To graph: The diagram that illustrates the effect of the confidence level on the width of the interval.

Question

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Answer 1

Question

Chapter 6.1, Problem 14E

Section 1

To determine

To graph: The diagram that illustrates the effect of the confidence level on the width of the interval.

Section 1

Expert Solution

Explanation of Solution

Calculation: The confidence intervals based on different sample sizes need to be determined in order to draw the required diagram. The lower and upper bounds of the confidence interval for the population mean $μ$ , at a significance level of $α$ is calculated using the following formula:

$Lower bound = x¯−m=x¯−Zα2×σn$

And,

$Upper bound = x¯+m=x¯+Zα2×σn$

Where $x¯$ and m represent the sample mean and margin of error respectivelyand $σ$ denotes the population standard deviation and n is the size of the sample. $Zα2$ is the critical value of the standard normal distribution at a significance level of $α$ , whose value is obtained from TableA provided in the book. For a confidence level of $80%$ , the level of significance $α=0.20$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.20/2×2849=73−1.282×2849=67.87$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.20/2×2849=73+1.282×2849=78.13$

Therefore, the required confidence interval is $(67.87, 78.13)$ .

For a confidence level of $90%$ , the level of significance $α=0.10$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.10/2×2849=73−1.645×2849=66.42$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.10/2×2849=73+1.645×2849=79.58$

Therefore, the required confidence interval is $(66.42, 79.58)$ .

For a confidence level of $95%$ , the level of significance $α=0.05$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.05/2×2849=73−1.96×2849=65.16$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.05/2×2849=73+1.96×2849=80.84$

Therefore, the required confidence interval is $(65.16, 80.84)$ .

For a confidence level of $99%$ , the level of significance $α=0.01$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.01/2×2849=73−2.58×2849=62.68$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.01/2×2849=73+2.58×2849=83.32$

Therefore, the required confidence interval is $(62.68, 83.32)$ . Hence, the diagram which shows the effect of the confidence level on the width of the interval is shown below:

Graph:

Section 2

To determine

To explain: The results.

Section 2

Expert Solution

Answer to Problem 14E

Solution: The width of the confidence interval expands due to hike in the level of confidence.

Explanation of Solution

The margin of error rises with a rise in the confidence level because there is a direct relationship between the level of confidence and margin of error. This leads to an expansion in the width of the confidence interval. Hence, it can be concluded that as level of confidence increases, the width of the confidence interval increases.

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Answer 2

Question

Chapter 6.1, Problem 14E

Section 1

To determine

To graph: The diagram that illustrates the effect of the confidence level on the width of the interval.

Section 1

Expert Solution

Explanation of Solution

Calculation: The confidence intervals based on different sample sizes need to be determined in order to draw the required diagram. The lower and upper bounds of the confidence interval for the population mean $μ$ , at a significance level of $α$ is calculated using the following formula:

$Lower bound = x¯−m=x¯−Zα2×σn$

And,

$Upper bound = x¯+m=x¯+Zα2×σn$

Where $x¯$ and m represent the sample mean and margin of error respectivelyand $σ$ denotes the population standard deviation and n is the size of the sample. $Zα2$ is the critical value of the standard normal distribution at a significance level of $α$ , whose value is obtained from TableA provided in the book. For a confidence level of $80%$ , the level of significance $α=0.20$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.20/2×2849=73−1.282×2849=67.87$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.20/2×2849=73+1.282×2849=78.13$

Therefore, the required confidence interval is $(67.87, 78.13)$ .

For a confidence level of $90%$ , the level of significance $α=0.10$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.10/2×2849=73−1.645×2849=66.42$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.10/2×2849=73+1.645×2849=79.58$

Therefore, the required confidence interval is $(66.42, 79.58)$ .

For a confidence level of $95%$ , the level of significance $α=0.05$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.05/2×2849=73−1.96×2849=65.16$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.05/2×2849=73+1.96×2849=80.84$

Therefore, the required confidence interval is $(65.16, 80.84)$ .

For a confidence level of $99%$ , the level of significance $α=0.01$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.01/2×2849=73−2.58×2849=62.68$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.01/2×2849=73+2.58×2849=83.32$

Therefore, the required confidence interval is $(62.68, 83.32)$ . Hence, the diagram which shows the effect of the confidence level on the width of the interval is shown below:

Graph:

Section 2

To determine

To explain: The results.

