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

Does there appear to be a positive or negative relationship between price and screen size? Use a scatter plot to examine the relationship. How to take snapshots: if you use a MacBook, press Command+ Shift+4 to take snapshots. If you are using Windows, use the Snipping Tool to take snapshots. Question 1: Determine and interpret the correlation coefficient between the two variables. In your interpretation, discuss the direction of the relationship (positive, negative, or zero relationship). Also discuss the strength of the relationship. Value of correlation coefficient: Direction of the relationship (positive, negative, or zero relationship): Strength of the relationship (strong/moderate/weak): Question 2: Estimate the relationship between screen size and price using a simple linear regression model and interpret the estimated coefficients. In your interpretation, tell the dollar amount by which price will change for each unit of increase in screen size. (The answer for the…

In this problem, we consider a Brownian motion (W+) t≥0. We consider a stock model (St)t>0 given (under the measure P) by d.St 0.03 St dt + 0.2 St dwt, with So 2. We assume that the interest rate is r = 0.06. The purpose of this problem is to price an option on this stock (which we name cubic put). This option is European-type, with maturity 3 months (i.e. T = 0.25 years), and payoff given by F = (8-5)+ (a) Write the Stochastic Differential Equation satisfied by (St) under the risk-neutral measure Q. (You don't need to prove it, simply give the answer.) (b) Give the price of a regular European put on (St) with maturity 3 months and strike K = 2. (c) Let X = S. Find the Stochastic Differential Equation satisfied by the process (Xt) under the measure Q. (d) Find an explicit expression for X₁ = S3 under measure Q. (e) Using the results above, find the price of the cubic put option mentioned above. (f) Is the price in (e) the same as in question (b)? (Explain why.)

Problem 4. Margrabe formula and the Greeks (20 pts) In the homework, we determined the Margrabe formula for the price of an option allowing you to swap an x-stock for a y-stock at time T. For stocks with initial values xo, yo, common volatility σ and correlation p, the formula was given by Fo=yo (d+)-x0Þ(d_), where In (±² Ꭲ d+ õ√T and σ = σ√√√2(1 - p). дго (a) We want to determine a "Greek" for ỡ on the option: find a formula for θα (b) Is дго θα positive or negative? (c) We consider a situation in which the correlation p between the two stocks increases: what can you say about the price Fo? (d) Assume that yo< xo and p = 1. What is the price of the option?

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.