To construct a scatter plot .

Question

Want to see more full solutions like this?

Answer 1

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 23, Problem 23.30E

a)

To determine

To construct a scatter plot.

a)

Expert Solution

Explanation of Solution

Given:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 1

Scatter plot for Actual vs 3D volume construction is,

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 2

Regression Analysis

	r²	0.954	n	30
	r	0.977	k	1
	Std. Error	0.649	Dep. Var.	Actual

ANOVA table
Source	SS	df	MS	F	p-value
Regression	242.8360	1	242.8360	576.93	3.17E-20
Residual	11.7855	28	0.4209
Total	254.6214	29


Regression output					confidence interval
variables	coefficients	std. error	t (df=28)	p-value	95% lower	95% upper
Intercept	0.4196	0.4671	0.898	.3767	-0.5373	1.3764
3D	2.4752	0.1031	24.019	3.17E-20	2.2641	2.6863

Therefore, least square regression equation is,

y^=0.4196+2.4752x

b)

To determine

To verify the conditions for inference.

b)

Expert Solution

Answer to Problem 23.30E

All the conditions satisfied.

Explanation of Solution

Given:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 3

The scatter plot shows linearly increasing trend. Therefore, the relationship is clearly linear, the scatterplot shows no unusual pattern that would indicate not Normally distributed residuals or residuals without a constant standard deviation, and the observations are independent.

c)

To determine

To test whether the linear relationship is statistically significant.

c)

Expert Solution

Answer to Problem 23.30E

There is sufficient evidence to conclude that the linear relationship between two variables is statistically significant.

Explanation of Solution

Given:

Regression Analysis

	r²	0.954	n	30
	r	0.977	k	1
	Std. Error	0.649	Dep. Var.	Actual

ANOVA table
Source	SS	df	MS	F	p-value
Regression	242.8360	1	242.8360	576.93	3.17E-20
Residual	11.7855	28	0.4209
Total	254.6214	29


Regression output					confidence interval
variables	coefficients	std. error	t (df=28)	p-value	95% lower	95% upper
Intercept	0.4196	0.4671	0.898	.3767	-0.5373	1.3764
3D	2.4752	0.1031	24.019	3.17E-20	2.2641	2.6863

Null and alternative hypotheses:

H0:β=0

H0:β≠0

Test statistic is,

t = 24.019

P-value = 0.0000

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that the linear relationship between two variables is statistically significant.

Therefore, 95% confidence interval for slope is,

(2.2641, 2.6863)

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

Questions An insurance company's cumulative incurred claims for the last 5 accident years are given in the following table: Development Year Accident Year 0 2018 1 2 3 4 245 267 274 289 292 2019 255 276 288 294 2020 265 283 292 2021 263 278 2022 271 It can be assumed that claims are fully run off after 4 years. The premiums received for each year are: Accident Year Premium 2018 306 2019 312 2020 318 2021 326 2022 330 You do not need to make any allowance for inflation. 1. (a) Calculate the reserve at the end of 2022 using the basic chain ladder method. (b) Calculate the reserve at the end of 2022 using the Bornhuetter-Ferguson method. 2. Comment on the differences in the reserves produced by the methods in Part 1.

To help consumers in purchasing a laptop computer, Consumer Reports calculates an overall test score for each computer tested based upon rating factors such as ergonomics, portability, performance, display, and battery life. Higher overall scores indicate better test results. The following data show the average retail price and the overall score for ten 13-inch models (Consumer Reports website, October 25, 2012). Brand & Model Price ($) Overall Score Samsung Ultrabook NP900X3C-A01US 1250 83 Apple MacBook Air MC965LL/A 1300 83 Apple MacBook Air MD231LL/A 1200 82 HP ENVY 13-2050nr Spectre XT 950 79 Sony VAIO SVS13112FXB 800 77 Acer Aspire S5-391-9880 Ultrabook 1200 74 Apple MacBook Pro MD101LL/A 1200 74 Apple MacBook Pro MD313LL/A 1000 73 Dell Inspiron I13Z-6591SLV 700 67 Samsung NP535U3C-A01US 600 63 a. Select a scatter diagram with price as the independent variable. b. What does the scatter diagram developed in part (a) indicate about the relationship…

