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# 1. Given the linear correlation coefficient r and the sample

1. Given the linear correlation coefficient r and the sample size n, determine the critical values of r and use your finding to state whether or not the given r represents a significant linear correlation. Use a significance level of 0.05.
r = 0.75, n = 9 (Points : 5)
Critical values: r = ±0.666, no significant linear correlation
Critical values: r = 0.666, no significant linear correlation
Critical values: r = -0.666, no significant linear correlation
Critical values: r = ±0.666, significant linear correlation
3. Find the value of the linear correlation coefficient r.
x 57 53 59 61 53 56 60
y 156 164 163 177 159 175 151
(Points : 5)
-0.054
0.214
0.109
-0.078
4. Suppose you will perform a test to determine whether there is sufficient evidence to support a claim of a linear correlation between two variables. Find the critical values of r given the number of pairs of data n and the significance level alpha.
n = 11, Alpha = 0.01 (Points : 5)
r = 0.765
r = ± 0.602
r = 0.735
r = ± 0.735
5. Use the given data to find the best predicted value of the response variable.
Six pairs of data yield r = 0.789 and the regression equation ŷ=4x-2. Also Ῡ=19.0. What is the best predicted value of y for x = 5? (Points : 5)
22.0
18.0
18.5
19.0
6. Use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary.
x 1 3 5 7 9
y 143 116 100 98 90
Choose A, B, C, or D. (Points : 5)
A. ŷ= -150.7+6.8x
B. ŷ= 150.7 - 6.8x
C. ŷ= 140.4 - 6.2x
D. ŷ= -140.4 + 6.2x
8. Use the given information to find the coefficient of determination.
A regression equation is obtained for a collection of paired data. It is found that the total variation is 24.488, the explained variation is 15.405, and the unexplained variation is 9.083. Find the coefficient of determination. (Points : 5)
0.629
0.590
1.590
0.371

9. Use the computer display to answer the question.
A collection of paired data consists of the number of years that students have studied Spanish and their scores on a Spanish language proficiency test. A computer program was used to obtain the least squares linear regression line and the computer output is shown below. Along with the paired sample data, the program was also given an x value of 2 (years of study) to be used for predicting test score.
The regression equation is
Score = 31.55 + 10.90 Years
Predictor Coef StDev T P
Constant 31.55 6.360 4.96 0.000
Years 10.90 1.744 6.25 0.000
S = 5.651 R-Sq = 83.0% R-Sq (Adj) = 82.7%
Predicted Values
Fit StDev Fit 95.0% CI 95.0% PI
53.35 3.168 (42.72, 63.98) (31.61, 75.09)
(Points : 5)
82.7%
17.0%
91.1%
83.0%
10. Find the explained variation for the paired data.
The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is ŷ = 44.8447 + 3.52427x. Find the explained variation.
x Hours of preparation 5 2 9 6 10
y Test of score 64 48 72 73 80
(Points : 5)
498.103
511.724
87.4757
599.2
11. Find the unexplained variation for the paired data.
The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is ŷ = 44.8447 + 3.52427x. Find the unexplained variation.
x Hours of preparation 5 2 9 6 10
y Test of score 64 48 72 73 80
(Points : 5)
511.724
87.4757
96.103
599.2
12. Find the total variation for the paired data.
The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is ŷ = 44.8447 + 3.52427x. Find the total variation.
x Hours of preparation 5 2 9 6 10
y Test of score 64 48 72 73 80
(Points : 5)
599.2
498.103
511.724
87.4757
13. Find the standard error of estimate for the paired data.
The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is ŷ = 44.8447 + 3.52427x. Find the the standard error of estimate.
x Hours of preparation 5 2 9 6 10
y Test of score 64 48 72 73 80
(Points : 5)
4.1097
7.1720
5.3999
13.060
14. Construct the indicated prediction interval for an individual y.
The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is ŷ = 44.845 + 3.524x and the standard error of estimate is Se = 5.40. Find the 99% prediction interval for the test score of a person who spent 7 hours preparing for the test.
x Hours of preparation 5 2 9 6 10
y Test of score 64 48 72 73 80
(Points : 5)
58 < y < 82
62 < y < 78
35 < y < 104
32 < y < 107

