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 38.7 cm. How does the result compare to the actual circumference of 121.6 cm? Use a significance level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball Diameter 7.3 24.1 4.2 21.7 7.0 4.0 21.4 Circumference 22.9 75.7 13.2 68.2 22.0 12.6 67.2 Click the icon to view the critical values of the Pearson correlation coefficient The regression equation is ĝ= + x. (Round to five decimal places as needed.) cm. The best predicted circumference for a diameter of 38.7 cm is (Round to one decimal place as needed.) How does the result compare to the actual circumference of 121.6 cm? O A. Since 38.7 cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference. O B. Even though 38.7 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference. O C. Even though 38.7 cm is within the scope of the sample diameters, the predicted value yields a very different circumference. OD. Since 38.7 cm is within the scope of the sample diameters, the predicted value yields the actual circumference.

Critical Values of the Pearson Correlation Coefficient r

n	α=0.05	α=0.01	​NOTE: To test H0​: ρ=0 against H1​: ρ≠​0, reject H0 if the absolute value of r is greater than the critical value in the table.
4	0.950	0.990
5	0.878	0.959
6	0.811	0.917
7	0.754	0.875
8	0.707	0.834
9	0.666	0.798
10	0.632	0.765
11	0.602	0.735
12	0.576	0.708
13	0.553	0.684
14	0.532	0.661
15	0.514	0.641
16	0.497	0.623
17	0.482	0.606
18	0.468	0.590
19	0.456	0.575
20	0.444	0.561
25	0.396	0.505
30	0.361	0.463
35	0.335	0.430
40	0.312	0.402
45	0.294	0.378
50	0.279	0.361
60	0.254	0.330
70	0.236	0.305
80	0.220	0.286
90	0.207	0.269
100	0.196	0.256
n	α=0.05	α=0.01

Answer 1

For the given data using Regression in Excel we get output as

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.9999996
R Square	0.9999992
Adjusted R Square	0.99999904
Standard Error	0.027980746
Observations	7
ANOVA
	df	SS	MS	F	Significance F
Regression	1	4894.513228	4894.513	6251596	1.94234E-16
Residual	5	0.003914611	0.000783
Total	6	4894.517143
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	0.00923744	0.019260396	0.479608	0.65175	-0.040272985	0.058747866
x	3.140862184	0.001256184	2500.319	1.94E-16	3.137633059	3.144091309

From the above output