show the leverage, studentized residual, and influence for each of the five observations in a small dataset. Table Data. ID X Y h R D A 1 2 0.39 -1.02 0.40 B 2 3 0.27 -0.56 0.06 C 3 5 0.21 0.89 0.11 D 4 6 0.20 1.22 0.19 E 8 7 0.73 -1.68 8.86 h is the leverage, R is the studentized residual, and D is Cook's measure of influence.

Run a regression analysis on the following bivariate set of data with y as the response variable. x y 61.1 78.9 58.6 84.5 70.2 58.1 62.3 65.4 64.2 68.9 58 84.1 63.7 70.2 64.6 71.9 68.3 71.5 61.7 77.5 59.3 77.4 64.6 71.2 Find the correlation coefficient and report it accurate to three decimal places.r = What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place.  (If the answer is 0.84471, then it would be 84.5%...you would enter 84.5 without the percent symbol.)r² = %Based on the data, calculate the regression line (each value to three decimal places)y =  x + Predict what value (on average) for the response variable will be obtained from a value of 61.9 as the explanatory variable. Use a significance level of α=0.05α=0.05 to assess the strength of the linear correlation.What is the predicted response value?  (Report answer accurate to one decimal place.)

Run a regression analysis on the following bivariate set of data with y as the response variable. x y 85.1 61.8 40.6 -11.4 36.9 -15.1 72.5 47.9 62 41.8 74.2 19.4 91.3 55.4 57.7 -2.1 84.5 52.1 94.2 55.2 87.8 83.3 51.3 7.1 58.4 -4.9 72.3 16.7 72.6 35.4 71.3 19.2 64.4 16.7 Verify that the correlation is significant at an α=0.05α=0.05. If the correlation is indeed significant, predict what value (on average) for the explanatory variable will give you a value of 72.4 on the response variable.What is the predicted explanatory value?x =

Graph the scatterplot and find the equation of the regression line GDP.       CO2 emissions (in millions metric tons) 1.7.          552.6 1.2.          462.3 2.5.          475.4 2.8.          374.3 3.6.          748.5 2.2.          400.9 0.8.          253.0 1.5.           318.6 2.4.           496.8 5.9.           1180.6

Run a regression analysis on the following data set with y as the response variable. y 114 67.1 40.3 87.2 89 91.4 26.5 92.3 9.8 94.9 22.2 88.3 -0.5 What is the predicted value of the response variable when x = 82.5? y= || Round to 1 decimal place. xin

Here is data with y as the response variable. a. Make a scatter plot of this data. Which point is an outlier? Enter as an ordered pair, e.g., (x,y). (x, y) b. Find the regression equation for the data set without the outlier. Enter the equation of the form mx + b rounded three decimal places. y wo X 48.4 77.3 88 67.1 108.3 75.9 64.8 -161.3 61.2 67.9 ŷw y 45.7 40.3 27.3 54.5 -63.2 36.3 73.2 96.5 82.2 119.8 = -- c. Find the regression equation for the data set with the outlier. Enter the equation of the form mx + b rounded to three decimal places.

Run a regression analysis on the following bivariate set of data with y as the response variable. y 56.3 -257.6 14.5 670.8 49.4 431 60.7 200.3 60.1 70.3 52.6 -162.4 46.2 -406.9 45.1 -3.6 39.6 292.4 51.2 274.8 53.1 -77.4 50 496.6 46.9 251.5 48.5 -188.7 79.8 233 33.5 -415.6 49.2 431.3 download link provided here: Download CSV Predict what value (on average) for the response variable will be obtained from a value of 32.1 as the explanatory variable. Use a significance level of a = 0.05 to assess the strength of the linear correlation. What is the predicted response value? y = 169.94127 (Report answer accurate to one decimal place.)

Next Run a regression analysis on the following bivariate set of data with y as the response variable. X y 48.1 41 63.7 62.6 40.1 33.4 49.4 46.3 16.3 -29.1 39.9 27.3 37.1 33.3 48.2 45.4 32.9 11.1 34.9 27.9 44.6 25.5 25.1 Z 23.2 Find the correlation coefficient and report it accurate to three decimal places. What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place. (If the answer is 0.84471, then it would be 84.5%...you would enter 84.5 without the percent symbol.) r² = % Submit Question