To the Internal Revenue Service, the reasonableness of total itemized deductions depends on the taxpayer’s adjusted gross income. Large deductions, which include charity and medical deductions, are more reasonable for taxpayers with large adjusted gross incomes. If a taxpayer claims larger than average itemized deductions for a given level of income, the chances of an IRS audit are increased. Data (in thousands of dollars) on adjusted gross income and the average or reasonable amount of itemized deductions follow. Adjusted Gross Income ($1000s) Reasonable Amount ofItemized Deductions ($1000s) 22 9.6 27 9.6 32 10.1 48 11.1 65 13.5 85 17.7 120 25.5 Compute b1 and b0 (to 4 decimals).b1 b0 Complete the estimated regression equation (to 2 decimals). = + x Predict a reasonable level of total itemized deductions for a taxpayer with an adjusted gross income of $52.5 thousand (to 2 decimals). thousand dollarsWhat is the value, in dollars, of…

Answer 2

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 23, Problem 23.30E

a)

To determine

To construct a scatter plot.

a)

Expert Solution

Explanation of Solution

Given:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 1

Scatter plot for Actual vs 3D volume construction is,

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 2

Regression Analysis

	r²	0.954	n	30
	r	0.977	k	1
	Std. Error	0.649	Dep. Var.	Actual

ANOVA table
Source	SS	df	MS	F	p-value
Regression	242.8360	1	242.8360	576.93	3.17E-20
Residual	11.7855	28	0.4209
Total	254.6214	29


Regression output					confidence interval
variables	coefficients	std. error	t (df=28)	p-value	95% lower	95% upper
Intercept	0.4196	0.4671	0.898	.3767	-0.5373	1.3764
3D	2.4752	0.1031	24.019	3.17E-20	2.2641	2.6863

Therefore, least square regression equation is,

y^=0.4196+2.4752x

b)

To determine

To verify the conditions for inference.

b)

Expert Solution

Answer to Problem 23.30E

All the conditions satisfied.

Explanation of Solution

Given:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 3

The scatter plot shows linearly increasing trend. Therefore, the relationship is clearly linear, the scatterplot shows no unusual pattern that would indicate not Normally distributed residuals or residuals without a constant standard deviation, and the observations are independent.

c)

To determine

To test whether the linear relationship is statistically significant.

c)

Expert Solution

Answer to Problem 23.30E

There is sufficient evidence to conclude that the linear relationship between two variables is statistically significant.

Explanation of Solution

Given:

Regression Analysis

	r²	0.954	n	30
	r	0.977	k	1
	Std. Error	0.649	Dep. Var.	Actual

ANOVA table
Source	SS	df	MS	F	p-value
Regression	242.8360	1	242.8360	576.93	3.17E-20
Residual	11.7855	28	0.4209
Total	254.6214	29


Regression output					confidence interval
variables	coefficients	std. error	t (df=28)	p-value	95% lower	95% upper
Intercept	0.4196	0.4671	0.898	.3767	-0.5373	1.3764
3D	2.4752	0.1031	24.019	3.17E-20	2.2641	2.6863

Null and alternative hypotheses:

H0:β=0

H0:β≠0

Test statistic is,

t = 24.019

P-value = 0.0000

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that the linear relationship between two variables is statistically significant.

