The degrees of freedom for the provided statistic.

Question

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Answer 1

Question

Chapter 12.1, Problem 11E

(a)

Section 1:

To determine

To find: The degrees of freedom for the provided statistic.

(a)

Section 1:

Expert Solution

Answer to Problem 11E

Solution: The degrees of freedom for groups is 4 and for observations is 25.

Explanation of Solution

Calculation: The degrees of freedom for groups is calculated as,

DFG=I−1=5−1=4

And the degrees of freedom for errors is computed as,

DFE=N−I=(5×6)−5=25

Section 2:

To determine

The critical values corresponding to calculated degrees of freedom.

Section 2:

Expert Solution

Answer to Problem 11E

Solution: The F statistic value will lie in between 4.18<F<6.49_.

Explanation of Solution

From the table E, provided in Appendix of the textbook the F-statistic value will lie in between 4.18 and 6.49.

(b)

To determine

To graph: A distribution plot.

(b)

Expert Solution

Explanation of Solution

Graph: To draw a probability distribution curve follow the below mentioned steps in Minitab,

Step 1: In a Minitab worksheet go to ‘Graph’ and click on Probability Distribution Plot.

Step 2: In the dialogue box that appears select ‘View Probability’ option and click OK.

Step 3: Next select the distribution as F and enter the corresponding numerator and denominator degrees of freedom in the respective fields and click on OK to obtain the required graph.

The graph obtained is attached below,

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 11E

(c)

To determine

The p-value.

(c)

Expert Solution

Answer to Problem 11E

Solution: The p-value will lie in between 0.001<P<0.01_.

Explanation of Solution

From part (a) it is observed that the F-statistic value will lie in between 4.18 and 6.49. From table E provided in Appendix of the textbook the p- value corresponding to 4.18<F<6.49 will fall in the range 0.001<P<0.01. Thus, corresponding to the provided test statistic the p-value will lie in between 0.001 and 0.01.

(d)

To determine

Weather all pairs of group means differ significantly or not.

(d)

Expert Solution

Answer to Problem 11E

Solution: It is not necessary that all the group means are different. There may be a single mean that differs.

Explanation of Solution

The p-value obtained in part (c) is very small to reject the null hypothesis. The null hypothesis assumes all the group means to be equal. If the null hypothesis is rejected then it does not implies that all the group means are unequal, there may be only one or more unequal means.

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Answer 2

Question

Chapter 12.1, Problem 11E

(a)

Section 1:

To determine

To find: The degrees of freedom for the provided statistic.

(a)

Section 1:

Expert Solution

Answer to Problem 11E

Solution: The degrees of freedom for groups is 4 and for observations is 25.

Explanation of Solution

Calculation: The degrees of freedom for groups is calculated as,

DFG=I−1=5−1=4

And the degrees of freedom for errors is computed as,

DFE=N−I=(5×6)−5=25

Section 2:

To determine

The critical values corresponding to calculated degrees of freedom.

Section 2:

Expert Solution

Answer to Problem 11E

Solution: The F statistic value will lie in between 4.18<F<6.49_.

Explanation of Solution

From the table E, provided in Appendix of the textbook the F-statistic value will lie in between 4.18 and 6.49.

(b)

To determine

To graph: A distribution plot.

(b)

Expert Solution

Explanation of Solution

Graph: To draw a probability distribution curve follow the below mentioned steps in Minitab,

Step 1: In a Minitab worksheet go to ‘Graph’ and click on Probability Distribution Plot.

Step 2: In the dialogue box that appears select ‘View Probability’ option and click OK.

Step 3: Next select the distribution as F and enter the corresponding numerator and denominator degrees of freedom in the respective fields and click on OK to obtain the required graph.

The graph obtained is attached below,

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 11E

(c)

To determine

The p-value.

(c)

Expert Solution

Answer to Problem 11E

Solution: The p-value will lie in between 0.001<P<0.01_.

Explanation of Solution

From part (a) it is observed that the F-statistic value will lie in between 4.18 and 6.49. From table E provided in Appendix of the textbook the p- value corresponding to 4.18<F<6.49 will fall in the range 0.001<P<0.01. Thus, corresponding to the provided test statistic the p-value will lie in between 0.001 and 0.01.

