To construct scatter plot and conclude about the correlation .

Question

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Answer 1

Question

Chapter 23, Problem 23.1AYK

a)

To determine

To construct scatter plot and conclude about the correlation.

a)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

The scatter plot:

PRACT STAT W/ ACCESS 6MO LOOSELEAF, Chapter 23, Problem 23.1AYK

Yes, there is strong relationship between two variables.

b)

To determine

To find slope and y-intercept.

b)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

Regression analysis:

Regression Analysis

	r²	0.862	n	6
	r	-0.928	k	1
	Std. Error	0.443	Dep. Var.	Y

ANOVA table
Source	SS	df	MS	F	p-value
Regression	4.8893	1	4.8893	24.89	.0075
Residual	0.7857	4	0.1964
Total	5.6750	5


Regression output
variables	coefficients	std. error	t (df=4)	p-value
Intercept	-1.30	0.41	-3.151	.0345
X	-0.53	0.11	-4.989	.0075

Slope = β = -0.53

Y-intercept = α = -1.30

Regression equation is,

y^=−1.30+(−0.53)x

c)

To determine

To find the residuals and standard error.

c)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

Slope = β = -0.53

Y-intercept = α = -1.30

Regression equation is,

y^=−1.30+(−0.53)x

The residuals are:

X	Y	Predicted	Residual	(Residual)2
1	-1.90	-1.83	-0.07	0.005
2	-2.50	-2.36	-0.14	0.020
3	-2.20	-2.89	0.69	0.470
4	-3.90	-3.41	-0.49	0.236
5	-4.10	-3.94	-0.16	0.025
6	-4.30	-4.47	0.17	0.029
Total			0.00	0.786

So, SSE = 0.786

Therefore, standard error is,

σ=SSEn−2=0.7866−2=0.443

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Answer 2

Question

Chapter 23, Problem 23.1AYK

a)

To determine

To construct scatter plot and conclude about the correlation.

a)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

The scatter plot:

PRACT STAT W/ ACCESS 6MO LOOSELEAF, Chapter 23, Problem 23.1AYK

Yes, there is strong relationship between two variables.

b)

To determine

To find slope and y-intercept.

b)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

Regression analysis:

Regression Analysis

	r²	0.862	n	6
	r	-0.928	k	1
	Std. Error	0.443	Dep. Var.	Y

ANOVA table
Source	SS	df	MS	F	p-value
Regression	4.8893	1	4.8893	24.89	.0075
Residual	0.7857	4	0.1964
Total	5.6750	5


Regression output
variables	coefficients	std. error	t (df=4)	p-value
Intercept	-1.30	0.41	-3.151	.0345
X	-0.53	0.11	-4.989	.0075

Slope = β = -0.53

Y-intercept = α = -1.30

Regression equation is,

y^=−1.30+(−0.53)x

c)

To determine

To find the residuals and standard error.

c)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

Slope = β = -0.53

Y-intercept = α = -1.30

Regression equation is,

y^=−1.30+(−0.53)x

The residuals are:

X	Y	Predicted	Residual	(Residual)2
1	-1.90	-1.83	-0.07	0.005
2	-2.50	-2.36	-0.14	0.020
3	-2.20	-2.89	0.69	0.470
4	-3.90	-3.41	-0.49	0.236
5	-4.10	-3.94	-0.16	0.025
6	-4.30	-4.47	0.17	0.029
Total			0.00	0.786

So, SSE = 0.786

Therefore, standard error is,

σ=SSEn−2=0.7866−2=0.443

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Answer 3

Question

Chapter 23, Problem 23.1AYK

a)

To determine

To construct scatter plot and conclude about the correlation.

a)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

The scatter plot:

PRACT STAT W/ ACCESS 6MO LOOSELEAF, Chapter 23, Problem 23.1AYK

Yes, there is strong relationship between two variables.

b)

To determine

To find slope and y-intercept.

b)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

Regression analysis:

Regression Analysis

	r²	0.862	n	6
	r	-0.928	k	1
	Std. Error	0.443	Dep. Var.	Y

ANOVA table
Source	SS	df	MS	F	p-value
Regression	4.8893	1	4.8893	24.89	.0075
Residual	0.7857	4	0.1964
Total	5.6750	5


