To find t statistic and p-value.

Question

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Answer 1

Question

Chapter 23, Problem 23.5AYK

a)

To determine

To find t statistic and p-value.

a)

Expert Solution

Answer to Problem 23.5AYK

Test statistic is -4.989 and p-value is 0.0038

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

	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

Null and alternative hypotheses:

H0:β=0

H0:β≠0

Test statistic is,

t = -4.989

Therefore, P-value of one tailed test is half of the two-tailed test.

P−value=0.00752=0.0038

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.

b)

To determine

To find correlation coefficient and find p-value for this one-tailed test.

b)

Expert Solution

Answer to Problem 23.5AYK

Test statistic is -4.981 and p-value is 0.0038

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

	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

Formula:

t=rn−21−r2

Correlation coefficient is r = -0.928

Null and alternative hypotheses:

H0:ρ=0

H0:ρ<0

Test statistic is

t=−0.9286−21−(−0.928)2t=−4.981

Therefore, P-value of one tailed test is.

P−value=0.0038

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.

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Answer 2

Question

Chapter 23, Problem 23.5AYK

a)

To determine

To find t statistic and p-value.

a)

Expert Solution

Answer to Problem 23.5AYK

Test statistic is -4.989 and p-value is 0.0038

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

	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

Null and alternative hypotheses:

H0:β=0

H0:β≠0

Test statistic is,

t = -4.989

Therefore, P-value of one tailed test is half of the two-tailed test.

P−value=0.00752=0.0038

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.

b)

To determine

To find correlation coefficient and find p-value for this one-tailed test.

b)

Expert Solution

Answer to Problem 23.5AYK

Test statistic is -4.981 and p-value is 0.0038

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

	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

Formula:

t=rn−21−r2

Correlation coefficient is r = -0.928

Null and alternative hypotheses:

H0:ρ=0

H0:ρ<0

Test statistic is

t=−0.9286−21−(−0.928)2t=−4.981

Therefore, P-value of one tailed test is.

P−value=0.0038

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.

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Answer 3

Question

Chapter 23, Problem 23.5AYK

a)

To determine

To find t statistic and p-value.

a)

Expert Solution

Answer to Problem 23.5AYK

Test statistic is -4.989 and p-value is 0.0038

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

	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

Null and alternative hypotheses:

H0:β=0

H0:β≠0

Test statistic is,

t = -4.989

Therefore, P-value of one tailed test is half of the two-tailed test.

P−value=0.00752=0.0038

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.

b)

To determine

To find correlation coefficient and find p-value for this one-tailed test.

b)

Expert Solution

Answer to Problem 23.5AYK

Test statistic is -4.981 and p-value is 0.0038

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

	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

Formula:

t=rn−21−r2

Correlation coefficient is r = -0.928

Null and alternative hypotheses:

H0:ρ=0

H0:ρ<0

Test statistic is

t=−0.9286−21−(−0.928)2t=−4.981

Therefore, P-value of one tailed test is.

P−value=0.0038

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.

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Answer 4

Question

Chapter 23, Problem 23.5AYK

a)

To determine

To find t statistic and p-value.

a)

Expert Solution

Answer to Problem 23.5AYK

Test statistic is -4.989 and p-value is 0.0038

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

	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

Null and alternative hypotheses:

H0:β=0

H0:β≠0

Test statistic is,

t = -4.989

Therefore, P-value of one tailed test is half of the two-tailed test.

P−value=0.00752=0.0038

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.

b)

To determine

To find correlation coefficient and find p-value for this one-tailed test.

b)

Expert Solution

Answer to Problem 23.5AYK

Test statistic is -4.981 and p-value is 0.0038

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

	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

Formula:

t=rn−21−r2

Correlation coefficient is r = -0.928

Null and alternative hypotheses:

H0:ρ=0

H0:ρ<0

Test statistic is

t=−0.9286−21−(−0.928)2t=−4.981

Therefore, P-value of one tailed test is.

P−value=0.0038

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.

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Answer 5

Chapter 23, Problem 23.5AYK

a)

To determine

To find t statistic and p-value.

a)

Expert Solution

Answer to Problem 23.5AYK

Test statistic is -4.989 and p-value is 0.0038

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

	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

Null and alternative hypotheses:

H0:β=0

H0:β≠0

Test statistic is,

t = -4.989

Therefore, P-value of one tailed test is half of the two-tailed test.

P−value=0.00752=0.0038

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.

b)

To determine

To find correlation coefficient and find p-value for this one-tailed test.

b)

Expert Solution

Answer to Problem 23.5AYK

Test statistic is -4.981 and p-value is 0.0038

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

	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

Formula:

t=rn−21−r2

Correlation coefficient is r = -0.928

Null and alternative hypotheses:

H0:ρ=0

H0:ρ<0

Test statistic is

t=−0.9286−21−(−0.928)2t=−4.981

Therefore, P-value of one tailed test is.

P−value=0.0038

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.

Answer 6

Chapter 23, Problem 23.5AYK

a)

To determine

To find t statistic and p-value.

a)

Expert Solution

Answer to Problem 23.5AYK

Test statistic is -4.989 and p-value is 0.0038

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

	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

Null and alternative hypotheses:

H0:β=0

H0:β≠0

Test statistic is,

t = -4.989

Therefore, P-value of one tailed test is half of the two-tailed test.

P−value=0.00752=0.0038

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.

Answer 7

Chapter 23, Problem 23.5AYK

a)

To determine

To find t statistic and p-value.

Answer 8

a)

Expert Solution

Answer to Problem 23.5AYK

Test statistic is -4.989 and p-value is 0.0038

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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

	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

Null and alternative hypotheses:

H0:β=0

H0:β≠0

Test statistic is,

t = -4.989

Therefore, P-value of one tailed test is half of the two-tailed test.

P−value=0.00752=0.0038

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.

Answer 13

b)

To determine

To find correlation coefficient and find p-value for this one-tailed test.

b)

Expert Solution

Answer to Problem 23.5AYK

Test statistic is -4.981 and p-value is 0.0038

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

	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

Formula:

t=rn−21−r2

Correlation coefficient is r = -0.928

Null and alternative hypotheses:

H0:ρ=0

H0:ρ<0

Test statistic is

t=−0.9286−21−(−0.928)2t=−4.981

Therefore, P-value of one tailed test is.

P−value=0.0038

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.

Answer 14

b)

To determine

To find correlation coefficient and find p-value for this one-tailed test.

Answer 15

b)

Expert Solution

Answer to Problem 23.5AYK

Test statistic is -4.981 and p-value is 0.0038

Answer 16

b)

Expert Solution

Answer 17

b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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

	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