CV
Class Interval
F
M
fM2
|M-x̄ |
f|M-x̄ |
f|M-x̄ |2
11-22
3
16.5
2,450.25
36.4
109.2
3,974.88
23-34
5
28.5
20,306.25
24.4
122
2,976.8
35-46
11
40.5
198,470.25
12.4
136.4
1,691.36
47-58
19
52.5
995,006.25
0.4
7.6
3.04
59-70
14
64.5
815,409
11.6
162.4
1,883.84
71-82
6
76.5
210,681
23.6
141.6
3,341.76
83-94
2
88.5
31,329
35.6
71.2
2,534.72
x̄ =52.9
∑fM2=184,311
Range:
R= 94.5-10.5
R=84
∑f|M- x̄ |=750.4
∑f|M- x̄ |2=16,406.4
Quartile Deviation
QD=
𝑄3−𝑄1
QD=
61.93−42.14
2
2
QD=9.9
Mean Average / Absolute Deviation
MAD=
∑f|M− x̄ |
MAD=
750.4
𝑛
60
MAD= 12.61
Standard Deviation
Method I
Method II
∑f|M− x̄|2
SD=√
60
16,406.4
SD=√
60
SD= √
∑fM2
𝑛
− (x̄ )2
184,311
SD=√
60
− (52.9)2
SD=√273.44
SD=√3,071.85 − 2,798.41
SD=16.54
SD=√273.44
SD=16.54
Variance
V=273.44
Coefficient of variation
𝑆𝐷
CV=( x̄ )(100)
16.54
CV=( 59.2 )(100)
CV=31.27
B. Use the data below:
(PUPCET Grades)→ X: 80, 95, 87, 84, 70,82,93;
(HS Average Grades)→ Y:88, 92, 90, 80, 85, 84, 87.
8. Compute Correlation Coefficient using the Pearson r Product Moment Correlation.
x
y
xy
x2
y2
80
88
7,040
6,400
7,744
95
92
8,740
9,025
8,464
87
90
7,830
7,569
8,100
84
80
6,720
7,056
6,400
70
85
5,950
4,900
7,225
82
84
6,888
6,724
7,056
93
87
8,091
8,649
7,569
∑x=591
r=
r=
r=
r=
r=
∑y=606
∑xy=51,259
𝑛∑xy−∑x∑y
√[𝑛∑𝑥 2 −(∑𝑥)2 ][𝑛∑𝑦 2 −(∑𝑦)2 ]
7(51,259)−(591)(606)
√[(7)(50,323)−(591)2 ][(7)(52,558)−(606)2 ]
358,813−358,146
√(352,261−349,281)(367,906−367,236)
667
√(2,980)(670)
667
1,413.01
r=0.47
∑x2=50,323
∑y2=52,558
9. Compute Correlation Coefficient using the Spearman “rho”.
x
y
Rx
Ry
D
D2
80
88
2
5
3
9
95
92
7
7
0
0
87
90
5
6
1
1
84
80
4
1
3
9
70
85
1
3
2
4
82
84
3
2
1
1
93
87
6
4
2
4
∑D2=28
p= 1p= 1p= 1-
6∑2
𝑛(𝑛2 −1)
6(28)
7(72 −1)
168
7(49−1)
p= 1p= 1-
168
7(48)
168
336
p= 1- 0.5
p= 0.5
10. Regression Line Equation to predict ‘HS Average Grades from PUPCET Grades’.
b=
𝑛∑𝑥𝑦− ∑𝑥∑𝑦
b=
7(51,259)−(591)(606)
b=
358,813−358,146
b=
𝑛∑𝑥 2 − (∑𝑥)2
7(50,323)−(591)2
352,261−349,281
667
2,980
b= 0.22
The LRE to predict ‘HS Average Grades from PUPCET Grades’ is y=68 + 0.22x
11. Regression Line Equation to predict ‘PUPCET Grades from HS Average Grades’.
b=
𝑛∑𝑥𝑦− ∑𝑥∑𝑦
b=
7(51,259)−(591)(606)
b=
358,813−358,146
b=
667
𝑛∑𝑦 2 − (∑𝑦)2
7(52,558)−(606)2
367,906−367,236
670
b= 1
The LRE to predict ‘PUPCET Grades from HS Average Grades’ is x= (-2.14) + y
12. If PUPCET Grade is 93, what is the predicted HS Average Grade?
y= 68 - 0.22x
y= 68 - (0.22)(93)
y= 68 - 20.46
y= 88.46
13. If HS Average grade is 83, what is the predicted PUPCET grade?
x= (-2.14) + y
x= (-2.14) + 83
x= 80.86
*** E N D ***