15. Solve the problem.
A confidence interval for the sl

1) D
Critical values: r = ±0.666, significant linear correlation

3. ) C (0.109)

4.) D (r = ± 0.735)

5.) B (18)

6.) C ŷ= 140.4 - 6.2x

8.) A (0.629)

9.)
Note, what is the question here? if the question is find the coefficient of determination (also known as r^2,) the answer is
D (83.0%)

but if the question is find the coefficient of correlation, also known as r, the answer is

C (91.1%)

10.) B (511.724)

11) B (87.4757)

12.) A (599.2)

13.) C (5.3999)

14.) A (58 < y < 82)

15.) Incomplete questions

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Customer: replied 4 years ago.

1. Given the linear correlation coefficient r and the sample size n, determine the critical values of r and use your finding to state whether or not the given r represents a significant linear correlation. Use a significance level of 0.05.
r = 0.75, n = 9 (Points : 5)
Critical values: r = ±0.666, no significant linear correlation
Critical values: r = 0.666, no significant linear correlation
Critical values: r = -0.666, no significant linear correlation
Critical values: r = ±0.666, significant linear correlation
3. Find the value of the linear correlation coefficient r.
x 57 53 59 61 53 56 60
y 156 164 163 177 159 175 151
(Points : 5)
-0.054
0.214
0.109
-0.078
4. Suppose you will perform a test to determine whether there is sufficient evidence to support a claim of a linear correlation between two variables. Find the critical values of r given the number of pairs of data n and the significance level alpha.
n = 11, Alpha = 0.01 (Points : 5)
r = 0.765
r = ± 0.602
r = 0.735
r = ± 0.735
5. Use the given data to find the best predicted value of the response variable.
Six pairs of data yield r = 0.789 and the regression equation ŷ=4x-2. Also Ῡ=19.0. What is the best predicted value of y for x = 5? (Points : 5)
22.0
18.0
18.5
19.0
6. Use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary.
x 1 3 5 7 9
y 143 116 100 98 90
Choose A, B, C, or D. (Points : 5)
A. ŷ= -150.7+6.8x
B. ŷ= 150.7 - 6.8x
C. ŷ= 140.4 - 6.2x
D. ŷ= -140.4 + 6.2x
8. Use the given information to find the coefficient of determination.
A regression equation is obtained for a collection of paired data. It is found that the total variation is 24.488, the explained variation is 15.405, and the unexplained variation is 9.083. Find the coefficient of determination. (Points : 5)
0.629
0.590
1.590
0.371

9. Use the computer display to answer the question.
A collection of paired data consists of the number of years that students have studied Spanish and their scores on a Spanish language proficiency test. A computer program was used to obtain the least squares linear regression line and the computer output is shown below. Along with the paired sample data, the program was also given an x value of 2 (years of study) to be used for predicting test score.
The regression equation is
Score = 31.55 + 10.90 Years
Predictor Coef StDev T P
Constant 31.55 6.360 4.96 0.000
Years 10.90 1.744 6.25 0.000
S = 5.651 R-Sq = 83.0% R-Sq (Adj) = 82.7%
Predicted Values
Fit StDev Fit 95.0% CI 95.0% PI
53.35 3.168 (42.72, 63.98) (31.61, 75.09)
(Points : 5)
82.7%
17.0%
91.1%
83.0%
10. Find the explained variation for the paired data.
The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is ŷ = 44.8447 + 3.52427x. Find the explained variation.
x Hours of preparation 5 2 9 6 10
y Test of score 64 48 72 73 80
(Points : 5)
498.103
511.724
87.4757
599.2
11. Find the unexplained variation for the paired data.
The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is ŷ = 44.8447 + 3.52427x. Find the unexplained variation.
x Hours of preparation 5 2 9 6 10
y Test of score 64 48 72 73 80
(Points : 5)
511.724
87.4757
96.103
599.2
12. Find the total variation for the paired data.
The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is ŷ = 44.8447 + 3.52427x. Find the total variation.
x Hours of preparation 5 2 9 6 10
y Test of score 64 48 72 73 80
(Points : 5)
599.2
498.103
511.724
87.4757
13. Find the standard error of estimate for the paired data.
The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is ŷ = 44.8447 + 3.52427x. Find the the standard error of estimate.
x Hours of preparation 5 2 9 6 10
y Test of score 64 48 72 73 80
(Points : 5)
4.1097
7.1720
5.3999
13.060
14. Construct the indicated prediction interval for an individual y.
The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is ŷ = 44.845 + 3.524x and the standard error of estimate is Se = 5.40. Find the 99% prediction interval for the test score of a person who spent 7 hours preparing for the test.
x Hours of preparation 5 2 9 6 10
y Test of score 64 48 72 73 80
(Points : 5)
58 < y < 82
62 < y < 78
35 < y < 104
32 < y < 107

