Provide an appropriate response.The data below are the average one-way commute times (in minutes) for selectedstudents and the number of absences for those students during the term. Find theequation of the regression line for the given data. What would be the predictednumber of absences if the commute time was 95 minutes? Is this a reasonablequestion? Round the predicted number of absences to the nearest whole number.Round the regression line values to the nearest hundredth.Commute time (min), xNumber of absences, y7285919088987510080371010815415A)0.45x - 30.27; 12 absences; No, it is not reasonable. 95 minutes is welloutside the scope of the model.B)y = 0.45x - 30.27; 12 absences; Yes, it is reasonable.C) y = 0.45x + 30.27; 73 absences; Yes, it is reasonable.D)y = 0.45x + 30.27; 73 absences; No, it is not reasonable. 95 minutes is welloutside the scope of the model.

The independent variable is Commute time. The dependent variable is Number of absences. We have to…

Answered: The data below are the average one-way…

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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r = -0.579 Calculate the coefficient of determination. (Round to three decimal places as needed.) What does this tell you about the explained variation of the data about the regression line? % of the variation can be explained by the regression line. (Round to one decimal place as needed.) About the unexplained variation? % of the variation is unexplained and is due to other factors or to sampling error. (Round to one decimal place as needed.)
Using your favorite statistics software package, you generate a scatter plot with a regression equation and correlation coefficient. The regression equation is reported asy=20.41x+15.59y=20.41x+15.59and the r=0.272r=0.272.What proportion of the variation in y can be explained by the variation in the values of x?r² = %Report answer as a percentage accurate to one decimal place.
You generate a scatter plot using Excel. You then have Excel plot the trend line and report the equation and the r2r2 value. The regression equation is reported asy=60.71x+33.51y=60.71x+33.51and the r2=0.1764r2=0.1764.What is the correlation coefficient for this data set?r =
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 = 96.61x + 36.88 and the r = 0.568. What proportion of the variation in y can be explained by the variation in the values of x? r² = % Report answer as a percentage accurate to one decimal place.
Using your favorite statistics software package, you generate a scatter plot with a regression equation and correlation coefficient. The regression equation is reported asy=98.15x+95.04y=98.15x+95.04and the r=0.73r=0.73.What proportion of the variation in y can be explained by the variation in the values of x?r² = %Report answer as a percentage accurate to one decimal place.
Using your favorite statistics software package, you generate a scatter plot with a regression equation and correlation coefficient. The regression equation is reported asy=43.25x+19.44y=43.25x+19.44and the r=0.98r=0.98.What proportion of the variation in y can be explained by the variation in the values of x?r² = %
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 = - 13.31a + 55.86 and the r = 0.819. What proportion of the variation in y can be explained by the variation in the values of x? r2 = % Report answer as a percentage accurate to one decimal place. Submit Question M hp % & 7 08 16 5. %23 3m
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=16.02x+83.71and the r=0.845 What proportion of the variation in y can be explained by the variation in the values of x? r² = Report answer as a percentage accurate to one decimal place.
Solve for the slope, y-intercept, equation of the trend line, the correlation coefficient, and the coefficient of determination for the following the data set. Show your work and explain the steps you used to solve. Round your answer to the nearest hundredth.
Please help me! I have no idea where to even start. Conclude that the mean height is not equal to 64 inches because the P-value is less than 0.05. a. Find the correlation coefficient between the height and the weight. b. Construct the equation of the regression line. c. Predict the weight of a student who is 68 inches tall.
= Question Help The data show the chest size and weight of several bears. Find the regression equation, letting chest size be the independent (x) variable. Then find the best predicted weight of a bear with a chest size of 48 inches. Is the result close to the actual weight of 218 pounds? Use a significance level of 0.05. Chest size (inches) Weight (pounds) 58 49 58 43 57 348 46 355 266 332 177 246 = Click the icon to view the critical values of the Pearson correlation coefficient r. What is the regression equation? y= x (Round to one decimal place as needed.) +. What is the best predicted weight of a bear with a chest size of 48 inches? The best predicted weight for a bear with a chest size of 48 inches is (Round to one decimal place as needed.) pounds. Is the result close to the actual weight of 218 pounds? O A. This result is not very close to the actual weight of the bear. O B. This result is very close to the actual weight of the bear. O C. This result is close to the actual…
Please help me. Thankyou.

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