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3 months ago

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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:

tester [12.3K]3 months ago

7 0

Answer:

1. 3.767

2. 0.145

Step-by-step explanation:

Define X as the exam scores and Y as 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

Thus, the sample covariance of exam scores and energy drink consumption 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 for the relationship between exam scores and energy drink consumption is 0.145.

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The simplest form of 1.666….. is?

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El valor más simple sería 5/3 en forma de fracción.

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3 months ago

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An ANOVA procedure is applied to data obtained from 6 samples where each sample contains 20 observations. The degrees of freedom

PIT_PIT [12445]

Answer:

C. 5 degrees of freedom for the numerator and 114 for the denominator

Step-by-step explanation:

Analysis of variance (ANOVA) is utilized to examine the variations among group means within a sample.

The sum of squares represents the cumulative square of variation, which refers to the deviation of each individual value from the grand mean.

Assuming there are $6$ groups and each group contains $j=1,\dots,20$ individuals, the variation can be calculated using the following formulas:

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2$

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2$

This also has the property

$SST=SS_{between}+SS_{within}$

The numerator's degrees of freedom in this case is given by $df_{num}=df_{within}=k-1=6-1=5$ where k = 6 represents the number of groups.

The denominator's degrees of freedom in this scenario is indicated by $df_{den}=df_{between}=N-K=6*20-6=114$ .

The total degrees of freedom would be $df=N-1=6*20 -1 =119$ .

Thus, the appropriate answer would be 5 degrees of freedom for the numerator and 119 degrees of freedom for the denominator.

C. 5 numerator and 114 denominator degrees of freedom

8 0

3 months ago

One of the industrial robots designed by a leading producer of servomechanisms has four major components. Components’ reliabilit

tester [12383]

Answer:

a) Robot Reliability = 0.7876

b1) Component 1: 0.8034

Component 2: 0.8270

Component 3: 0.8349

Component 4: 0.8664

b2) To maximize overall reliability, Component 4 should be backed up.

c) To achieve the highest reliability of 0.8681, backup for Component 4 with a reliability of 0.92 should be implemented.

Step-by-step explanation:

Component Reliabilities:

Component 1 (R1): 0.98

Component 2 (R2): 0.95

Component 3 (R3): 0.94

Component 4 (R4): 0.90

a) The reliability of the robot can be determined by calculating the reliabilities of the individual components that constitute the robot.

Robot Reliability = R1 x R2 x R3 x R4

= 0.98 x 0.95 x 0.94 x 0.90

Robot Reliability = 0.787626 ≅ 0.7876

b1) As only a single backup can be used at once, and its reliability matches that of the original, we evaluate each component's backup sequentially:

Robot Reliability with Component 1 backup is calculated by first assessing the failure probability of the component plus its backup:

Failure probability = 1 - R1

= 1 - 0.98

= 0.02

Combined failure probability for Component 1 and backup = 0.02 x 0.02 = 0.0004

Thus, reliability of combined Component 1 and backup (R1B) = 1 - 0.0004 = 0.9996

Robot Reliability = R1B x R2 x R3 x R4

= 0.9996 x 0.95 x 0.94 x 0.90

Robot Reliability = 0.8034

To determine reliability of Component 2:

Failure probability for Component 2 = 1 - 0.95 = 0.05

Combined failure probability of Component 2 and backup = 0.05 x 0.05 = 0.0025

Reliability of Component 2 with backup (R2B) = 1 - 0.0025 = 0.9975

Robot Reliability = R1 x R2B x R3 x R4

= 0.98 x 0.9975 x 0.94 x 0.90

Robot Reliability = 0.8270

Robot Reliability with backup of Component 3 calculates as follows:

Failure probability for Component 3 = 1 - 0.94 = 0.06

Combined failure probability of Component 3 and backup = 0.06 x 0.06 = 0.0036

Reliability for Component 3 with backup (R3B) = 1 - 0.0036 = 0.9964

Robot Reliability = R1 x R2 x R3B x R4

= 0.98 x 0.95 x 0.9964 x 0.90

Robot Reliability = 0.8349

Robot Reliability with Component 4 backup calculates as:

Failure probability for Component 4 = 1 - 0.90 = 0.10

Combined failure probability of Component 4 and backup = 0.10 x 0.10 = 0.01

Reliability for Component 4 and backup (R4B) = 1 - 0.01 = 0.99

Robot Reliability = R1 x R2 x R3 x R4B

= 0.98 x 0.95 x 0.94 x 0.99

Robot Reliability = 0.8664

b2) The best reliability is achieved with the backup of Component 4, yielding a value of 0.8664. Thus, Component 4 is the best candidate for backup to optimize reliability.

c) A reliability of 0.92 indicates a failure probability of = 1 - 0.92 = 0.08

We can compute the probability of failure for each component along with its backup:

Component 1 = 0.02 x 0.08 = 0.0016

Component 2 = 0.05 x 0.08 = 0.0040

Component 3 = 0.06 x 0.08 = 0.0048

Component 4 = 0.10 x 0.08 = 0.0080

Thus, the reliabilities for each component and its backup become:

Component 1 (R1BB) = 1 - 0.0016 = 0.9984

Component 2 (R2BB) = 1 - 0.0040 = 0.9960

Component 3 (R3BB) = 1 - 0.0048 = 0.9952

Component 4 (R4BB) = 1 - 0.0080 = 0.9920

Reliability of robot including backups for each of the components can be calculated as:

Reliability with Backup for Component 1 = R1BB x R2 x R3 x R4

= 0.9984 x 0.95 x 0.94 x 0.90

Reliability with Backup for Component 1 = 0.8024

Reliability with Backup for Component 2 = R1 x R2BB x R3 x R4

= 0.98 x 0.9960 x 0.94 x 0.90

Reliability with Backup for Component 2 = 0.8258

Reliability with Backup for Component 3 = R1 x R2 x R3BB x R4

= 0.98 x 0.95 x 0.9952 x 0.90

Reliability with Backup for Component 3 = 0.8339

Reliability with Backup for Component 4 = R1 x R2 x R3 x R4BB

= 0.98 x 0.95 x 0.94 x 0.9920

Reliability with Backup for Component 4 = 0.8681

To maximize overall reliability, Component 4 should be backed up at a reliability of 0.92, achieving an overall reliability of 0.8681.

4 0

3 months ago