15. Solve the problem.
A confidence interval for the slope ß1 for a regression line y = ß0 + ß1x can be found by evaluating the limits in the interval below:
b1 - E < ß1 < b1 + E,
Where E = (tα/2)ˢe/√∑x² - (∑x)²/n
The critical value tα/2 is found from the t-table using n - 2 degrees of freedom and b1 is calculated in the usual way from the sample data.
Use the data below to obtain a 95% confidence interval estimate of ß1
x(hours studied) 2.5 4.5 5.1 7.9 11.6
y(score on test) 66 70 60 83 93
(Points : 5)
1.936 < B1 < 4.874
0.686 < B1 < 6.124
0.322 < B1 < 6.488
0.134 < B1 < 6.676
16. Use computer software to find the multiple regression equation. Can the equation be used for prediction?
An anti-smoking group used data in the table to relate the carbon monoxide of various brands of cigarettes to their tar and nicotine content.
CO TAR NIC
15 1.2 16
15 1.2 16
17 1.0 16
6 0.8 9
1 0.1 1
8 0.8 8
10 0.8 10
17 1.0 16
15 1.2 15
11 0.7 9
18 1.4 18
16 1.0 15
10 0.8 9
7 0.5 5
18 1.1 16
(Points : 5)
CO = 1.37 - 5.53TAR + 1.33NIC; Yes, because the R2 is high
CO = 1.37 + 5.50TAR - 1.38NIC; Yes, because the P-value is high
CO = 1.3 + 5.5TAR - 1.3NIC; Yes, because the adjusted R2 is high
CO = 1.25 + 1.55TAR - 5.79NIC; Yes, because the P-value is too low
18. Use computer software to obtain the multiple regression equation. Use the estimated equation to find the predicted value.
A health specialist gathered the data in the table to see if pulse rates can be explained by exercise and smoking. For exercise, he assigns 1 for yes, 2 for no. For smoking, he assigns 1 for yes, 2 for no. He then used his results to predict the pulse rate of a person whose exercise value was 1 and whose smoking value was 1.
PULSE EXERCISE SMOKE
97 2 2
88 1 2
69 1 2
67 1 2
83 1 2
77 1 2
66 2 2
78 2 2
73 1 1
67 1 1
55 1 2
82 1 1
70 1 2
55 1 2
76 1 2
(Points : 5)
70 beats/min
74 beats/min
81 beats/min
77 beats/min

