Question

An article gave a scatter plot along with the least squares line of x = rainfall volume (m3) and y = runoff volume (m3) for a particular location. The accompanying values were read from the plot.

c) Calculate a point estimate of the true average runoff volume when rainfall volume is 51. (Round your answer to four decimal places.)

(d) Calculate a point estimate of the standard deviation . (Round your answer to two decimal places.)

(e) What proportion of the observed variation in runoff volume can be attributed to the simple linear regression relationship between runoff and rainfall? (Round your answer to four decimal places.)

x 6 12 14 16 23 30 40 52 55 67 72 81 96 112 127

y 4 10 13 14 15 25 27 48 38 46 53 72 82 99 100

Answer 1

y = 0.834X - 1.58015

Slope = 0.8340 ; Intercept = - 1.5802

y = 40.9539

19.93

0.9765

Step-by-step explanation:

X: Rainfall volume

6

12

14

16

23

30

40

52

55

67

72

81

96

112

127

Y : Runoff

4

10

13

14

15

25

27

48

38

46

53

72

82

99

100

The scatterplot shows a reasonable linear trend between the Rainfall volume and run off.

The estimated regression equation obtained using a linear regression calculator is :

y = 0.834X - 1.58015

y = Runoff ; x = Rainfall volume

Slope = 0.8340 ; Intercept = - 1.5802

Point estimate for Runoff, when, x = 51

y = 0.834X - 1.58015

y = 0.834(51) - 1.58015

y = 40.95385

y = 40.9539

d.)

Point estimate for standard deviation :

s = 5.145

σ = s * √n

σ = √15 * 5.145

= 19.93

e.)

r² = Coefficient of determination gives the proportion of explained variance in Runoff due to the regression line. From the model output, the r² value = 0.9765. Which means That about 97.65% Runoff is due to Rainfall volume.