Therefore, 95% confidence interval for slope is,

(2.2641, 2.6863)

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

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 23, Problem 23.30E

a)

To determine

To construct a scatter plot.

a)

Expert Solution

Explanation of Solution

Given:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 1

Scatter plot for Actual vs 3D volume construction is,

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 2

Regression Analysis

	r²	0.954	n	30
	r	0.977	k	1
	Std. Error	0.649	Dep. Var.	Actual

ANOVA table
Source	SS	df	MS	F	p-value
Regression	242.8360	1	242.8360	576.93	3.17E-20
Residual	11.7855	28	0.4209
Total	254.6214	29


Regression output					confidence interval
variables	coefficients	std. error	t (df=28)	p-value	95% lower	95% upper
Intercept	0.4196	0.4671	0.898	.3767	-0.5373	1.3764
3D	2.4752	0.1031	24.019	3.17E-20	2.2641	2.6863

Therefore, least square regression equation is,

y^=0.4196+2.4752x

b)

To determine

To verify the conditions for inference.

b)

Expert Solution

Answer to Problem 23.30E

All the conditions satisfied.

Explanation of Solution

Given:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 3

The scatter plot shows linearly increasing trend. Therefore, the relationship is clearly linear, the scatterplot shows no unusual pattern that would indicate not Normally distributed residuals or residuals without a constant standard deviation, and the observations are independent.

c)

To determine

To test whether the linear relationship is statistically significant.

c)

Expert Solution

Answer to Problem 23.30E

There is sufficient evidence to conclude that the linear relationship between two variables is statistically significant.

Explanation of Solution

Given:

Regression Analysis

	r²	0.954	n	30
	r	0.977	k	1
	Std. Error	0.649	Dep. Var.	Actual

ANOVA table
Source	SS	df	MS	F	p-value
Regression	242.8360	1	242.8360	576.93	3.17E-20
Residual	11.7855	28	0.4209
Total	254.6214	29


Regression output					confidence interval
variables	coefficients	std. error	t (df=28)	p-value	95% lower	95% upper
Intercept	0.4196	0.4671	0.898	.3767	-0.5373	1.3764
3D	2.4752	0.1031	24.019	3.17E-20	2.2641	2.6863

Null and alternative hypotheses:

H0:β=0

H0:β≠0

Test statistic is,

t = 24.019

P-value = 0.0000

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that the linear relationship between two variables is statistically significant.

Therefore, 95% confidence interval for slope is,

(2.2641, 2.6863)

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

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 23, Problem 23.30E

a)

To determine

To construct a scatter plot.

a)

Expert Solution

Explanation of Solution

Given:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 1

Scatter plot for Actual vs 3D volume construction is,

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 2

Regression Analysis

	r²	0.954	n	30
	r	0.977	k	1
	Std. Error	0.649	Dep. Var.	Actual

ANOVA table
Source	SS	df	MS	F	p-value
Regression	242.8360	1	242.8360	576.93	3.17E-20
Residual	11.7855	28	0.4209
Total	254.6214	29


Regression output					confidence interval
variables	coefficients	std. error	t (df=28)	p-value	95% lower	95% upper
Intercept	0.4196	0.4671	0.898	.3767	-0.5373	1.3764
3D	2.4752	0.1031	24.019	3.17E-20	2.2641	2.6863

Therefore, least square regression equation is,

y^=0.4196+2.4752x

b)

To determine

To verify the conditions for inference.

b)

Expert Solution

Answer to Problem 23.30E

All the conditions satisfied.

Explanation of Solution

Given:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 3

The scatter plot shows linearly increasing trend. Therefore, the relationship is clearly linear, the scatterplot shows no unusual pattern that would indicate not Normally distributed residuals or residuals without a constant standard deviation, and the observations are independent.

c)

To determine

To test whether the linear relationship is statistically significant.

c)

Expert Solution

Answer to Problem 23.30E

There is sufficient evidence to conclude that the linear relationship between two variables is statistically significant.

