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Homework: Section 10.2 Homework Question 8, 10.2.27 HW Score: 66.67%, 6 of 9 points Part 2 of 3 O Points: 0 of 1 Save Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a marble with a diameter of 1.4 cm. How does the result compare to the actual circumference of 4.4 cm? Use significance level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball Diameter 7.3 23.6 4.3 21.7 6.9 4.0 21.2 Circumference 22.9 74.1 13.5 68.2 21.7 12.6 66.6 - X Critical Values of the Pearson Correlation Coefficient r Click the icon to view the critical values of the Pearson correlation coefficient r. Critical Values of the Pearson Correlation Coefficient r n or = 0.05 = 0.01 NOTE To test Ho: p = 0 0.950 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 The regression equation is y = 0.0101 + 3.14078 x. 0.754 0.875 value in the table. (Round to five decimal places as needed.) 0.707 0.834 0.666 0.798 The best predicted circumference for a diameter of 1.4 cm is cm. 10 0.632 0.765 (Round to one decimal place as needed.) 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 0.463 35 0.335 0.430 40 0.312 0.402 45 0.294 10 378 Help me solve this View an