Find the regression equation letting the diameter be the pre
Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a beach ball with a diameter of 47.4 cm. How does the result compare to the actual circumference of 148.9 cm? Use a significance level of 0.05.
The regression equation is y = ______ + ______x. (Round to five decimal places as needed.
The best predicted circumference for a diameter of 47.4cm is _____cm (Round to one decimal place as needed.)
How does the result compare to the actual circumference of 148.9 cm?
A. Since 47.4cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference.
B. Since 47.4cm is within the scope of the sample diameters, the predicted value yields the actual circumference.
C. Even though 47.4cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference.
D. Even though 47.4cm is within the scope of the sample diameters, the predicted value yields a very different circumference.
| Baseball | Basketball | Golf | Soccer | Tennis | Ping Pong | Volleyball | |
| Diameter | 7.3 | 23.8 | 4.3 | 22.1 | 7.0 | 3.9 | 20.7 |
| Circumference | 22.9 | 74.8 | 13.5 | 69.4 | 22.0 | 12.3 | 65.0 |
Solution
Find do scatter plot and graph
When x=47.4 cm
y = 0.008+3.141(47.4)
= 148.8914
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Actual =148.9 cm
C. Even though 47.4cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference.