Section 2

Expert Solution

Answer to Problem 14E

Solution: The width of the confidence interval expands due to hike in the level of confidence.

Explanation of Solution

The margin of error rises with a rise in the confidence level because there is a direct relationship between the level of confidence and margin of error. This leads to an expansion in the width of the confidence interval. Hence, it can be concluded that as level of confidence increases, the width of the confidence interval increases.

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Answer 3

Question

Chapter 6.1, Problem 14E

Section 1

To determine

To graph: The diagram that illustrates the effect of the confidence level on the width of the interval.

Section 1

Expert Solution

Explanation of Solution

Calculation: The confidence intervals based on different sample sizes need to be determined in order to draw the required diagram. The lower and upper bounds of the confidence interval for the population mean $μ$ , at a significance level of $α$ is calculated using the following formula:

$Lower bound = x¯−m=x¯−Zα2×σn$

And,

$Upper bound = x¯+m=x¯+Zα2×σn$

Where $x¯$ and m represent the sample mean and margin of error respectivelyand $σ$ denotes the population standard deviation and n is the size of the sample. $Zα2$ is the critical value of the standard normal distribution at a significance level of $α$ , whose value is obtained from TableA provided in the book. For a confidence level of $80%$ , the level of significance $α=0.20$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.20/2×2849=73−1.282×2849=67.87$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.20/2×2849=73+1.282×2849=78.13$

Therefore, the required confidence interval is $(67.87, 78.13)$ .

For a confidence level of $90%$ , the level of significance $α=0.10$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.10/2×2849=73−1.645×2849=66.42$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.10/2×2849=73+1.645×2849=79.58$

Therefore, the required confidence interval is $(66.42, 79.58)$ .

For a confidence level of $95%$ , the level of significance $α=0.05$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.05/2×2849=73−1.96×2849=65.16$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.05/2×2849=73+1.96×2849=80.84$

Therefore, the required confidence interval is $(65.16, 80.84)$ .

For a confidence level of $99%$ , the level of significance $α=0.01$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.01/2×2849=73−2.58×2849=62.68$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.01/2×2849=73+2.58×2849=83.32$

Therefore, the required confidence interval is $(62.68, 83.32)$ . Hence, the diagram which shows the effect of the confidence level on the width of the interval is shown below:

Graph:

Section 2

To determine

To explain: The results.

Section 2

Expert Solution

Answer to Problem 14E

Solution: The width of the confidence interval expands due to hike in the level of confidence.

Explanation of Solution

The margin of error rises with a rise in the confidence level because there is a direct relationship between the level of confidence and margin of error. This leads to an expansion in the width of the confidence interval. Hence, it can be concluded that as level of confidence increases, the width of the confidence interval increases.

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Answer 4

Question

Chapter 6.1, Problem 14E

Section 1

To determine

To graph: The diagram that illustrates the effect of the confidence level on the width of the interval.

Section 1

Expert Solution

Explanation of Solution

Calculation: The confidence intervals based on different sample sizes need to be determined in order to draw the required diagram. The lower and upper bounds of the confidence interval for the population mean $μ$ , at a significance level of $α$ is calculated using the following formula:

$Lower bound = x¯−m=x¯−Zα2×σn$

And,

$Upper bound = x¯+m=x¯+Zα2×σn$

Where $x¯$ and m represent the sample mean and margin of error respectivelyand $σ$ denotes the population standard deviation and n is the size of the sample. $Zα2$ is the critical value of the standard normal distribution at a significance level of $α$ , whose value is obtained from TableA provided in the book. For a confidence level of $80%$ , the level of significance $α=0.20$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.20/2×2849=73−1.282×2849=67.87$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.20/2×2849=73+1.282×2849=78.13$

Therefore, the required confidence interval is $(67.87, 78.13)$ .

For a confidence level of $90%$ , the level of significance $α=0.10$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.10/2×2849=73−1.645×2849=66.42$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.10/2×2849=73+1.645×2849=79.58$

Therefore, the required confidence interval is $(66.42, 79.58)$ .

For a confidence level of $95%$ , the level of significance $α=0.05$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.05/2×2849=73−1.96×2849=65.16$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.05/2×2849=73+1.96×2849=80.84$

Therefore, the required confidence interval is $(65.16, 80.84)$ .

