To find t statistic and p-value.

Question

Want to see more full solutions like this?

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.

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

Cape Fear Community Colle X ALEKS ALEKS - Dorothy Smith - Sec X www-awu.aleks.com/alekscgi/x/Isl.exe/10_u-IgNslkr7j8P3jH-IQ1w4xc5zw7yX8A9Q43nt5P1XWJWARE... Section 7.1,7.2,7.3 HW 三 Question 21 of 28 (1 point) | Question Attempt: 5 of Unlimited The proportion of phones that have more than 47 apps is 0.8783 Part: 1 / 2 Part 2 of 2 (b) Find the 70th The 70th percentile of the number of apps. Round the answer to two decimal places. percentile of the number of apps is Try again Skip Part Recheck Save 2025 Mc

Hi, I need to sort out where I went wrong. So, please us the data attached and run four separate regressions, each using the Recruiters rating as the dependent variable and GMAT, Accept Rate, Salary, and Enrollment, respectively, as a single independent variable. Interpret this equation. Round your answers to four decimal places, if necessary. If your answer is negative number, enter "minus" sign. Equation for GMAT: Ŷ = _______ + _______ GMAT Equation for Accept Rate: Ŷ = _______ + _______ Accept Rate Equation for Salary: Ŷ = _______ + _______ Salary Equation for Enrollment: Ŷ = _______ + _______ Enrollment

Question 21 of 28 (1 point) | Question Attempt: 5 of Unlimited Dorothy ✔ ✓ 12 ✓ 13 ✓ 14 ✓ 15 ✓ 16 ✓ 17 ✓ 18 ✓ 19 ✓ 20 = 21 22 > How many apps? According to a website, the mean number of apps on a smartphone in the United States is 82. Assume the number of apps is normally distributed with mean 82 and standard deviation 30. Part 1 of 2 (a) What proportion of phones have more than 47 apps? Round the answer to four decimal places. The proportion of phones that have more than 47 apps is 0.8783 Part: 1/2 Try again kip Part ی E Recheck == == @ W D 80 F3 151 E R C レ Q FA 975 % T B F5 10 の 000 园 Save For Later Submit Assignment © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility Y V& U H J N * 8 M I K O V F10 P = F11 F12 . d

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.

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

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.

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

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.

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 23.5AYK

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