Regression output
variables	coefficients	std. error	t (df=4)	p-value
Intercept	-1.30	0.41	-3.151	.0345
X	-0.53	0.11	-4.989	.0075

Slope = β = -0.53

Y-intercept = α = -1.30

Regression equation is,

y^=−1.30+(−0.53)x

c)

To determine

To find the residuals and standard error.

c)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

Slope = β = -0.53

Y-intercept = α = -1.30

Regression equation is,

y^=−1.30+(−0.53)x

The residuals are:

X	Y	Predicted	Residual	(Residual)2
1	-1.90	-1.83	-0.07	0.005
2	-2.50	-2.36	-0.14	0.020
3	-2.20	-2.89	0.69	0.470
4	-3.90	-3.41	-0.49	0.236
5	-4.10	-3.94	-0.16	0.025
6	-4.30	-4.47	0.17	0.029
Total			0.00	0.786

So, SSE = 0.786

Therefore, standard error is,

σ=SSEn−2=0.7866−2=0.443

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Answer 4

Question

Chapter 23, Problem 23.1AYK

a)

To determine

To construct scatter plot and conclude about the correlation.

a)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

The scatter plot:

PRACT STAT W/ ACCESS 6MO LOOSELEAF, Chapter 23, Problem 23.1AYK

Yes, there is strong relationship between two variables.

b)

To determine

To find slope and y-intercept.

b)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

Regression analysis:

Regression Analysis

	r²	0.862	n	6
	r	-0.928	k	1
	Std. Error	0.443	Dep. Var.	Y

ANOVA table
Source	SS	df	MS	F	p-value
Regression	4.8893	1	4.8893	24.89	.0075
Residual	0.7857	4	0.1964
Total	5.6750	5


Regression output
variables	coefficients	std. error	t (df=4)	p-value
Intercept	-1.30	0.41	-3.151	.0345
X	-0.53	0.11	-4.989	.0075

Slope = β = -0.53

Y-intercept = α = -1.30

Regression equation is,

y^=−1.30+(−0.53)x

c)

To determine

To find the residuals and standard error.

c)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

Slope = β = -0.53

Y-intercept = α = -1.30

Regression equation is,

y^=−1.30+(−0.53)x

The residuals are:

X	Y	Predicted	Residual	(Residual)2
1	-1.90	-1.83	-0.07	0.005
2	-2.50	-2.36	-0.14	0.020
3	-2.20	-2.89	0.69	0.470
4	-3.90	-3.41	-0.49	0.236
5	-4.10	-3.94	-0.16	0.025
6	-4.30	-4.47	0.17	0.029
Total			0.00	0.786

So, SSE = 0.786

Therefore, standard error is,

σ=SSEn−2=0.7866−2=0.443

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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 23.1AYK

a)

To determine

To construct scatter plot and conclude about the correlation.

a)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

The scatter plot:

PRACT STAT W/ ACCESS 6MO LOOSELEAF, Chapter 23, Problem 23.1AYK

Yes, there is strong relationship between two variables.

b)

To determine

To find slope and y-intercept.

b)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

Regression analysis:

Regression Analysis

	r²	0.862	n	6
	r	-0.928	k	1
	Std. Error	0.443	Dep. Var.	Y

ANOVA table
Source	SS	df	MS	F	p-value
Regression	4.8893	1	4.8893	24.89	.0075
Residual	0.7857	4	0.1964
Total	5.6750	5


Regression output
variables	coefficients	std. error	t (df=4)	p-value
Intercept	-1.30	0.41	-3.151	.0345
X	-0.53	0.11	-4.989	.0075

Slope = β = -0.53

Y-intercept = α = -1.30

Regression equation is,

y^=−1.30+(−0.53)x

c)

To determine

To find the residuals and standard error.

c)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

Slope = β = -0.53

Y-intercept = α = -1.30

Regression equation is,

y^=−1.30+(−0.53)x

The residuals are:

