Please answer C6, C7, C8, and C9. Please show formulas used and how you came to
ID: 3360605 • Letter: P
Question
Please answer C6, C7, C8, and C9. Please show formulas used and how you came to your answer. Thank you.
For a ner pcoject in a specific location, ABC Constauction Co. studied the aunoff vohume) produced by rainfall volume (m). The results are shown below fall ofF C2 What is the slope of the line? Interpret the slope C1 Construct a scatter plot the data; insert the trend line 0.827 12 14 C3 Interpret th e slope in context of the problem. 40 47 27 46 38 46 53 70 82 C4 What is the coefficient of determination? 67 72 81 96 C5 Interpret the coefficient of determination? 127 100 C6 What is the correlation coefficient? C7 Interpret the correlation coefficien C8 Complete an appropriate hypothesis test to see if the relationship between the variables is significant. Ho Ha Test Statistic CV DecissIOn: C9 Predict the runoff for a rainfall of 90 mExplanation / Answer
Part C6
Solution:
Here, we have to find the correlation coefficient between two variables X and Y.
Formula for correlation coefficient is given as below:
Correlation coefficient = r = [nxy - xy]/sqrt[(nx^2 – (x)^2)*(ny^2 – (y)^2)]
Calculation table is given as below:
No.
X
Y
X^2
Y^2
XY
1
5
4
25
16
20
2
12
10
144
100
120
3
14
13
196
169
182
4
17
15
289
225
255
5
23
15
529
225
345
6
30
25
900
625
750
7
40
27
1600
729
1080
8
47
46
2209
2116
2162
9
55
38
3025
1444
2090
10
67
46
4489
2116
3082
11
72
53
5184
2809
3816
12
81
70
6561
4900
5670
13
96
82
9216
6724
7872
14
112
99
12544
9801
11088
15
127
100
16129
10000
12700
Total
798
643
63040
41999
51232
Mean
53.2
42.86667
X = 798
Y = 643
X^2 = 63040
Y^2 = 41999
XY = 51232
n = 15
Now, plug all values in the formula
r = [nxy - xy]/sqrt[(nx^2 – (x)^2)*(ny^2 – (y)^2)]
r = [15*51232 – 798*643]/sqrt[(15*63040 – 798^2)*(15*41999– 643^2)]
r = 0.987557
Correlation coefficient = r = 0.987557
Part C7
Given correlation indicate that there is a very strong positive linear relationship or association exists between the two variables rainfall and runoff.
Part C8
Here, we have to use t test for checking whether the relationship between the variables is significant or not.
The null and alternative hypothesis for this test is given as below:
Null hypothesis: H0: The relationship between the variables is not statistically significant.
Alternative hypothesis: Ha: The relationship between the variables is statistically significant.
H0: = 0 versus Ha: 0
This is a two tailed test.
Assume level of significance as 5% or = 0.05.
The test statistic formula is given as below:
t = r*sqrt(n – 2) / sqrt(1 – r2)
t = 0.987557*sqrt(15 – 2) / sqrt(1 - 0.987557^2)
t = 22.63534209
DF = n – 2 = 15 – 2 = 13
CV = critical value = 2.160368652
P-value = 0.00
(CV and P-value are calculated by using t-table or excel)
P-value < = 0.05
So, we reject the null hypothesis that the relationship between the variables is not statistically significant.
There is sufficient evidence to conclude that the relationship between the variables is statistically significant.
Part C9
We have to predict the runoff for rainfall = 90 m3.
Regression equation is given as below:
Runoff = -1.1283 + 0.826973*Rainfall
Runoff = -1.1283 + 0.826973*90
Runoff = 73.29927 m3
No.
X
Y
X^2
Y^2
XY
1
5
4
25
16
20
2
12
10
144
100
120
3
14
13
196
169
182
4
17
15
289
225
255
5
23
15
529
225
345
6
30
25
900
625
750
7
40
27
1600
729
1080
8
47
46
2209
2116
2162
9
55
38
3025
1444
2090
10
67
46
4489
2116
3082
11
72
53
5184
2809
3816
12
81
70
6561
4900
5670
13
96
82
9216
6724
7872
14
112
99
12544
9801
11088
15
127
100
16129
10000
12700
Total
798
643
63040
41999
51232
Mean
53.2
42.86667
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