Question

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

marblemarble

with a diameter of

1.7

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

5.3

​cm? Use a significance level of

0.05

_   Diameter   Circumference
Baseball   7.4   23.2
Basketball   24.4   76.7
Golf   4.2   13.2
Soccer   21.9   68.8
Tennis   7.0   22.0
Ping-Pong   4.0   12.6
Volleyball   20.9   65.7

The regression equation is

ModifyingAbove y with caretyequals=nothingplus+nothingx.

​(Round to five decimal places as​ needed.)

The best predicted circumference for a diameter of

1.7

cm is

nothing

cm.

​(Round to one decimal place as​ needed.)

How does the result compare to the actual circumference of

5.3

​cm?

A.Since

1.7

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

B.Since

1.7

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

C.Even though

1.7

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

D.Even though

1.7

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

Click to select your answer(s).

Answer 1

The regression equation is y^ = -.006+3.143x

The predicted circumference for a diameter 1.7 is 5.3

On comparing with actual circumference of 5.3,C.Even though 1.7 cm is beyond the scope of the sample​ diameters, the predicted value yields the actual circumference