(d)

To determine

Weather all pairs of group means differ significantly or not.

(d)

Expert Solution

Answer to Problem 11E

Solution: It is not necessary that all the group means are different. There may be a single mean that differs.

Explanation of Solution

The p-value obtained in part (c) is very small to reject the null hypothesis. The null hypothesis assumes all the group means to be equal. If the null hypothesis is rejected then it does not implies that all the group means are unequal, there may be only one or more unequal means.

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Answer 3

Question

Chapter 12.1, Problem 11E

(a)

Section 1:

To determine

To find: The degrees of freedom for the provided statistic.

(a)

Section 1:

Expert Solution

Answer to Problem 11E

Solution: The degrees of freedom for groups is 4 and for observations is 25.

Explanation of Solution

Calculation: The degrees of freedom for groups is calculated as,

DFG=I−1=5−1=4

And the degrees of freedom for errors is computed as,

DFE=N−I=(5×6)−5=25

Section 2:

To determine

The critical values corresponding to calculated degrees of freedom.

Section 2:

Expert Solution

Answer to Problem 11E

Solution: The F statistic value will lie in between 4.18<F<6.49_.

Explanation of Solution

From the table E, provided in Appendix of the textbook the F-statistic value will lie in between 4.18 and 6.49.

(b)

To determine

To graph: A distribution plot.

(b)

Expert Solution

Explanation of Solution

Graph: To draw a probability distribution curve follow the below mentioned steps in Minitab,

Step 1: In a Minitab worksheet go to ‘Graph’ and click on Probability Distribution Plot.

Step 2: In the dialogue box that appears select ‘View Probability’ option and click OK.

Step 3: Next select the distribution as F and enter the corresponding numerator and denominator degrees of freedom in the respective fields and click on OK to obtain the required graph.

The graph obtained is attached below,

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 11E

(c)

To determine

The p-value.

(c)

Expert Solution

Answer to Problem 11E

Solution: The p-value will lie in between 0.001<P<0.01_.

Explanation of Solution

From part (a) it is observed that the F-statistic value will lie in between 4.18 and 6.49. From table E provided in Appendix of the textbook the p- value corresponding to 4.18<F<6.49 will fall in the range 0.001<P<0.01. Thus, corresponding to the provided test statistic the p-value will lie in between 0.001 and 0.01.

(d)

To determine

Weather all pairs of group means differ significantly or not.

(d)

Expert Solution

Answer to Problem 11E

Solution: It is not necessary that all the group means are different. There may be a single mean that differs.

Explanation of Solution

The p-value obtained in part (c) is very small to reject the null hypothesis. The null hypothesis assumes all the group means to be equal. If the null hypothesis is rejected then it does not implies that all the group means are unequal, there may be only one or more unequal means.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 12.1, Problem 11E

(a)

Section 1:

To determine

To find: The degrees of freedom for the provided statistic.

(a)

Section 1:

Expert Solution

Answer to Problem 11E

Solution: The degrees of freedom for groups is 4 and for observations is 25.

Explanation of Solution

Calculation: The degrees of freedom for groups is calculated as,

DFG=I−1=5−1=4

And the degrees of freedom for errors is computed as,

DFE=N−I=(5×6)−5=25

Section 2:

To determine

The critical values corresponding to calculated degrees of freedom.

Section 2:

Expert Solution

Answer to Problem 11E

Solution: The F statistic value will lie in between 4.18<F<6.49_.

Explanation of Solution

From the table E, provided in Appendix of the textbook the F-statistic value will lie in between 4.18 and 6.49.

(b)

To determine

To graph: A distribution plot.

(b)

Expert Solution

Explanation of Solution

Graph: To draw a probability distribution curve follow the below mentioned steps in Minitab,

Step 1: In a Minitab worksheet go to ‘Graph’ and click on Probability Distribution Plot.

Step 2: In the dialogue box that appears select ‘View Probability’ option and click OK.

Step 3: Next select the distribution as F and enter the corresponding numerator and denominator degrees of freedom in the respective fields and click on OK to obtain the required graph.

The graph obtained is attached below,

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 11E

(c)

To determine

The p-value.

