The degrees of freedom for the provided statistic.

Question

Want to see more full solutions like this?

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,

Introduction to the Practice of Statistics, 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.

Want to see more full solutions like this?

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

Students have asked these similar questions

An Arts group holds a raffle. Each raffle ticket costs $2 and the raffle consists of 2500 tickets. The prize is a vacation worth $3,000. a. Determine your expected value if you buy one ticket. b. Determine your expected value if you buy five tickets. How much will the Arts group gain or lose if they sell all the tickets?

Please show as much work as possible to clearly show the steps you used to find each solution. If you plan to use a calculator, please be sure to clearly indicate your strategy. Consider the following game. It costs $3 each time you roll a six-sided number cube. If you roll a 6 you win $15. If you roll any other number, you receive nothing. a) Find the expected value of the game. b) If you play this game many times, will you expect to gain or lose money?

= 12:02 WeBWorK / 2024 Fall Rafeek MTH23 D02 / 9.2 Testing the Mean mu / 3 38 WEBWORK Previous Problem Problem List Next Problem 9.2 Testing the Mean mu: Problem 3 (1 point) Test the claim that the population of sophomore college students has a mean grade point average greater than 2.2. Sample statistics include n = 71, x = 2.44, and s = 0.9. Use a significance level of a = 0.01. The test statistic is The P-Value is between : The final conclusion is < P-value < A. There is sufficient evidence to support the claim that the mean grade point average is greater than 2.2. ○ B. There is not sufficient evidence to support the claim that the mean grade point average is greater than 2.2. Note: You can earn partial credit on this problem. Note: You are in the Reduced Scoring Period. All work counts for 50% of the original. Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining. . Oli wwm01.bcc.cuny.edu

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,

Introduction to the Practice of Statistics, 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.

Want to see more full solutions like this?

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

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,

Introduction to the Practice of Statistics, 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.

Want to see more full solutions like this?

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,

Introduction to the Practice of Statistics, 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.

Want to see more full solutions like this?

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,

Introduction to the Practice of Statistics, 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,

Introduction to the Practice of Statistics, 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,

Introduction to the Practice of Statistics, 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.