To construct a scatter plot .

Question

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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:

Practice of Statistics in the Life Sciences, Chapter 23, Problem 23.30E , additional homework tip 1

Scatter plot for Actual vs 3D volume construction is,

Practice of Statistics in the Life Sciences, 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:

Practice of Statistics in the Life Sciences, 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)

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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:

Practice of Statistics in the Life Sciences, Chapter 23, Problem 23.30E , additional homework tip 1

Scatter plot for Actual vs 3D volume construction is,

Practice of Statistics in the Life Sciences, 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:

Practice of Statistics in the Life Sciences, 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)

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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:

Practice of Statistics in the Life Sciences, Chapter 23, Problem 23.30E , additional homework tip 1

Scatter plot for Actual vs 3D volume construction is,

Practice of Statistics in the Life Sciences, 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:

Practice of Statistics in the Life Sciences, 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)

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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:

Practice of Statistics in the Life Sciences, Chapter 23, Problem 23.30E , additional homework tip 1

Scatter plot for Actual vs 3D volume construction is,

Practice of Statistics in the Life Sciences, 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:

Practice of Statistics in the Life Sciences, 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)

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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 23, Problem 23.30E

a)

To determine

To construct a scatter plot.

a)

Expert Solution

Explanation of Solution

Given:

Practice of Statistics in the Life Sciences, Chapter 23, Problem 23.30E , additional homework tip 1

Scatter plot for Actual vs 3D volume construction is,

Practice of Statistics in the Life Sciences, 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:

Practice of Statistics in the Life Sciences, 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:

Practice of Statistics in the Life Sciences, Chapter 23, Problem 23.30E , additional homework tip 1

Scatter plot for Actual vs 3D volume construction is,

Practice of Statistics in the Life Sciences, 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:

Practice of Statistics in the Life Sciences, Chapter 23, Problem 23.30E , additional homework tip 1

Scatter plot for Actual vs 3D volume construction is,

Practice of Statistics in the Life Sciences, 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:

Practice of Statistics in the Life Sciences, 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:

Practice of Statistics in the Life Sciences, 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