nd the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a marble with a diameter of 1.5 cm. How does the result compare to the actual circumference of 4.7 cm? Use a significance Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball ameter 7.4 24.3 4.2 21.9 7.1 ircumference 23.2 76.3 13.2 68.8 22.3 4.1 21.2 12.9 66.6 Click the icon to view the critical values of the Pearson correlation coefficient r he regression equation is y=+K Cound to five decimal places as needed.) he best predicted circumference for a diameter of 1.5 cm is c cm. Cound to one decimal place as needed.) ow does the result compare to the actual circumference of 4.7 cm? A. Since 1.5 cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference. B. Even though 1.5 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference. OC. Even though 1.5 cm is within the scope of the sample diameters, the predicted value yields a very different circumference. D. Since 1.5 cm is within the scope of the sample diameters, the predicted value yields the actual circumference.

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**Educational Website Transcription**

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### Regression Analysis: Predicting Circumference from Diameter

#### Problem Statement:
Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a marble with a diameter of 1.5 cm. How does the result compare to the actual circumference of 4.7 cm? Use a significance level of 0.05.

| Sport       | Baseball | Basketball | Golf | Soccer | Tennis | Ping-Pong | Volleyball |
|-------------|----------|------------|------|--------|--------|-----------|------------|
| **Diameter** | 7.4      | 24.3       | 4.2  | 21.9   | 7.1    | 4.1       | 21.2       |
| **Circumference** | 23.2     | 76.3       | 13.2 | 68.8   | 22.3   | 12.9      | 66.6       |

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

---

#### Regression Equation
The regression equation is \( \hat{y} = b_0 + b_1 x \).  
(Round to five decimal places as needed.)

---

#### Prediction
The best predicted circumference for a diameter of 1.5 cm is \(\_\_\_\_\_) cm.  
(Round to one decimal place as needed.)

---

#### Comparison to Actual Circumference
How does the result compare to the actual circumference of 4.7 cm?

1. **Option A**: Since 1.5 cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference.
2. **Option B**: Even though 1.5 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference.
3. **Option C**: Even though 1.5 cm is within the scope of the sample diameters, the predicted value yields a very different circumference.
4. **Option D**: Since 1.5 cm is within the scope of the sample diameters, the predicted value yields the actual circumference.

---
Transcribed Image Text:**Educational Website Transcription** --- ### Regression Analysis: Predicting Circumference from Diameter #### Problem Statement: Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a marble with a diameter of 1.5 cm. How does the result compare to the actual circumference of 4.7 cm? Use a significance level of 0.05. | Sport | Baseball | Basketball | Golf | Soccer | Tennis | Ping-Pong | Volleyball | |-------------|----------|------------|------|--------|--------|-----------|------------| | **Diameter** | 7.4 | 24.3 | 4.2 | 21.9 | 7.1 | 4.1 | 21.2 | | **Circumference** | 23.2 | 76.3 | 13.2 | 68.8 | 22.3 | 12.9 | 66.6 | **Click the icon to view the critical values of the Pearson correlation coefficient r.** --- #### Regression Equation The regression equation is \( \hat{y} = b_0 + b_1 x \). (Round to five decimal places as needed.) --- #### Prediction The best predicted circumference for a diameter of 1.5 cm is \(\_\_\_\_\_) cm. (Round to one decimal place as needed.) --- #### Comparison to Actual Circumference How does the result compare to the actual circumference of 4.7 cm? 1. **Option A**: Since 1.5 cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference. 2. **Option B**: Even though 1.5 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference. 3. **Option C**: Even though 1.5 cm is within the scope of the sample diameters, the predicted value yields a very different circumference. 4. **Option D**: Since 1.5 cm is within the scope of the sample diameters, the predicted value yields the actual circumference. ---
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