X	Y	Predicted	Residual	(Residual)2
1	-1.90	-1.83	-0.07	0.005
2	-2.50	-2.36	-0.14	0.020
3	-2.20	-2.89	0.69	0.470
4	-3.90	-3.41	-0.49	0.236
5	-4.10	-3.94	-0.16	0.025
6	-4.30	-4.47	0.17	0.029
Total			0.00	0.786

So, SSE = 0.786

Therefore, standard error is,

σ=SSEn−2=0.7866−2=0.443

Answer 6

Chapter 23, Problem 23.1AYK

a)

To determine

To construct scatter plot and conclude about the correlation.

a)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

The scatter plot:

PRACT STAT W/ ACCESS 6MO LOOSELEAF, Chapter 23, Problem 23.1AYK

Yes, there is strong relationship between two variables.

Answer 7

Chapter 23, Problem 23.1AYK

a)

To determine

To construct scatter plot and conclude about the correlation.

Answer 8

a)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

The scatter plot:

PRACT STAT W/ ACCESS 6MO LOOSELEAF, Chapter 23, Problem 23.1AYK

Yes, there is strong relationship between two variables.

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

b)

To determine

To find slope and y-intercept.

b)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

Regression analysis:

Regression Analysis

	r²	0.862	n	6
	r	-0.928	k	1
	Std. Error	0.443	Dep. Var.	Y

ANOVA table
Source	SS	df	MS	F	p-value
Regression	4.8893	1	4.8893	24.89	.0075
Residual	0.7857	4	0.1964
Total	5.6750	5


Regression output
variables	coefficients	std. error	t (df=4)	p-value
Intercept	-1.30	0.41	-3.151	.0345
X	-0.53	0.11	-4.989	.0075

Slope = β = -0.53

Y-intercept = α = -1.30

Regression equation is,

y^=−1.30+(−0.53)x

Answer 13

b)

To determine

To find slope and y-intercept.

Answer 14

b)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

Regression analysis:

Regression Analysis

	r²	0.862	n	6
	r	-0.928	k	1
	Std. Error	0.443	Dep. Var.	Y

ANOVA table
Source	SS	df	MS	F	p-value
Regression	4.8893	1	4.8893	24.89	.0075
Residual	0.7857	4	0.1964
Total	5.6750	5


Regression output
variables	coefficients	std. error	t (df=4)	p-value
Intercept	-1.30	0.41	-3.151	.0345
X	-0.53	0.11	-4.989	.0075

Slope = β = -0.53

Y-intercept = α = -1.30

Regression equation is,

y^=−1.30+(−0.53)x

Answer 15

b)

Expert Solution

Answer 16

b)

Expert Solution

Answer 17

Expert Solution

Answer 18

c)

To determine

To find the residuals and standard error.

c)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

Slope = β = -0.53

Y-intercept = α = -1.30

Regression equation is,

y^=−1.30+(−0.53)x

The residuals are:

X	Y	Predicted	Residual	(Residual)2
1	-1.90	-1.83	-0.07	0.005
2	-2.50	-2.36	-0.14	0.020
3	-2.20	-2.89	0.69	0.470
4	-3.90	-3.41	-0.49	0.236
5	-4.10	-3.94	-0.16	0.025
6	-4.30	-4.47	0.17	0.029
Total			0.00	0.786

So, SSE = 0.786

Therefore, standard error is,

σ=SSEn−2=0.7866−2=0.443

Answer 19

c)

To determine

To find the residuals and standard error.

Answer 20

c)

Expert Solution

Explanation of Solution

Given:

Number of D.magna grazers	Net growth rate of G. semen
1	-1.9
2	-2.5
3	-2.2
4	-3.9
5	-4.1
6	-4.3

Slope = β = -0.53

Y-intercept = α = -1.30

Regression equation is,

y^=−1.30+(−0.53)x

The residuals are:

X	Y	Predicted	Residual	(Residual)2
1	-1.90	-1.83	-0.07	0.005
2	-2.50	-2.36	-0.14	0.020
3	-2.20	-2.89	0.69	0.470
4	-3.90	-3.41	-0.49	0.236
5	-4.10	-3.94	-0.16	0.025
6	-4.30	-4.47	0.17	0.029
Total			0.00	0.786