13. Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a marble with a diameter of 1.2 cm. How does the result compare to the actual circumference of 3.8 cm? Use a significance level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball 20.7 7.5 24.4 4.2 22.3 7.1 4.0 Diameter Circumference 23.6 76.7 13.2 70.1 22.3 12.6 65.0 6. Click the icon to view the critical values of the Pearson correlation coefficient r. The regression equation is y = (Round to five decimal places as needed.) X. The best predicted circumference for a diameter of 1.2 cm is (Round to one decimal place as needed.) ст. How does the result compare to the actual circumference of 3.8 cm? O A. Even though 1.2 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference. O B. Since 1.2 cm is within the scope of the sample diameters, the predicted value yields the actual circumference. O C. Since 1.2 cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference. O D. Even though 1.2 cm is within the scope of the sample diameters, the predicted value yields a very different circumference.

13. Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a marble with a diameter of 1.2 cm. How does the result compare to the actual circumference of 3.8 cm? Use a significance level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball 20.7 7.5 24.4 4.2 22.3 7.1 4.0 Diameter Circumference 23.6 76.7 13.2 70.1 22.3 12.6 65.0 6. Click the icon to view the critical values of the Pearson correlation coefficient r. The regression equation is y = (Round to five decimal places as needed.) X. The best predicted circumference for a diameter of 1.2 cm is (Round to one decimal place as needed.) ст. How does the result compare to the actual circumference of 3.8 cm? O A. Even though 1.2 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference. O B. Since 1.2 cm is within the scope of the sample diameters, the predicted value yields the actual circumference. O C. Since 1.2 cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference. O D. Even though 1.2 cm is within the scope of the sample diameters, the predicted value yields a very different circumference.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Using your favorite statistics software package, you generate a scatter plot with a regression equation and correlation coefficient. The regression equation is reported as y=−66.75x+57.0 and the r=−0.504. Report answer as a percentage accurate to one decimal place.
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The best predicted circumference for a diameter 1.2 cm is?
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Calculate the least square regression equation for the data. Include the slope and y intercept. Round to the nearest hundredth.
Find the regression equation, letting overhead width be the predictor (x) variable. Find the best predicted weight of a seal if the overhead width measured from a photograph is 2.1 cm. Can the prediction be correct? What is wrong with predicting the weight in this case? Use a significance level of 0.05. Overhead Width (cm) 8.1 7.6 8.9 7.7 8.6 9.8 Weight (kg) 156 166 213 138 197 251 Click the icon to view the critical values of the Pearson correlation coefficient r. The regression equation is y = (Round to one decimal place as needed.)
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Need help answering A, B and C. B) Perform the linear regression calculation and provide the linear regression equation that describes the relationship between Y (number of air conditioning units sold) and X (outside temperature in degrees Fahrenheit. Calculate: Coefficient of determination (r2) and the coefficient of correlation (r).  SST, SSE and SSR for this linear regression.  Estimate for the variance (σ2) and the standard deviation for the linear regression model you have developed. C) Using the linear regression equation that you developed in topic (b), calculate the estimated sales for a day that will reach 72 degrees F and for a day that will reach 94 F and for both temperature levels calculate the error “e” when comparing the estimated value against the actual data provided. At which of the two temperatures, is your model more accurate? Justify.
Section 10.2 Question #12 Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a marblewith a diameter of 1.2 cm. How does the result compare to the actual circumference of 3.8 cm? Use a significance level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball Diameter 7.5 24.3 4.2 21.7 6.9 4.1 20.7 Circumference 23.6 76.3 13.2 68.2 21.7 12.9 65.0 View the critical values of the Pearson correlation coefficient r. 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…
The arm span and foot length were both measured (in centimeters) for each of 20 students in a biology class. The computer output displays the regression analysis. Predictor Coef SE Coef t-ratio Constant Arm span -7.611 2.567 2.965 0.046 0.186 0.035 5.377 0.000 S = 1.61 R-Sq = 63.0% R-Sq (Adj) = 64.9% Which of the following is the best interpretation of the standard deviation of the residuals? The typical arm span is 161 centimeters. The typical foot length is 16.1 centimeters. The typical distance between the observed and predicted arm spans is 1.61 centimeters. The typical distance between the observed and predicted foot lengths is 1.61 centimeters. Op