Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a beachball with a diameter of 46.4 cm. How does the result compare to the actual circumference of 145.8 cm? Use a significance level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball O Diameter 7.5 23.7 4.2 21.6 7.0 3.9 20.7 Circumference 23.6 74.5 13.2 67.9 22.0 12.3 65.0 Click the icon to view the critical values of the Pearson correlation coefficient r. The regression equation is y =+x. (Round to five decimal places as needed.) The best predicted circumference for a diameter of 46.4 cm is cm. (Round to one decimal place as needed.) How does the result compare to the actual circumference of 145.8 cm? O A. Since 46.4 cm is within the scope of the sample diameters, the predicted value yields the actual circumference. O B. Even though 46.4 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference. OC. Even though 46.4 cm is within the scope of the sample diameters, the predicted value yields a very different circumference. O D. Since 46.4 cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference.
Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a beachball with a diameter of 46.4 cm. How does the result compare to the actual circumference of 145.8 cm? Use a significance level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball O Diameter 7.5 23.7 4.2 21.6 7.0 3.9 20.7 Circumference 23.6 74.5 13.2 67.9 22.0 12.3 65.0 Click the icon to view the critical values of the Pearson correlation coefficient r. The regression equation is y =+x. (Round to five decimal places as needed.) The best predicted circumference for a diameter of 46.4 cm is cm. (Round to one decimal place as needed.) How does the result compare to the actual circumference of 145.8 cm? O A. Since 46.4 cm is within the scope of the sample diameters, the predicted value yields the actual circumference. O B. Even though 46.4 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference. OC. Even though 46.4 cm is within the scope of the sample diameters, the predicted value yields a very different circumference. O D. Since 46.4 cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference.
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Correlation
Correlation defines a relationship between two independent variables. It tells the degree to which variables move in relation to each other. When two sets of data are related to each other, there is a correlation between them.
Linear Correlation
A correlation is used to determine the relationships between numerical and categorical variables. In other words, it is an indicator of how things are connected to one another. The correlation analysis is the study of how variables are related.
Regression Analysis
Regression analysis is a statistical method in which it estimates the relationship between a dependent variable and one or more independent variable. In simple terms dependent variable is called as outcome variable and independent variable is called as predictors. Regression analysis is one of the methods to find the trends in data. The independent variable used in Regression analysis is named Predictor variable. It offers data of an associated dependent variable regarding a particular outcome.
Question
Transcribed Image Text:Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a beachball with a diameter of 46.4 cm. How
does the result compare to the actual circumference of 145.8 cm? Use a significance level of 0.05.
Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball
Diameter
7.5
23.7
4.2
21.6
7.0
3.9
20.7
Circumference
23.6
74.5
13.2
67.9
22.0
12.3
65.0
Click the icon to view the critical values of the Pearson correlation coefficient r.
The regression equation is y =
x.
+
(Round to five decimal places as needed.)
The best predicted circumference for a diameter of 46.4 cm is
cm.
(Round to one decimal place as needed.)
How does the result compare to the actual circumference of 145.8 cm?
O A. Since 46.4 cm is within the scope of the sample diameters, the predicted value yields the actual circumference.
B. Even though 46.4 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference.
C. Even though 46.4 cm is within the scope of the sample diameters, the predicted value yields a very different circumference.
D. Since 46.4 cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference.
Transcribed Image Text:Critical Values of the Pearson Correlation Coefficientr
a = 0.05
0.950
0.878
0.811
0.754
0.707
0.666
0.632
0.602
0.576
0.553
0.532
0.514
0.497
0.482
0.468
0.456
0.444
0.396
0.361
0.335
0.312
0.294
0.279
In
a = 0.01
NOTE: To test Ho: p= 0
against H,: p+0, reject H,
if the absolute value of r is
greater than the critical
value in the table.
0.990
0.959
0.917
0.875
0.834
0.798
0.765
0.735
0.708
0.684
0.661
0.641
0.623
0.606
0.590
0.575
0.561
0.505
0.463
0.430
0.402
0.378
0.361
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