(c)

Expert Solution

Answer to Problem 11E

Solution: The p-value will lie in between 0.001<P<0.01_.

Explanation of Solution

From part (a) it is observed that the F-statistic value will lie in between 4.18 and 6.49. From table E provided in Appendix of the textbook the p- value corresponding to 4.18<F<6.49 will fall in the range 0.001<P<0.01. Thus, corresponding to the provided test statistic the p-value will lie in between 0.001 and 0.01.

(d)

To determine

Weather all pairs of group means differ significantly or not.

(d)

Expert Solution

Answer to Problem 11E

Solution: It is not necessary that all the group means are different. There may be a single mean that differs.

Explanation of Solution

The p-value obtained in part (c) is very small to reject the null hypothesis. The null hypothesis assumes all the group means to be equal. If the null hypothesis is rejected then it does not implies that all the group means are unequal, there may be only one or more unequal means.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 12.1, Problem 11E

(a)

Section 1:

To determine

To find: The degrees of freedom for the provided statistic.

(a)

Section 1:

Expert Solution

Answer to Problem 11E

Solution: The degrees of freedom for groups is 4 and for observations is 25.

Explanation of Solution

Calculation: The degrees of freedom for groups is calculated as,

DFG=I−1=5−1=4

And the degrees of freedom for errors is computed as,

DFE=N−I=(5×6)−5=25

Section 2:

To determine

The critical values corresponding to calculated degrees of freedom.

Section 2:

Expert Solution

Answer to Problem 11E

Solution: The F statistic value will lie in between 4.18<F<6.49_.

Explanation of Solution

From the table E, provided in Appendix of the textbook the F-statistic value will lie in between 4.18 and 6.49.

(b)

To determine

To graph: A distribution plot.

(b)

Expert Solution

Explanation of Solution

Graph: To draw a probability distribution curve follow the below mentioned steps in Minitab,

Step 1: In a Minitab worksheet go to ‘Graph’ and click on Probability Distribution Plot.

Step 2: In the dialogue box that appears select ‘View Probability’ option and click OK.

Step 3: Next select the distribution as F and enter the corresponding numerator and denominator degrees of freedom in the respective fields and click on OK to obtain the required graph.

The graph obtained is attached below,

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 11E

(c)

To determine

The p-value.

(c)

Expert Solution

Answer to Problem 11E

Solution: The p-value will lie in between 0.001<P<0.01_.

Explanation of Solution

From part (a) it is observed that the F-statistic value will lie in between 4.18 and 6.49. From table E provided in Appendix of the textbook the p- value corresponding to 4.18<F<6.49 will fall in the range 0.001<P<0.01. Thus, corresponding to the provided test statistic the p-value will lie in between 0.001 and 0.01.

(d)

To determine

Weather all pairs of group means differ significantly or not.

(d)

Expert Solution

Answer to Problem 11E

Solution: It is not necessary that all the group means are different. There may be a single mean that differs.

Explanation of Solution

The p-value obtained in part (c) is very small to reject the null hypothesis. The null hypothesis assumes all the group means to be equal. If the null hypothesis is rejected then it does not implies that all the group means are unequal, there may be only one or more unequal means.

Answer 6

Chapter 12.1, Problem 11E

(a)

Section 1:

To determine

To find: The degrees of freedom for the provided statistic.

(a)

Section 1:

Expert Solution

Answer to Problem 11E

Solution: The degrees of freedom for groups is 4 and for observations is 25.

Explanation of Solution

Calculation: The degrees of freedom for groups is calculated as,

DFG=I−1=5−1=4

And the degrees of freedom for errors is computed as,

DFE=N−I=(5×6)−5=25

Answer 7

Chapter 12.1, Problem 11E

(a)

Section 1:

To determine

To find: The degrees of freedom for the provided statistic.

Answer 8

(a)

Section 1:

Expert Solution

Answer to Problem 11E

Solution: The degrees of freedom for groups is 4 and for observations is 25.

Answer 9

(a)

Section 1:

Expert Solution

Answer 10

(a)

Section 1:

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation: The degrees of freedom for groups is calculated as,

DFG=I−1=5−1=4

And the degrees of freedom for errors is computed as,

DFE=N−I=(5×6)−5=25

Answer 13

Section 2:

To determine

The critical values corresponding to calculated degrees of freedom.

