The document presents data on the math and English scores of 5 students. It asks to compute the Spearman Rank Correlation Coefficient and Pearson's Correlation Coefficient between the scores. The Spearman Rank Correlation Coefficient is -0.96, indicating a very strong negative correlation between ranks. Pearson's Correlation Coefficient is -0.99, also indicating a very strong negative correlation. The document also provides price and age data for 8 cars and asks to find the regression line and make predictions based on the line.

1. “ Spearman Rank Correlation Coefficient”
Presented by : Eng. Waleed Alzaghal
You Tube Channel : Waleed Alzaghal

2. The scores of five students in Math and English are shown in below table.
Compute the Spearman Rank Correlation Coefficient for these data. Consider the
rank of the score is as the following:
• 90 – 100 Excellent
• 80 – 89 Very Good
• 70 – 79 Good
• 60 – 69 Weak
Student Math Score (x) English Score (y) Results(x) Results (y)
1 95 97 Excellent Excellent
2 90 92 Excellent Excellent
3 75 80 Good Very Good
4 85 78 Very Good Good
5 60 68 Weak Weak

3. 1- Display the Scatter Plot.
2- Find the Equation of the Regression Line.
3- Draw the Regression Line.
4- Compute the Pearson’s Correlation Coefficient.
5- Compute the Spearman Rank Correlation Coefficient.
6- Predict the value of ŷ for x = 1 , x = 4
7- Compute the Error of Prediction for x = 9
8- Compute the Least Squares of Errors.
9- Compute the Sum of Squares of Errors for the following Equation:
ŷ = - 36.00 x + 322.44
Obs. Age (x) Price(y)
1 8 45
2 3 210
3 6 100
4 9 33
5 2 267
6 5 134
7 6 109
8 3 235
The price of 8 cars (in hundred dollars) with different ages (in year) are in below table:

4. 1- Display the Scatter Plot.

5. ŷ = mx + b
m =
𝑛 𝑥𝑦 −( 𝑥)( 𝑦)
𝑛 𝑥2
−( 𝑥)2
m =
8 4450 −(42)(1133)
8 264 − 42 2
m =
−11,986
348
m ≈ − 34.44 (Slope of the Regression Line)
Obs. Age (x) Price(y) x
2
y
2
xy
1 8 45 64 2025 360
2 3 210 9 44100 630
3 6 100 36 10000 600
4 9 33 81 1089 297
5 2 267 4 71289 534
6 5 134 25 17956 670
7 6 109 36 11881 654
8 3 235 9 55225 705
Sum 42 1133 264 213565 4450
2- Find the Equation of the Regression Line.

6. ŷ = mx + b
b =
𝑦 −𝑚 𝑥
𝑛
b =
1133 − (−34.44)(42)
8
b =
2,579.48
8
b = 322.44
ŷ = − 34.44x + 322.44
Obs. Age (x) Price(y) x
2
y
2
xy
1 8 45 64 2025 360
2 3 210 9 44100 630
3 6 100 36 10000 600
4 9 33 81 1089 297
5 2 267 4 71289 534
6 5 134 25 17956 670
7 6 109 36 11881 654
8 3 235 9 55225 705
Sum 42 1133 264 213565 4450

7. 3- Draw the Regression Line.

8. Obs. Age (x) Price(y) x
2
y
2
xy
1 8 45 64 2025 360
2 3 210 9 44100 630
3 6 100 36 10000 600
4 9 33 81 1089 297
5 2 267 4 71289 534
6 5 134 25 17956 670
7 6 109 36 11881 654
8 3 235 9 55225 705
Sum 42 1133 264 213565 4450
𝑟 =
𝑛 𝑥𝑦 − ( 𝑥)( 𝑦)
(𝑛 𝑥2 − ( 𝑥)2). (𝑛 𝑦2 − ( 𝑦)2
𝑟 =
8 4450 − (42)(1133)
( 8 264 − 42 2)( 8 213565 − 1133 2)
𝑟 =
−11,986
(348)(424,831)
𝑟 =
−11986
12,159.00
r ≈ −0.99
(Interpretation: A very Strong Negative Correlation)
4- Compute the Pearson’s Correlation Coefficient.

9. rs = 1 −
6 𝑑2
𝑛(𝑛2
−1)
rs = 1 −
(6)(165)
8(82
−1)
rs = 1 −
990
8(64 −1)
rs = 1 −
990
504
rs = − 0.96
(Interpretation: a very strong negative correlation).
5- Compute the Spearman Rank Correlation Coefficient.
Obs. Age (x) Price(y) Rank (Age) Rank (Price) d d
2
1 8 45 7 2 5.0 25.00
2 3 210 2.5 6 -3.5 12.25
3 6 100 5.5 3 2.5 6.25
4 9 33 8 1 7.0 49.00
5 2 267 1 8 -7.0 49.00
6 5 134 4 5 -1.0 1.00
7 6 109 5.5 4 1.5 2.25
8 3 235 2.5 7 -4.5 20.25
165Sum

10. 6- Predict the value of ŷ for x = 1
Ŷ = (-34.44)(1) + 322.44 = 288

11. Ŷ = (-34.44)(4) + 322.44 = 184.68
6- Predict the value of ŷ for x = 4

12. 7- Compute the Error of Prediction for x=9
Ŷ = (-34.44)(9) + 322.44 = 12.48
E = Y – Ŷ = 33 – 12.48 = 20.52

13. 8- Compute the Sum (Least) Squares of Errors.
Obs. Age (x) Price(y) ŷ=-34.44x+322.44 E=y-ŷ E
2
=(y-ŷ)
2
1 8 45 46.92 -1.92 3.69
2 3 210 219.12 -9.12 83.17
3 6 100 115.80 -15.80 249.64
4 9 33 12.48 20.52 421.07
5 2 267 253.56 13.44 180.63
6 5 134 150.24 -16.24 263.74
7 6 109 115.80 -6.80 46.24
8 3 235 219.12 15.88 252.17
1500.36Sum
Least Squares of Error = ((E1)2 + (E2)2 + (E3)2 + (E4)2 + (E5)2 + (E6)2 + (E7)2 + (E8)2)
Least Squares of Error = 3.69 + 83.17 + ……. + 46.24 + 252.17 = 1500.36

14. 9- Compute the Squares of Errors for the following Equation: ŷ = -36.00 x+322.44
Obs. Age (x) Price(y) ŷ=-36.00 x+322.44 E=y-ŷ E
2
=(y-ŷ)
2
1 8 45 34.44 10.56 111.51
2 3 210 214.44 -4.44 19.71
3 6 100 106.44 -6.44 41.47
4 9 33 -1.56 34.56 1194.39
5 2 267 250.44 16.56 274.23
6 5 134 142.44 -8.44 71.23
7 6 109 106.44 2.56 6.55
8 3 235 214.44 20.56 422.71
2141.83Sum
Sum of Squares of Error = ((E1)2 + (E2)2 + (E3)2 + (E4)2 + (E5)2 + (E6)2 + (E7)2 + (E8)2)
Sum of Squares of Error = 111.51 + 19.71 + ……. + 6.55 + 422.71 = 2141.83