Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a

marble

with a diameter of

1.6

cm. How does the result compare to the actual circumference of

5.0

cm? Use a significance level of

0.05.

	Baseball	Basketball	Golf	Soccer	Tennis	Ping-Pong	Volleyball
Diameter	7.3	24.1	4.3	21.7	7.1	4.1	21.1
Circumference	22.9	75.7	13.5	68.2	22.3	12.9	66.3

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Click the icon to view the critical values of the Pearson correlation coefficient r.

The regression equation is

y=nothing+nothingx.

(Round to five decimal places as needed.)

The best predicted circumference for a diameter of

1.6

cm is

nothing

cm.

(Round to one decimal place as needed.)

How does the result compare to the actual circumference of

5.0

cm?

A.

Even though

1.6

cm is within the scope of the sample diameters, the predicted value yields a very different circumference.

B.

Since

1.6

cm is within the scope of the sample diameters, the predicted value yields the actual circumference.

C.

Even though

1.6

cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference.

D.

Since

1.6

cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference.