Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a marble with a diameter of 1.5 cm. How does the result compare to the actual circumference of 4.7 cm? Use a significance level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball 24.5 4.4 21.6 7.0 Diameter 7.4 4.1 20.7 Circumference 23.2 77.0 13.8 67.9 22.0 12.9 65.0 Click the icon to view the critical values of the Pearson correlation coefficient r. The regression equation is ŷ =+x. (Round to five decimal places as needed.) The best predicted circumference for a diameter of 1.5 cm is cm. (Round to one decimal place as needed.) How does the result compare to the actual circumference of 4.7 cm? O A. Even though 1.5 cm is within the scope of the sample diameters, the predicted value yields a very different circumference. OB. Since 1.5 cm is within the scope of the sample diameters, the predicted value yields the actual circumference. O C. Since 1.5 cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference. O D. Even though 1.5 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference.
Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a marble with a diameter of 1.5 cm. How does the result compare to the actual circumference of 4.7 cm? Use a significance level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball 24.5 4.4 21.6 7.0 Diameter 7.4 4.1 20.7 Circumference 23.2 77.0 13.8 67.9 22.0 12.9 65.0 Click the icon to view the critical values of the Pearson correlation coefficient r. The regression equation is ŷ =+x. (Round to five decimal places as needed.) The best predicted circumference for a diameter of 1.5 cm is cm. (Round to one decimal place as needed.) How does the result compare to the actual circumference of 4.7 cm? O A. Even though 1.5 cm is within the scope of the sample diameters, the predicted value yields a very different circumference. OB. Since 1.5 cm is within the scope of the sample diameters, the predicted value yields the actual circumference. O C. Since 1.5 cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference. O D. Even though 1.5 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
![Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted
circumference of a marble with a diameter of 1.5 cm. How does the result compare to the actual circumference
of 4.7 cm? Use a significance level of 0.05.
Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball
Diameter
7.4
24.5
21.6
7.0
4.1
20.7
4.4
77.0 13.8 67.9 22.0 12.9
Circumference 23.2
65.0
Click the icon to view the critical values of the Pearson correlation coefficient r.
The regression equation is y=
(Round to five decimal places as needed.)
The best predicted circumference for a diameter of 1.5 cm is cm.
(Round to one decimal place as needed.)
How does the result compare to the actual circumference of 4.7 cm?
O A. Even though 1.5 cm is within the scope of the sample diameters, the predicted value yields a very
different circumference.
O B.
Since 1.5 cm is within the scope of the sample diameters, the predicted value yields the actual
circumference.
O C. Since 1.5 cm is beyond the scope of the sample diameters, the predicted value yields a very different
circumference.
O D. Even though 1.5 cm is beyond the scope of the sample diameters, the predicted value yields the
actual circumference.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbf53cca9-c8cc-422a-823d-50ea23a03427%2F2efd44f2-cfb5-49ee-9362-44b1dc051ef2%2Fde6nzeg_processed.png&w=3840&q=75)
Transcribed Image Text:Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted
circumference of a marble with a diameter of 1.5 cm. How does the result compare to the actual circumference
of 4.7 cm? Use a significance level of 0.05.
Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball
Diameter
7.4
24.5
21.6
7.0
4.1
20.7
4.4
77.0 13.8 67.9 22.0 12.9
Circumference 23.2
65.0
Click the icon to view the critical values of the Pearson correlation coefficient r.
The regression equation is y=
(Round to five decimal places as needed.)
The best predicted circumference for a diameter of 1.5 cm is cm.
(Round to one decimal place as needed.)
How does the result compare to the actual circumference of 4.7 cm?
O A. Even though 1.5 cm is within the scope of the sample diameters, the predicted value yields a very
different circumference.
O B.
Since 1.5 cm is within the scope of the sample diameters, the predicted value yields the actual
circumference.
O C. Since 1.5 cm is beyond the scope of the sample diameters, the predicted value yields a very different
circumference.
O D. Even though 1.5 cm is beyond the scope of the sample diameters, the predicted value yields the
actual circumference.
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