Question

Data collected on the discharge of the Colorado River and the speed are given in the table:

Discharge (ft3) Speed
1.3 2.3
2.2 0.99
5.8 3.5
11 5
12 16
14 22
16 27
21 14
49 33
51 35


What is the value of the correlation coefficient and its interpretation?
0.76; There is a strong, positive association between discharge and speed.
0.87; There is a strong, positive association between discharge and speed.
−0.76; There is a strong, negative association between discharge and speed.
−0.87; There is a strong, negative association between discharge and speed.

Answer 1

To calculate the correlation coefficient between the discharge and speed of the Colorado River, we need to calculate the mean and standard deviation of both variables. Using the provided data, we find that the correlation coefficient is 0.497, indicating a moderate positive association between discharge and speed.

Step-by-step explanation:

To calculate the correlation coefficient between two variables, we need to first calculate the mean and standard deviation of both variables. The correlation coefficient is a measure of the strength and direction of the linear relationship between the two variables. In this case, the variables are the discharge and speed of the Colorado River.

Step 1: Calculate the mean of the discharge and speed variables:

Discharge mean = (1.3 + 2.2 + 5.8 + 11 + 12 + 14 + 16 + 21 + 49 + 51) / 10 = 18.4
Speed mean = (2.3 + 0.99 + 3.5 + 5 + 16 + 22 + 27 + 14 + 33 + 35) / 10 = 16.79

Step 2: Calculate the standard deviation of the discharge and speed variables:

Discharge standard deviation = sqrt(sum((x - mean)^2) / (n-1))
= sqrt(( (1.3-18.4)^2 + (2.2-18.4)^2 + (5.8-18.4)^2 + (11-18.4)^2 + (12-18.4)^2 + (14-18.4)^2 + (16-18.4)^2 + (21-18.4)^2 + (49-18.4)^2 + (51-18.4)^2) / (10-1))
= sqrt(9267.04 / 9)
= sqrt(1029.67)
= 32.08

Speed standard deviation = sqrt(sum((x - mean)^2) / (n-1))
= sqrt(( (2.3-16.79)^2 + (0.99-16.79)^2 + (3.5-16.79)^2 + (5-16.79)^2 + (16-16.79)^2 + (22-16.79)^2 + (27-16.79)^2 + (14-16.79)^2 + (33-16.79)^2 + (35-16.79)^2) / (10-1))
= sqrt(267.28 / 9)
= sqrt(29.6978)
= 5.45

Step 3: Calculate the correlation coefficient using the formula:

Correlation coefficient = (1/n) * sum((x - mean of x) * (y - mean of y) / (standard deviation of x * standard deviation of y))
= (1/10) * ((1.3-18.4) * (2.3-16.79) / (32.08 * 5.45) + (2.2-18.4) * (0.99-16.79) / (32.08 * 5.45) + (5.8-18.4) * (3.5-16.79) / (32.08 * 5.45) + (11-18.4) * (5-16.79) / (32.08 * 5.45) + (12-18.4) * (16-16.79) / (32.08 * 5.45) + (14-18.4) * (22-16.79) / (32.08 * 5.45) + (16-18.4) * (27-16.79) / (32.08 * 5.45) + (21-18.4) * (14-16.79) / (32.08 * 5.45) + (49-18.4) * (33-16.79) / (32.08 * 5.45) + (51-18.4) * (35-16.79) / (32.08 * 5.45))
= (1/10) * ((-17.1) * (-14.49) / (32.08 * 5.45) + (-16.2) * (-15.8) / (32.08 * 5.45) + (-12.6) * (-13.29) / (32.08 * 5.45) + (-7.4) * (-11.79) / (32.08 * 5.45) + (-6.4) * (-0.79) / (32.08 * 5.45) + (-4.4) * (5.21) / (32.08 * 5.45) + (-2.4) * (10.21) / (32.08 * 5.45) + (2.6) * (-2.79) / (32.08 * 5.45) + (30.6) * (16.21) / (32.08 * 5.45) + (32.6) * (18.21) / (32.08 * 5.45))
= (1/10) * (49.7083) = 4.97

The correlation coefficient is 0.497, which indicates a moderate positive association between the discharge and speed of the Colorado River.

Answer 2

The value of the correlation coefficient is 0.87, which means there is a strong, positive association between discharge and speed. A positive correlation means that as one variable increases, the other variable also increases. In this case, as the discharge of the Colorado River increases, the speed of the river also increases. The strength of the association is indicated by the magnitude of the correlation coefficient, which ranges from -1 to 1. A value of 0.87 is considered a strong positive correlation, indicating that there is a strong relationship between the two variables.