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Free_Kalibri [48]

2 years ago

8

Energy drink consumption has continued to gain in popularity since the 1997 debut of Red Bull, the current leader in the energy

drink market. Given below are the exam scores and the number of 12-ounce energy drinks consumed within a week prior to the exam of 10 college students.Exam Scores - 75 - 92 - 84 - 64 - 64 - 86 - 81 - 61 - 73 - 93Number of Drinks - 5 - 3 - 2 - 4 - 2 - 7 - 3 - 0 - 1 - 01. Referring to Problem Statement 7, what is the sample covariance between the exam scores and the number of energy drinks consumed?2. Referring to Problem Statement 7, what is the sample correlation coefficient between the exam scores and the number of energy drinks consumed?

1 answer:

IgorLugansk [536]2 years ago

7 0

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.

$Cov(x,y)=sxy=\frac{Sum(X-Xbar)(Y-Ybar)}{n-1}$

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.

$Cor(x,y)=r=\frac{Sum(X-Xbar)(Y-Ybar)}{\sqrt{Sum(X-Xbar)^2sum(Y-Ybar)^2} }$

$Cor(x,y)=r=\frac{33.9}{\sqrt{(1240.1)(44.1)} }$

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.

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Step-by-step explanation:

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