Explanation of Solution

Given:

Regression Analysis

	r²	0.954	n	30
	r	0.977	k	1
	Std. Error	0.649	Dep. Var.	Actual

ANOVA table
Source	SS	df	MS	F	p-value
Regression	242.8360	1	242.8360	576.93	3.17E-20
Residual	11.7855	28	0.4209
Total	254.6214	29


Regression output					confidence interval
variables	coefficients	std. error	t (df=28)	p-value	95% lower	95% upper
Intercept	0.4196	0.4671	0.898	.3767	-0.5373	1.3764
3D	2.4752	0.1031	24.019	3.17E-20	2.2641	2.6863

Null and alternative hypotheses:

H0:β=0

H0:β≠0

Test statistic is,

t = 24.019

P-value = 0.0000

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that the linear relationship between two variables is statistically significant.

Therefore, 95% confidence interval for slope is,

(2.2641, 2.6863)

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 23, Problem 23.30E

a)

To determine

To construct a scatter plot.

a)

Expert Solution

Explanation of Solution

Given:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 1

Scatter plot for Actual vs 3D volume construction is,

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 2

Regression Analysis

	r²	0.954	n	30
	r	0.977	k	1
	Std. Error	0.649	Dep. Var.	Actual

ANOVA table
Source	SS	df	MS	F	p-value
Regression	242.8360	1	242.8360	576.93	3.17E-20
Residual	11.7855	28	0.4209
Total	254.6214	29


Regression output					confidence interval
variables	coefficients	std. error	t (df=28)	p-value	95% lower	95% upper
Intercept	0.4196	0.4671	0.898	.3767	-0.5373	1.3764
3D	2.4752	0.1031	24.019	3.17E-20	2.2641	2.6863

Therefore, least square regression equation is,

y^=0.4196+2.4752x

b)

To determine

To verify the conditions for inference.

b)

Expert Solution

Answer to Problem 23.30E

All the conditions satisfied.

Explanation of Solution

Given:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 3

The scatter plot shows linearly increasing trend. Therefore, the relationship is clearly linear, the scatterplot shows no unusual pattern that would indicate not Normally distributed residuals or residuals without a constant standard deviation, and the observations are independent.

c)

To determine

To test whether the linear relationship is statistically significant.

c)

Expert Solution

Answer to Problem 23.30E

There is sufficient evidence to conclude that the linear relationship between two variables is statistically significant.

Explanation of Solution

Given:

Regression Analysis

	r²	0.954	n	30
	r	0.977	k	1
	Std. Error	0.649	Dep. Var.	Actual

ANOVA table
Source	SS	df	MS	F	p-value
Regression	242.8360	1	242.8360	576.93	3.17E-20
Residual	11.7855	28	0.4209
Total	254.6214	29


Regression output					confidence interval
variables	coefficients	std. error	t (df=28)	p-value	95% lower	95% upper
Intercept	0.4196	0.4671	0.898	.3767	-0.5373	1.3764
3D	2.4752	0.1031	24.019	3.17E-20	2.2641	2.6863

Null and alternative hypotheses:

H0:β=0

H0:β≠0

Test statistic is,

t = 24.019

P-value = 0.0000

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that the linear relationship between two variables is statistically significant.

Therefore, 95% confidence interval for slope is,

(2.2641, 2.6863)

Answer 6

Chapter 23, Problem 23.30E

a)

To determine

To construct a scatter plot.

a)

Expert Solution

Explanation of Solution

Given:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 1

Scatter plot for Actual vs 3D volume construction is,

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 2

Regression Analysis

	r²	0.954	n	30
	r	0.977	k	1
	Std. Error	0.649	Dep. Var.	Actual

ANOVA table
Source	SS	df	MS	F	p-value
Regression	242.8360	1	242.8360	576.93	3.17E-20
Residual	11.7855	28	0.4209
Total	254.6214	29


Regression output					confidence interval
variables	coefficients	std. error	t (df=28)	p-value	95% lower	95% upper
Intercept	0.4196	0.4671	0.898	.3767	-0.5373	1.3764
3D	2.4752	0.1031	24.019	3.17E-20	2.2641	2.6863

Therefore, least square regression equation is,

y^=0.4196+2.4752x

Answer 7

Chapter 23, Problem 23.30E

a)

To determine

To construct a scatter plot.

