Math302 B004 Week 7 Test Wit Answers

1.0/ 1.0 PointsWeek 7 TestWeek 7 TestReturn to Assessment ListPart 1 of 7 - Linear Regression and Correlation; Outliers and Values of Correlation2.0/ 2.0 PointsQuestion 1 of 20The least squares regression line for a data set is y= -2.30.33x and the standard deviation of theresiduals is 0.26.Does a case with the values x = -3.33, y = -1.27 qualify as an outlier? A. Yes! B. No C. Cannot be determined with the given informationAnswer Key: Answer Key: BFeedback:Feedback:Plug in -3.33 for x.y = -2.3 -.33(-3.33)y = -1.2011Residual is y-given - y-predicted.-1.27 - (-1.2011)-1.27 + 1.2011 = -.2011 -> this is the residual value.To see if it is an outlier take -2 and multiply it by .26-2*.26 = -.52MATH302 B004 Spr 20 " # $ Tests & QuizzesTests & QuizzesTests & Quizzes1.0/ 1.0 Points1.0/ 1.0 Points-.2011 is greater than -.52, No, it is not an outlier because if it inside the range of the -2 to 2.Question 2 of 20The least squares regression line for a data set is y=5+0.3x and the standard deviation of theresiduals is 0.52.Does a case with the values x = -1.59, y = 5.78 qualify as an outlier? A. Cannot be determined with the given information