Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumferenceof a marble with a diameter of 1.8 cm. How does the result compare to the actual circumference of 5.7 cm? Use asignificance level of 0.05.

Part 1 of 3 Points: 0 of 1 level of 0.05. Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a marble with a diameter of 1.8 cm. How does the result compare Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball Diameter 7.3 23.8 4.3 21.6 7.0 3.9 21.2 Circumference 22.9 74.8 13.5 67.9 22.0 12.3 66.6 Click the icon to view the critical values of the Critical Values of the Pearson Correlation Coefficient r - X The regression equation is y =+x a = 0.05 a = 0.01 NOTE: To test Ho: p = 0 0.950 (Round to five decimal places as needed.) 0.990 against H,: p #0, reject Ho 0.878 0.959 if the absolute value of r is 0.811 0.917 greater than the critical 0.754 0.875 value in the table. 8 0.707 0.834 0.666 0.798 10 10.632 0.765 11 0.602 0.735 12 0.576 0.708 13 0.553 0.684 14 0.532 0.661 15 0.514 0.641 16 0.497 0.623 17 0.482 0.606 18 0.468 0.590 19 0.456 0.575 20 0.444 0.561 25 0.396 0.505 30 0.361 0.46 35 0.335 0.430 40 0.312 0.402 45 0.294 0.378 50 0.279 0.361 60 0.254 0.330 70 0.236 0.305 30 0.220 0.286 190 0.207 0.269 100 0.196 0.256 Help me solve this View an exan in O = 0.05 = 0.01