Respuesta :
Answer:
1. 3.767
2. 0.145
Step-by-step explanation:
Let X be the exam scores and Y be the number of drinks.
X Y X-Xbar Y-Ybar (X-Xbar)(Y-Ybar) (X-Xbar)² (Y-Ybar)²
75 5 -2.3 2.3 -5.29 5.29 5.29
92 3 14.7 0.3 4.41 216.09 0.09
84 2 6.7 -0.7 -4.69 44.89 0.49
64 4 -13.3 1.3 -17.29 176.89 1.69
64 2 -13.3 -0.7 9.31 176.89 0.49
86 7 8.7 4.3 37.41 75.69 18.49
81 3 3.7 0.3 1.11 13.69 0.09
61 0 -16.3 -2.7 44.01 265.69 7.29
73 1 -4.3 -1.7 7.31 18.49 2.89
93 0 15.7 -2.7 -42.39 246.49 7.29
sumx=773, sumy=27, sum(x-xbar)(y-ybar)= 33.9 , sum(X-Xbar)²= 1240.1 ,sum(Y-Ybar)²= 44.1
Xbar=sumx/n=773/10=77.3
Ybar=sumy/n=27/10=2.7
1.
[tex]Cov(x,y)=sxy=\frac{Sum(X-Xbar)(Y-Ybar)}{n-1}[/tex]
Cov(x,y)=33.9/9
Cov(x,y)=3.76667
The the sample co-variance between the exam scores and the number of energy drinks consumed is 3.767
2.
[tex]Cor(x,y)=r=\frac{Sum(X-Xbar)(Y-Ybar)}{\sqrt{Sum(X-Xbar)^2sum(Y-Ybar)^2} }[/tex]
[tex]Cor(x,y)=r=\frac{33.9}{\sqrt{(1240.1)(44.1)} }[/tex]
Cor(x,y)=r=33.9/233.85553
Cor(x,y)=r=0.14496
The sample correlation coefficient between the exam scores and the number of energy drinks consumed 0.145.