Find the regressionequation, letting the diameter be the predictor(x) variable. Find the best predicted circumference of a beachball with a diameter of 45.7 cm. How does the result compare to the actual circumference of 143.6 cm? Use a significance level of 0.05.

Baseball

Basketball

Golf

Soccer

Tennis

Ping-Pong

Volleyball

Diameter

7.5

23.8

4.4

21.5

6.9

4.0

20.9

Circumference

23.6

74.8

13.8

67.5

21.7

12.6

The regression equation is

y= what + what x. (Round to five decimal places asneeded.)

The best predicted circumference for a diameter of 45.7 cm is what cm.(Round to one decimal place asneeded.)

How does the result compare to the actual circumference of 143.6 cm?

A.

Even though 45.7 cm is beyond the scope of the samplediameters, the predicted value yields the actual circumference.

B.

Since 45.7 cm is within the scope of the samplediameters, the predicted value yields the actual circumference.

C.

Even though 45.7 cm is within the scope of the samplediameters, the predicted value yields a very different circumference.

D.

Since 45.7 cm is beyond the scope of the samplediameters, the predicted value yields a very different circumference.