For a confidence level of $99%$ , the level of significance $α=0.01$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.01/2×2849=73−2.58×2849=62.68$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.01/2×2849=73+2.58×2849=83.32$

Therefore, the required confidence interval is $(62.68, 83.32)$ . Hence, the diagram which shows the effect of the confidence level on the width of the interval is shown below:

Graph:

Section 2

To determine

To explain: The results.

Section 2

Expert Solution

Answer to Problem 14E

Solution: The width of the confidence interval expands due to hike in the level of confidence.

Explanation of Solution

The margin of error rises with a rise in the confidence level because there is a direct relationship between the level of confidence and margin of error. This leads to an expansion in the width of the confidence interval. Hence, it can be concluded that as level of confidence increases, the width of the confidence interval increases.

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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 6.1, Problem 14E

Section 1

To determine

To graph: The diagram that illustrates the effect of the confidence level on the width of the interval.

Section 1

Expert Solution

Explanation of Solution

Calculation: The confidence intervals based on different sample sizes need to be determined in order to draw the required diagram. The lower and upper bounds of the confidence interval for the population mean $μ$ , at a significance level of $α$ is calculated using the following formula:

$Lower bound = x¯−m=x¯−Zα2×σn$

And,

$Upper bound = x¯+m=x¯+Zα2×σn$

Where $x¯$ and m represent the sample mean and margin of error respectivelyand $σ$ denotes the population standard deviation and n is the size of the sample. $Zα2$ is the critical value of the standard normal distribution at a significance level of $α$ , whose value is obtained from TableA provided in the book. For a confidence level of $80%$ , the level of significance $α=0.20$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.20/2×2849=73−1.282×2849=67.87$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.20/2×2849=73+1.282×2849=78.13$

Therefore, the required confidence interval is $(67.87, 78.13)$ .

For a confidence level of $90%$ , the level of significance $α=0.10$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.10/2×2849=73−1.645×2849=66.42$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.10/2×2849=73+1.645×2849=79.58$

Therefore, the required confidence interval is $(66.42, 79.58)$ .

For a confidence level of $95%$ , the level of significance $α=0.05$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.05/2×2849=73−1.96×2849=65.16$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.05/2×2849=73+1.96×2849=80.84$

Therefore, the required confidence interval is $(65.16, 80.84)$ .

For a confidence level of $99%$ , the level of significance $α=0.01$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.01/2×2849=73−2.58×2849=62.68$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.01/2×2849=73+2.58×2849=83.32$

Therefore, the required confidence interval is $(62.68, 83.32)$ . Hence, the diagram which shows the effect of the confidence level on the width of the interval is shown below:

Graph:

Section 2

To determine

To explain: The results.

Section 2

Expert Solution

Answer to Problem 14E

Solution: The width of the confidence interval expands due to hike in the level of confidence.

Explanation of Solution

The margin of error rises with a rise in the confidence level because there is a direct relationship between the level of confidence and margin of error. This leads to an expansion in the width of the confidence interval. Hence, it can be concluded that as level of confidence increases, the width of the confidence interval increases.

Answer 6

Chapter 6.1, Problem 14E

Section 1

To determine

To graph: The diagram that illustrates the effect of the confidence level on the width of the interval.

Section 1

Expert Solution

Explanation of Solution

Calculation: The confidence intervals based on different sample sizes need to be determined in order to draw the required diagram. The lower and upper bounds of the confidence interval for the population mean $μ$ , at a significance level of $α$ is calculated using the following formula:

$Lower bound = x¯−m=x¯−Zα2×σn$

And,

$Upper bound = x¯+m=x¯+Zα2×σn$

Where $x¯$ and m represent the sample mean and margin of error respectivelyand $σ$ denotes the population standard deviation and n is the size of the sample. $Zα2$ is the critical value of the standard normal distribution at a significance level of $α$ , whose value is obtained from TableA provided in the book. For a confidence level of $80%$ , the level of significance $α=0.20$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.20/2×2849=73−1.282×2849=67.87$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.20/2×2849=73+1.282×2849=78.13$

Therefore, the required confidence interval is $(67.87, 78.13)$ .

For a confidence level of $90%$ , the level of significance $α=0.10$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.10/2×2849=73−1.645×2849=66.42$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.10/2×2849=73+1.645×2849=79.58$

Therefore, the required confidence interval is $(66.42, 79.58)$ .

For a confidence level of $95%$ , the level of significance $α=0.05$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.05/2×2849=73−1.96×2849=65.16$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.05/2×2849=73+1.96×2849=80.84$

Therefore, the required confidence interval is $(65.16, 80.84)$ .

