Question

Find the regression​ equation, letting the diameter be the predictor​ (x) variable. Find the best predicted circumference of a marble with a diameter of 1.9 cm. How does the result compare to the actual circumference of 6.0 ​cm? Use a significance level of 0.05.

_   Diameter   Circumference
Baseball   7.3   22.9
Basketball   23.6   74.1
Golf   4.2   13.2
Soccer   21.8   68.5
Tennis   6.9   21.7
Ping-Pong   4.0   12.6
Volleyball   20.8   65.3

The regression equation is y^=  + x.
​(Round to five decimal places as​ needed.)
The best-predicted circumference for a diameter of 1.9 cm is cm.
​(Round to one decimal place as​ needed.)
How does the result compare to the actual circumference of 6.0 ​cm?
A.
Since 1.9 cm is beyond the scope of the sample​ diameters, the predicted value yields a very different circumference.
B.
Even though 1.9 cm is within the scope of the sample​ diameters, the predicted value yields a very different circumference.
C.
Even though 1.9 cm is beyond the scope of the sample​ diameters, the predicted value yields the actual circumference.
D.
Since 1.9 cm is within the scope of the sample​ diameters, the predicted value yields the actual circumference.

Answer 1

The statistical software output for this problem is:

From above output:

Regression equation:

y = 0.01976 + 3.13952 x

Best predicted value = 6.0

Even though 1.9 cm is beyond the scope of the sample​ diameters, the predicted value yields the actual circumference.