20.573011/15 answeredQuestion 13You run a regression analysis on a bivariate set of data (nyou obtain the regression equation107). With i = 49.6 and y = 58.6,y = 1.646x + 6.149with a correlation coefficient of r = 0.187. You want to predict what value (on average) for theresponse variable will be obtained from a value of 50 as the explanatory variable.What is the predicted response value?(Report answer accurate to one decimal place.)Question Help: DPost to forumSubmit Questiona

ou run a regression analysis on a bivariate set of data (n rou obtain the regression equation 107). With i = 49.6 and y = 58.6, %3D %3D y = 1.646x + 6.149 0.187. You want to predict what value (on average) for the with a correlation coefficient of r response variable will be obtained from a value of 50 as the explanatory variable. %3| What is the predicted response value?

ou run a regression analysis on a bivariate set of data (n rou obtain the regression equation 107). With i = 49.6 and y = 58.6, %3D %3D y = 1.646x + 6.149 0.187. You want to predict what value (on average) for the with a correlation coefficient of r response variable will be obtained from a value of 50 as the explanatory variable. %3| What is the predicted response value?

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Use the value of the linear correlation coefficient to calculate the coefficient of determination. What does this tell you about the explained variation of the data about the regression line? About the unexplained variation? r = -0.338 Calculate the coefficient of determination. (Round to three decimal places as needed.) % of the variation can be explained by the regression line (Round to one decimal place as needed.) % of the variation is unexplained and is due to other factors or to sampling error. (Round to one decimal place as needed.)
Please do not give solution in image formate thanku. i am stuck on this question.. You run a regression analysis on a bivariate set of data ( n = 34 ). You obtain the regression equation y = - 2.112 x + 59.334 with a correlation coefficient of r = - 0.878 (which is significant at α = 0.01 ). You want to predict what value (on average) for the explanatory variable will give you a value of 70 on the response variable. What is the predicted explanatory value? x = (Report answer accurate to one decimal place.)
Using your favorite statistic software package, you generate a scatter plot with a regression equation and correlation coefficient. The regression equation is reported as y = - 10.71 x + 65.32 and the r = 0.044. What proportion of the variation in y can be explained by the variation in the values of x? r^2 = % % to one decimal
Listed below are the overhead widths (cm) of seals measured from photographs and weights (kg) of the seals. Find the regression equation, letting the overhead width be the predictor (x) variable. Find the best predicted weight of a seal if the overhead width measured from a photograph is 1.8 cm, using the regression equation. Can the prediction be correct? If not, what is wrong? Use a significance level of 0.05. Overhead Width (cm) 7.3 7.4 9.8 9.5 8.8 8.5 Weight (kg) 152 187 286 247 237 231 The regression equation is y =+ (x. (Round the y-intercept to the nearest integer as needed. Round the slope to one decimal place as needed.)
A trucking company considered a multiple regression model for relating the dependent variable y = total daily travel time for one of its drivers (hours) to the predictors x₁ = distance traveled (miles) and x₂ = the number of deliveries made. Suppose that the model equation is Y = -0.800+ 0.060x₁ +0.900x₂ + e (a) What is the mean value of travel time when distance traveled is 50 miles and four deliveries are made? hr (b) How would you interpret ₁ = 0.060, the coefficient of the predictor x₁? O When the number of deliveries is constant, the average change in travel time associated with a ten-mile (i.e. one unit) increase in distance traveled is 0.060 hours. O The total daily travel time increases by 0.060 hours when the distance traveled increases by 1. O When the number of deliveries is held fixed, the average change in travel time associated with a one-mile (i.e. one unit) increase in distance traveled is 0.060 hours. O The average change in travel time associated with a one-mile (i.e.…
Why this type of plot is used in analysing regression models
The money raised and spent (both in millions of dollars) by all congressional campaigns for 8 recent 2-year periods are shown in the table. The equation of the regression line is y = 0.942x +27.609. Find the standard error of estimate s, and interpret the result. 793.9 1042.3 957.7 1203.3 450.7 673.7 745.1 778.6 Money raised, x Money spent, y 734.8 1024.2 929.1 1160.6 448.6 697.9 735.7 751.2 Find the standard error of estimate s, and interpret the result. (Round to three decimal places as needed.) How can the standard error of estimate be interpreted? O A. The standard error of estimate of the money raised for a specific amount of money spent is about s, million dollars. O B. The standard error of estimate of the money spent for a specific amount of money raised is about s, million dollars.
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 63 inches. Is the result close to the actual weight of 562 pounds? Use a significance level of 0.05. Chest_size_(inches) Weight_ (pounds)58 41450 31265 49959 45059 45648 260 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 63 inches? ^y =____ pounds (round one decimal as needed)Is the result close to the actual weight of 452 pounds?(a) This result is very close to the actual weight of the bear.(b) This result is close to the actual weight of the bear.(c) This result is exactly…
Listed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. Find the regression equation, letting the right arm blood pressure be the predictor (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 90 mm Hg. Use a significance level of 0.05. Right Arm 101 100 92 75 75 O Left Arm 174 167 181 149 147 Click the icon to view the critical values of the Pearson correlation coefficient r The regression equation is y = + x. (Round to one decimal place as needed.) Given that the systolic blood pressure in the right arm is 90 mm Hg, the best predicted systolic blood pressure in the left arm is mm Hg. (Round to one decimal place as needed.)
Listed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. Find the regression equation, letting the right arm blood pressure be the predictor (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 80 mm Hg. Use a significance level of 0.05. Right Arm 100 99 93 77 77 Q Left Arm 174 168 148 148 146 Click the icon to view the critical values of the Pearson correlation coefficient r The regression equation is ŷ=+x. (Round to one decimal place as needed.) mm Hg. Given that the systolic blood pressure in the right arm is 80 mm Hg, the best predicted systolic blood pressure in the left arm is (Round to one decimal place as needed.) Data table Critical Values of the Pearson Correlation Coefficient r α = 0.05 α = 0.01 0.950 0.990 0.959 0.878 0.811 0.917 0.754 0.875 0.707 0.834 0.666 0.798 0.632 0.765 0.602 0.735 0.576 0.708 0.553 0.684 0.532 0.661 0.514 0.641 0.497 0.623 0.482…
You run a regression analysis and obtain the regression equation y 3.176x +124.355 with a correlation coefficient of r = - 0.748. You want to predict what value (on average) for the response variable will be obtained from a value of x = 150 as the explanatory variable. What is the predicted response value? y = (Report answer accurate to one decimal place.)
i submitted a picture of the question