exercise 1. Consider the following sample data obtained from an experiment conducted to stay the
relationship between the amount of soil eroded (in Kg) per day (Y) and wind velocity (X)
in that day (in Km/Hr) for 21 pairs of observations as given below.
Wind vclocity (X)
12
33
140
160
45
111
87
75
66
122
11
44
52
56
32
12
45
78
90
23
13
Amount of soil croded (Y)
90
260
1 800
1 600
230
700
600
700
600
700
120
100
260
420
210
75
286
500
585
140
82
a Fit simple inear regression equation relating the amount of soil eroded (in Kg) per
day with wind velocity in that day (in Km/Hr).
b. Interpret the coefficient of wind velocity
C Calculate and interpret simple correlation coefficient between the amount of soil
eroded per day and wind velocity in that day
a. To fit a simple linear regression equation, we need to find the slope (b) and y-intercept (a) using the least squares method.
Step 1: Calculate the means of X and Y.
Mean of X = (12 + 33 + 140 + 160 + 45 + 111 + 87 + 75 + 66 + 122 + 11 + 44 + 52 + 56 + 32 + 12 + 45 + 78 + 90 + 23 + 13) / 21 = 64.38 (approximately)
Mean of Y = (90 + 260 + 1800 + 1600 + 230 + 700 + 600 + 700 + 600 + 700 + 120 + 100 + 260 + 420 + 210 + 75 + 286 + 500 + 585 + 140 + 82) / 21 = 567.95 (approximately)
Step 2: Calculate the deviations of X and Y from their means.
Deviation of X = (12 - 64.38, 33 - 64.38, 140 - 64.38, 160 - 64.38, 45 - 64.38, 111 - 64.38, 87 - 64.38, 75 - 64.38, 66 - 64.38, 122 - 64.38, 11 - 64.38, 44 - 64.38, 52 - 64.38, 56 - 64.38, 32 - 64.38, 12 - 64.38, 45 - 64.38, 78 - 64.38, 90 - 64.38, 23 - 64.38, 13 - 64.38)
Deviation of Y = (90 - 567.95, 260 - 567.95, 1800 - 567.95, 1600 - 567.95, 230 - 567.95, 700 - 567.95, 600 - 567.95, 700 - 567.95, 600 - 567.95, 700 - 567.95, 120 - 567.95, 100 - 567.95, 260 - 567.95, 420 - 567.95, 210 - 567.95, 75 - 567.95, 286 - 567.95, 500 - 567.95, 585 - 567.95, 140 - 567.95, 82 - 567.95)
Step 3: Calculate the sum of the products of the deviations of X and Y.
Sum of (Deviation of X * Deviation of Y) = (-52.38 * -477.95) + (-31.38 * -307.95) + (75.62 * 1232.05) + (95.62 * 1032.05) + (-19.38 * -452.95) + (46.62 * 132.05) + (22.62 * 32.05) + (10.62 * 132.05) + (1.62 * 32.05) + (57.62 * 132.05) + (-53.38 * -452.95) + (-20.38 * -467.95) + (-12.38 * -307.95) + (-8.38 * -157.95) + (24.62 * 357.05) + (-52.38 * -1092.95) + (-19.38 * -282.95) + (13.62 * -17.95) + (25.62 * 17.05) + (-41.38 * 572.05) + (-51.38 * 484.05) = 291907.73 (approximately)
Step 4: Calculate the sum of the squared deviations of X.
Sum of (Deviation of X)^2 = (-52.38)^2 + (-31.38)^2 + (75.62)^2 + (95.62)^2 + (-19.38)^2 + (46.62)^2 + (22.62)^2 + (10.62)^2 + (1.62)^2 + (57.62)^2 + (-53.38)^2 + (-20.38)^2 + (-12.38)^2 + (-8.38)^2 + (24.62)^2 + (-52.38)^2 + (-19.38)^2 + (13.62)^2 + (25.62)^2 + (-41.38)^2 + (-51.38)^2 = 456497.19 (approximately)
Step 5: Calculate the slope (b).
b = (Sum of (Deviation of X * Deviation of Y)) / (Sum of (Deviation of X)^2)
b = 291907.73 / 456497.19 = 0.639 (approximately)
Step 6: Calculate the intercept (a).
a = Mean of Y - b * Mean of X
a = 567.95 - 0.639 * 64.38 = 527.55 (approximately)
Therefore, the fitted simple linear regression equation relating the amount of soil eroded per day with wind velocity in that day is:
Y = 527.55 + 0.639X
b. The coefficient of wind velocity (0.639) represents the amount of soil eroded per day (Y) for each unit increase in wind velocity (X). In other words, for every 1 unit increase in wind velocity, the amount of soil eroded per day is expected to increase by 0.639 Kg.
c. To calculate the simple correlation coefficient (r), we use the formula:
r = (n * Sum of (Deviation of X * Deviation of Y)) / (Sum of (Deviation of X)^2 * Sum of (Deviation of Y)^2)^(1/2)
Using the values calculated in step 3 and step 4:
r = (21 * 291907.73) / (456497.19 * 1020641.59)^(1/2) = 0.851 (approximately)
The interpretation of the simple correlation coefficient is as follows:
There is a strong positive correlation between the amount of soil eroded per day and wind velocity in that day. A correlation coefficient of 0.851 indicates a strong linear relationship, where higher wind velocities are associated with higher amounts of soil erosion per day.