Find the regression equation, letting the diameter be the predictor $(x)$ variable. Find the best predicted circumference of a marble with a diameter of $1.3 \mathrm{~cm}$. How does the result compare to the actual circumference of $4.1 \mathrm{~cm}$ ? Use a significance level of 0.05 .
\begin{tabular}{l|c|c|c|c|c|c|c}
\hline & Baseball & Basketball & Golf & Soccer & Tennis & Ping-Pong & Volleyball \\
\hline Diameter & 7.3 & 24.2 & 4.4 & 21.9 & 7.0 & 3.9 & 20.5 \\
\hline Circumference & 22.9 & 76.0 & 13.8 & 68.8 & 22.0 & 12.3 & 64.4 \\
\hline
\end{tabular}
Click the icon to view the critical values of the Pearson correlation coefficient $r$.
The regression equation is $\hat{y}=0.007+3.14070 x$.
(Round to five decimal places as needed.)
The best predicted circumference for a diameter of $1.3 \mathrm{~cm}$ is $4.1 \mathrm{~cm}$. (Round to one decimal place as needed.)
How does the result compare to the actual circumference of $4.1 \mathrm{~cm}$ ?
A. Even though $1.3 \mathrm{~cm}$ is within the scope of the sample diameters, the predicted value yields a very different circumference.
B. Even though $1.3 \mathrm{~cm}$ is beyond the scope of the sample diameters, the predicted value yields the actual circumference.
C. Since $1.3 \mathrm{~cm}$ is within the scope of the sample diameters, the predicted value yields the actual circumference.
D. Since $1.3 \mathrm{~cm}$ is beyond the scope of the sample diameters, the predicted value yields a very different circumference.
Answer
\(\boxed{\text{The correct answer is C. Since } 1.3 \text{ cm is within the scope of the sample diameters, the predicted value yields the actual circumference.}}\)
Step 1 ：Given the regression equation \(\hat{y}=0.007+3.14070 x\), where \(x\) is the diameter of the marble.
Step 2 ：Substitute \(x = 1.3\) into the equation to find the predicted circumference: \(\hat{y}=0.007+3.14070 \times 1.3\).
Step 3 ：Calculate the predicted circumference to get approximately \(4.08991\) cm.
Step 4 ：Compare the predicted circumference \(4.08991\) cm with the actual circumference \(4.1\) cm. They are very close.
Step 5 ：Given the diameters of several different types of balls, we can see that \(1.3\) cm is within the range of these diameters.
Step 6 ：Conclude that the predicted value yields the actual circumference and that \(1.3\) cm is within the scope of the sample diameters.
Step 7 ：\(\boxed{\text{The correct answer is C. Since } 1.3 \text{ cm is within the scope of the sample diameters, the predicted value yields the actual circumference.}}\)