19. Find the indicated multiple regression equation.
Below are performance and attitude ratings on employees.
Performance 59 63 65 69 58 77 76 69 70 64
Attitude 72 67 78 82 75 87 92 83 87 78
Managers also rate the same employees according to adaptability, and below are the results that correspond to those given above.
Adaptability: 50 52 54 60 46 67 66 59 62 55
Find the multiple regression equation that expresses performance in terms of attitude and adaptability.
(Points : 5)
P-hat = 14.09 + 0.213(Att.) + 0.895(Adapt.)
P-hat = 14.09 + 0.895(Att.) + 0.213(Adapt.)
P-hat = 14.09 + 0.907(Att.) + 0.014(Adapt.)
P-hat = 14.09 + 0.014(Att.) + 0.907(Adapt.)
20. Use computer software to find the best multiple regression equation to explain the variation in the dependent variable, Y, in terms of the independent variables, X1, X2, X3.
Y X1 X2
15 1.2 16
15 1.2 16
17 1.0 16
6 0.8 9 CORRELATION COEFFICIENT
1 0.1 1 Y/X1 = 0.886
8 0.8 8 Y/X2 = 0.965
10 0.8 10
17 1.0 16 COEFFICIENTS OF DETERMINATION
15 1.2 15 Y/X2 = 0.932
11 0.7 9 Y/X2, X1 = 0.943
18 1.4 18
16 1.0 15
10 0.8 9
7 0.5 5
18 1.1 16
(Points : 5)
Y-hat = 0.42 + 0.99X2
Y-hat = 1.38 - 5.53X1 + 1.33X2
Y-hat = 1.25 - 1.55X1 + 5.79X2
Y-hat = -0.49 + 14.07X1

1) D
Critical values: r = ±0.666, significant linear correlation

3.) C (0.109)

4.) D (r = ± 0.735)

5.) B (18)

6.) C ŷ= 140.4 - 6.2x

8.) A (0.629)

9.)
Note, what is the question here? if the question is find the coefficient of determination (also known as r^2,) the answer is
D (83.0%)

but if the question is find the coefficient of correlation, also known as r, the answer is

C (91.1%)

10.) B (511.724)

11) B (87.4757)

12.) A (599.2)

13.) C (5.3999)

14) A (58 < y < 82)

15.) C (0.322 < B1 < 6.488)

16.) A (CO = 1.37 - 5.53TAR + 1.33NIC; Yes, because the R2 is high)

18..) B (74 beats/min)

19..) D (P-hat = 14.09 + 0.014(Att.) + 0.907(Adapt.))

20.) B (Y-hat = 1.38 - 5.53X1 + 1.33X2)

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If you need my help asap, just write in the first part of your question: "For Trigeoscal only!" Thanks again

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Are you satisfied with the answers???

thanks

,

.

Customer: replied 4 years ago.

Apologies, information isn't copying over correctly.

9. What percentage of the total variation in test scores can be explained by the linear relationship between years of study and test scores?

17. Use computer software to obtain the multiple regression equation and identify R2, adjusted R2, and the P-value.
An anti-smoking group used data in the table to relate the carbon monoxide of various brands of cigarettes to their tar and nicotine content.
CO TAR NIC
15 1.2 16
15 1.2 16
17 1.0 16
6 0.8 9
1 0.1 1
8 0.8 8
10 0.8 10
17 1.0 16
15 1.2 15
11 0.7 9
18 1.4 18
16 1.0 15
10 0.8 9
7 0.5 5
18 1.1 16
(Points : 5)
0.976, 0.921, 0.002
0.943, 0.934, 0.000
0.931, 0.902, 0.000
0.861, 0.900, 0.015

9.) D (83.0%)

17.) B (0.943, 0.934, 0.000)

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Bonus is welcome!

Or if you are happy and satisfied with the answers, kindly post a reply that “you are satisfied with the answers posted”

If you still have some more questions, feel free to ask, you can request me by writing at the start of your question, "For Trigeoscal only" thanks

Please Leave a POSITIVE FEEDBACK and a HIGH RATING to Credit me. BONUS is welcome.Thanks.

If you need my help asap, just write in the first part of your question: "For Trigeoscal only!" Thanks again

Don't forget to accept my answers if you are satisfied only, so that I will be compensated in helping you, thanks again!

Are you satisfied with the answers???

thanks

,

.

Customer: replied 4 years ago.

Missed 4 out of 18.

Thanks for accepting my answers (for giving a high rating and also for the bonus, if there are any)

If you still have some more questions, feel free to ask, you can request me by writing at the start of your question, "For Trigeoscal only" thanks

(please do not forget to put my name at the start of the question)

Have a nice and wonderful day!

-Trigeoscal-

:)

,

Customer: replied 4 years ago.
Trigeoscal,
I submitted a new set of questions and I need them ASAP. You said I could rely on you..