Section 2:

Expert Solution

Answer to Problem 11E

Solution: The F statistic value will lie in between 4.18<F<6.49_.

Explanation of Solution

From the table E, provided in Appendix of the textbook the F-statistic value will lie in between 4.18 and 6.49.

Answer 14

Section 2:

To determine

The critical values corresponding to calculated degrees of freedom.

Answer 15

Section 2:

Expert Solution

Answer to Problem 11E

Solution: The F statistic value will lie in between 4.18<F<6.49_.

Answer 16

Section 2:

Expert Solution

Answer 17

Section 2:

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

From the table E, provided in Appendix of the textbook the F-statistic value will lie in between 4.18 and 6.49.

Answer 20

(b)

To determine

To graph: A distribution plot.

(b)

Expert Solution

Explanation of Solution

Graph: To draw a probability distribution curve follow the below mentioned steps in Minitab,

Step 1: In a Minitab worksheet go to ‘Graph’ and click on Probability Distribution Plot.

Step 2: In the dialogue box that appears select ‘View Probability’ option and click OK.

Step 3: Next select the distribution as F and enter the corresponding numerator and denominator degrees of freedom in the respective fields and click on OK to obtain the required graph.

The graph obtained is attached below,

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 11E

Answer 21

(b)

To determine

To graph: A distribution plot.

Answer 22

(b)

Expert Solution

Explanation of Solution

Graph: To draw a probability distribution curve follow the below mentioned steps in Minitab,

Step 1: In a Minitab worksheet go to ‘Graph’ and click on Probability Distribution Plot.

Step 2: In the dialogue box that appears select ‘View Probability’ option and click OK.

Step 3: Next select the distribution as F and enter the corresponding numerator and denominator degrees of freedom in the respective fields and click on OK to obtain the required graph.

The graph obtained is attached below,

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 11E

Answer 23

(b)

Expert Solution

Answer 24

(b)

Expert Solution

Answer 25

Expert Solution

Answer 26

(c)

To determine

The p-value.

(c)

Expert Solution

Answer to Problem 11E

Solution: The p-value will lie in between 0.001<P<0.01_.

Explanation of Solution

From part (a) it is observed that the F-statistic value will lie in between 4.18 and 6.49. From table E provided in Appendix of the textbook the p- value corresponding to 4.18<F<6.49 will fall in the range 0.001<P<0.01. Thus, corresponding to the provided test statistic the p-value will lie in between 0.001 and 0.01.

Answer 27

(c)

To determine

The p-value.

Answer 28

(c)

Expert Solution

Answer to Problem 11E

Solution: The p-value will lie in between 0.001<P<0.01_.

Answer 29

(c)

Expert Solution

Answer 30

(c)

Expert Solution

Answer 31

Expert Solution

Answer 32

Explanation of Solution

From part (a) it is observed that the F-statistic value will lie in between 4.18 and 6.49. From table E provided in Appendix of the textbook the p- value corresponding to 4.18<F<6.49 will fall in the range 0.001<P<0.01. Thus, corresponding to the provided test statistic the p-value will lie in between 0.001 and 0.01.

Answer 33

(d)

To determine

Weather all pairs of group means differ significantly or not.

(d)

Expert Solution

Answer to Problem 11E

Solution: It is not necessary that all the group means are different. There may be a single mean that differs.

Explanation of Solution

The p-value obtained in part (c) is very small to reject the null hypothesis. The null hypothesis assumes all the group means to be equal. If the null hypothesis is rejected then it does not implies that all the group means are unequal, there may be only one or more unequal means.

Answer 34

(d)

To determine

Weather all pairs of group means differ significantly or not.

Answer 35

(d)

Expert Solution

Answer to Problem 11E

Solution: It is not necessary that all the group means are different. There may be a single mean that differs.

Answer 36

(d)

Expert Solution

Answer 37

(d)

Expert Solution

Answer 38

Expert Solution

Answer 39

Explanation of Solution

The p-value obtained in part (c) is very small to reject the null hypothesis. The null hypothesis assumes all the group means to be equal. If the null hypothesis is rejected then it does not implies that all the group means are unequal, there may be only one or more unequal means.