Find the regression equation, letting the diameter be the predictor (x) variable

Question

Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a beachball with a diameter of 43.2 cm. How does the result compare to the actual circumference of 135.7 cm? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient t. The best predicted circumference for a diameter of 43.2 cm is cm. How does the result compare to the actual circumference of 135.7 cm? A. Since 43.2 cm within the scope of the sample diameters, the predicted value yields the actual circumference. B. Even though 43.2 cm is within the scope of the sample diameters. the predicted value yields a very different circumference. C. Since 43.2 cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference. D. Even though 43.2 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference.

Explanation / Answer

The regression analysis for the given data using the MICROSOFT EXCEL plugin in data analysis gives the following result.

Based on this we will answer the questions.

Best predicted circumference for a daimeter of 43.2 cm would be given by,

Circumference = 0.02+43.2*3.14 = 135.7

The best predicted circumference for a diameter of 43.2cm is 135.7 cm.

How does the result compare to the actual circumference of 135.7?

D. Even though 43.2 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference.

SUMMARY OUTPUT Regression Statistics Multiple R 0.999999639 R Square 1.00 Adjusted R Square 1.00 Standard Error 0.026 Observations 7 ANOVA df SS MS F Significance F Regression 1 4837.5 4837.5 6932861.0 0.0 Residual 5 0.0 0.0 Total 6 4837.5 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 0.02 0.02 1.27 0.26 -0.02 0.07 Diameter 3.14 0.00 2633.03 0.00 3.14 3.14

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