nespuise 1s \"and xl ne tris shown below to determine which predictor variable w
ID: 3322614 • Letter: N
Question
nespuise 1s "and xl ne tris shown below to determine which predictor variable would be used first to x6 are mat possible predictor variables. Use the correlat corelation Correlation Matrix y x1 x2 x3 x4 x5_x6_ which variable is used firs x6 1.00 x1 0.61 1.00 CIRCLE your answer x2 0.83 0.48 1.00 x3 0.09 0.00 0.11 1.00 x4 -0.13 -0.55 -0.03 0.08 1.00 x5 -0.78 -0.66 -0.63 0.19 0.26 1.00 x6 -0.32 -0.32 -0.17 -0.06 0.28 0.33 1.00 xl x2 3 x5 x6 You are told that the SLOPE is found to be 15.52 and the INTERCEPT is found to be-100.2 e residuals from the first regression equation were regressed against all 6 "x" variables with the results shown below: x4 x5 x6 x3 -0.004 0.366 0.000 xl x2 -0.594 -0 , 378 42.193 15.510 20.403 0.214 -0 . 425 0.000 Slope 0.168 Intercept8.382 .000 0.000 0.105 0.039 0.136 CIRCLE your answer Which variable would enter the xI x2 x3 x4 x5 x6 regression on the next step? Finally, using the SLOPE and INTERCEPT from the first step as well as the line of best fit the 2nd variable you selected, write the equation after the second step: shown for Fill in all 5 boxes YOU CANNOT SKIP THIS PROBLEMExplanation / Answer
!st we add an X2 variable which is highly correlated with Y.
hence we add X2 add regression line is ycap=15.52-100.2X2
after that, we add the X5 variable as it has a maximum R2.
hence, Add X5 the nd final regression line is
ycap=15.52-100.2X2-0.378X5
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