In the table above, the proportion of the variation in "Score received on the exam" that can be explained by the variation in attendance is given by Regression Statistics Multiple R R Square Standard Error Observations Intercept Attendance Intercept Standard error The P-value R Square 0.142620229 0.02034053 20.25979924 22 Coefficients Standard Error 39.39027309 37.24347659 0.340583573 0.52852452 T Stat 1.057642216 0.644404489 P-value 0.302826622 0.526635689

It appears that there is a significant relationship between cyberbullying (X) and internet trolling (Y).   Model Summary Model R R Square Adjusted R Square Std. Error of the Estimate 1 .889a .790 .748 1.50555 a. Predictors: (Constant), CyberBullying     ANOVAa Model Sum of Squares df Mean Square F Sig. 1 Regression 42.667 1 42.667 18.824 .007b Residual 11.333 5 2.267     Total 54.000 6       a. Dependent Variable: InternetTrolling b. Predictors: (Constant), CyberBullying     Coefficientsa Model Unstandardized Coefficients Standardized Coefficients t Sig. B Std. Error Beta 1 (Constant) .333 1.639   .203 .847 CyberBullying 1.333 .307 .889 4.339 .007 a. Dependent Variable: InternetTrolling       State hypothesis for the correlation: Is the correlation significant using α = .05? State the conclusion in APA format.…

A simple regression model developed for ten pairs of data resulted in a sum of squares of error, SSE = 125. The standard error of the estimate is 12.5 3.5 25 15.6 O 3.95

What mean physical health score would you expect in a group of 28-year-old women with a graduate degree?