For a confidence level of $99%$ , the level of significance $α=0.01$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.01/2×2849=73−2.58×2849=62.68$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.01/2×2849=73+2.58×2849=83.32$

Therefore, the required confidence interval is $(62.68, 83.32)$ . Hence, the diagram which shows the effect of the confidence level on the width of the interval is shown below:

Graph:

Answer 7

Chapter 6.1, Problem 14E

Section 1

To determine

To graph: The diagram that illustrates the effect of the confidence level on the width of the interval.

Answer 8

Section 1

Expert Solution

Explanation of Solution

Calculation: The confidence intervals based on different sample sizes need to be determined in order to draw the required diagram. The lower and upper bounds of the confidence interval for the population mean $μ$ , at a significance level of $α$ is calculated using the following formula:

$Lower bound = x¯−m=x¯−Zα2×σn$

And,

$Upper bound = x¯+m=x¯+Zα2×σn$

Where $x¯$ and m represent the sample mean and margin of error respectivelyand $σ$ denotes the population standard deviation and n is the size of the sample. $Zα2$ is the critical value of the standard normal distribution at a significance level of $α$ , whose value is obtained from TableA provided in the book. For a confidence level of $80%$ , the level of significance $α=0.20$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.20/2×2849=73−1.282×2849=67.87$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.20/2×2849=73+1.282×2849=78.13$

Therefore, the required confidence interval is $(67.87, 78.13)$ .

For a confidence level of $90%$ , the level of significance $α=0.10$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.10/2×2849=73−1.645×2849=66.42$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.10/2×2849=73+1.645×2849=79.58$

Therefore, the required confidence interval is $(66.42, 79.58)$ .

For a confidence level of $95%$ , the level of significance $α=0.05$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.05/2×2849=73−1.96×2849=65.16$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.05/2×2849=73+1.96×2849=80.84$

Therefore, the required confidence interval is $(65.16, 80.84)$ .

For a confidence level of $99%$ , the level of significance $α=0.01$ . Therefore,

$Lower bound = x¯−Zα/2×σn=73−Z0.01/2×2849=73−2.58×2849=62.68$

And,

$Upper bound = x¯+Zα/2×σn=73+Z0.01/2×2849=73+2.58×2849=83.32$

Therefore, the required confidence interval is $(62.68, 83.32)$ . Hence, the diagram which shows the effect of the confidence level on the width of the interval is shown below:

Graph:

Answer 9

Section 1

Expert Solution

Answer 10

Section 1

Expert Solution

Answer 11

Expert Solution

Answer 12

Section 2

To determine

To explain: The results.

Section 2

Expert Solution

Answer to Problem 14E

Solution: The width of the confidence interval expands due to hike in the level of confidence.

Explanation of Solution

The margin of error rises with a rise in the confidence level because there is a direct relationship between the level of confidence and margin of error. This leads to an expansion in the width of the confidence interval. Hence, it can be concluded that as level of confidence increases, the width of the confidence interval increases.

Answer 13

Section 2

To determine

To explain: The results.

Answer 14

Section 2

Expert Solution

Answer to Problem 14E

Solution: The width of the confidence interval expands due to hike in the level of confidence.

Answer 15

Section 2

Expert Solution

Answer 16

Section 2

Expert Solution

Answer 17

Expert Solution

Answer 18

Explanation of Solution

The margin of error rises with a rise in the confidence level because there is a direct relationship between the level of confidence and margin of error. This leads to an expansion in the width of the confidence interval. Hence, it can be concluded that as level of confidence increases, the width of the confidence interval increases.

Answer 19

Ch. 6.1 - Prob. 1UYK Ch. 6.1 - Prob. 2UYK Ch. 6.1 - Prob. 3UYK Ch. 6.1 - Prob. 4UYK Ch. 6.1 - Prob. 5UYK Ch. 6.1 - Prob. 6UYK Ch. 6.1 - Prob. 7UYK Ch. 6.1 - Prob. 8UYK Ch. 6.1 - Prob. 9UYK Ch. 6.1 - Prob. 10UYK

To graph: The diagram that illustrates the effect of the confidence level on the width of the interval.

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To graph: The diagram that illustrates the effect of the confidence level on the width of the interval.

Explanation of Solution

To explain: The results.

Answer to Problem 14E

Explanation of Solution

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