Answer 8

a)

Expert Solution

Explanation of Solution

Given:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 1

Scatter plot for Actual vs 3D volume construction is,

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 2

Regression Analysis

	r²	0.954	n	30
	r	0.977	k	1
	Std. Error	0.649	Dep. Var.	Actual

ANOVA table
Source	SS	df	MS	F	p-value
Regression	242.8360	1	242.8360	576.93	3.17E-20
Residual	11.7855	28	0.4209
Total	254.6214	29


Regression output					confidence interval
variables	coefficients	std. error	t (df=28)	p-value	95% lower	95% upper
Intercept	0.4196	0.4671	0.898	.3767	-0.5373	1.3764
3D	2.4752	0.1031	24.019	3.17E-20	2.2641	2.6863

Therefore, least square regression equation is,

y^=0.4196+2.4752x

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

b)

To determine

To verify the conditions for inference.

b)

Expert Solution

Answer to Problem 23.30E

All the conditions satisfied.

Explanation of Solution

Given:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 3

The scatter plot shows linearly increasing trend. Therefore, the relationship is clearly linear, the scatterplot shows no unusual pattern that would indicate not Normally distributed residuals or residuals without a constant standard deviation, and the observations are independent.

Answer 13

b)

To determine

To verify the conditions for inference.

Answer 14

b)

Expert Solution

Answer to Problem 23.30E

All the conditions satisfied.

Answer 15

b)

Expert Solution

Answer 16

b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Explanation of Solution

Given:

PRACT. OF STAT. IN LIFE SCI.W/ACHIEVE 1, Chapter 23, Problem 23.30E , additional homework tip 3

The scatter plot shows linearly increasing trend. Therefore, the relationship is clearly linear, the scatterplot shows no unusual pattern that would indicate not Normally distributed residuals or residuals without a constant standard deviation, and the observations are independent.

Answer 19

c)

To determine

To test whether the linear relationship is statistically significant.

c)

Expert Solution

Answer to Problem 23.30E

There is sufficient evidence to conclude that the linear relationship between two variables is statistically significant.

Explanation of Solution

Given:

Regression Analysis

	r²	0.954	n	30
	r	0.977	k	1
	Std. Error	0.649	Dep. Var.	Actual

ANOVA table
Source	SS	df	MS	F	p-value
Regression	242.8360	1	242.8360	576.93	3.17E-20
Residual	11.7855	28	0.4209
Total	254.6214	29


Regression output					confidence interval
variables	coefficients	std. error	t (df=28)	p-value	95% lower	95% upper
Intercept	0.4196	0.4671	0.898	.3767	-0.5373	1.3764
3D	2.4752	0.1031	24.019	3.17E-20	2.2641	2.6863

Null and alternative hypotheses:

H0:β=0

H0:β≠0

Test statistic is,

t = 24.019

P-value = 0.0000

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that the linear relationship between two variables is statistically significant.

Therefore, 95% confidence interval for slope is,

(2.2641, 2.6863)

Answer 20

c)

To determine

To test whether the linear relationship is statistically significant.

Answer 21

c)

Expert Solution

Answer to Problem 23.30E

There is sufficient evidence to conclude that the linear relationship between two variables is statistically significant.

Answer 22

c)

Expert Solution

Answer 23

c)

Expert Solution

Answer 24

Expert Solution

Answer 25

Explanation of Solution

Given:

Regression Analysis

	r²	0.954	n	30
	r	0.977	k	1
	Std. Error	0.649	Dep. Var.	Actual

ANOVA table
Source	SS	df	MS	F	p-value
Regression	242.8360	1	242.8360	576.93	3.17E-20
Residual	11.7855	28	0.4209
Total	254.6214	29


Regression output					confidence interval
variables	coefficients	std. error	t (df=28)	p-value	95% lower	95% upper
Intercept	0.4196	0.4671	0.898	.3767	-0.5373	1.3764
3D	2.4752	0.1031	24.019	3.17E-20	2.2641	2.6863