4. 0.7
To calculate the correlation coefficient, we first need to calculate the covariance and standard deviations of both variables.
First, we calculate the means of x and y:
Mean of x = (90 + 91 + 93 + 97 + 97 + 99) / 6 = 95.2
Mean of y = (7 + 17 + 9 + 16 + 34 + 27) / 6 = 18.3
Next, we calculate the covariance:
Covariance = Σ[(x_i - mean of x)(y_i - mean of y)] / (n-1)
Covariance = [(90-95.2)*(7-18.3) + (91-95.2)*(17-18.3) + (93-95.2)*(9-18.3) + (97-95.2)*(16-18.3) + (97-95.2)*(34-18.3) + (99-95.2)*(27-18.3)] / 5
Covariance = [(-5.2)*(-11.3) + (-4.2)*(-1.3) + (-2.2)*(-9.3) + 1.8*(-2.3) + 1.8*(15.7) + 3.8*(8.7)] / 5
Covariance = 58.12
Next, we calculate the standard deviations of x and y:
Standard deviation of x = √[Σ(x_i - mean of x)^2 / (n-1)]
Standard deviation of x = √[((-5.2)^2 + (-4.2)^2 + (-2.2)^2 + 1.8^2 + 1.8^2 + 3.8^2) / 5] = 3.94
Standard deviation of y = √[Σ(y_i - mean of y)^2 / (n-1)]
Standard deviation of y = √[((7-18.3)^2 + (17-18.3)^2 + (9-18.3)^2 + (16-18.3)^2 + (34-18.3)^2 + (27-18.3)^2) / 5] = 9.72
Finally, we calculate the correlation coefficient:
r = Covariance / (Standard deviation of x * Standard deviation of y)
r = 58.12 / (3.94 * 9.72) = 0.7003
Therefore, the correlation coefficient, r, of the data is 0.7.
Find the correlation coefficient, r, of the data described below. A movie theater manager is studying the snack buying habits of his customers. He expects that when longer movies are shown, people buy more snacks. Each week he calculates the average duration of all movies being shown and the percentage of customers who buy snacks. The manager calculates the average movie duration (in minutes), x, and the percentage of the customers who bought snacks, y.
Average movie duration - Percentage of customs that bought snacks
90 7
91 17
93 9
97 16
97 34
99 27
1. 0.2
2. 0.8
3. -0.1
4. 0.7
1 answer