Question
variable1 variable2 -0.21582 0.89369 0.56997 -0.72620 -0.54850 -0.09185 -0.12385 0.50086 0.06975 -0.73607 0.16327 0.88498 -0.72595 -0.27512 0.22500 0.62647 -0.40463 0.92432 0.67652 0.56368 -0.82322 0.73005 0.06747
variable1 | variable2 | ||
-0.21582 | 0.89369 | ||
0.56997 | -0.72620 | ||
-0.54850 | -0.09185 | ||
-0.12385 | 0.50086 | ||
0.06975 | -0.73607 | ||
0.16327 | 0.88498 | ||
-0.72595 | -0.27512 | ||
0.22500 | 0.62647 | ||
-0.40463 | 0.92432 | ||
0.67652 | 0.56368 | ||
-0.82322 | 0.73005 | ||
0.06747 | -0.74824 | ||
0.74055 | 0.79412 | ||
-0.71577 | -0.04509 | ||
-0.82231 | -0.70951 | ||
-0.47603 | 0.01573 | ||
0.58094 | 0.51169 | ||
-0.58573 | 0.10376 | ||
0.19003 | -0.90089 | ||
-0.49528 | 0.04767 | ||
0.93083 | -0.16886 | ||
0.61389 | -0.65529 | ||
-0.91742 | 0.25296 | ||
-0.60957 | -0.24747 | ||
Linear regression is not applicable because it appears that there are two linear patterns indicating that the data come from two populations.
Linear regression is not applicable because the point pattern is curvilinear (has a curve). You need more information before deciding to use linear regression.
The linear regression equation will be very useful because the points have a strong linear pattern.
Linear regression is not useful because the points have no discernible pattern
QUESTION:You are required to setup a predictive equation involving variable 1 and variable 2. First, you plot the DATA to determine if linear regression applies. You decide
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