Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predictedcircumference of a marble with a diameter of 1.5 cm. How does the result compare to the actual circumferenceof 4.7 cm? Use a significance level of 0.05.Baseball Basketball Golf Soccer Tennis Ping-Pong VolleyballDiameter7.424.521.67.04.120.74.477.0 13.8 67.9 22.0 12.9Circumference 23.265.0Click 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 verydifferent circumference.O B.Since 1.5 cm is within the scope of the sample diameters, the predicted value yields the actualcircumference.O C. Since 1.5 cm is beyond the scope of the sample diameters, the predicted value yields a very differentcircumference.O D. Even though 1.5 cm is beyond the scope of the sample diameters, the predicted value yields theactual circumference.

Answered: Find the regression equation, letting…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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