Construct a scatterplot and identify the mathematical model that best fits the data. Assume that the model is to be used only for the scope of the given data and consider only linear, quadratic, logarithmic, exponential, and power models. Use a calculator or computer to obtain the regression equation of the model that best fits the data. You may need to fit several
models and compare the values of R²
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\begin{array} { c c c c } \mathrm { Y } & \mathrm { X } _ { 1 } & \mathrm { X } _ { 2 } & \\ 98.6 & 87.4 & 108.5 & \\ 101.2 & 97.6 & 110.1 & \\ 102.4 & 96.7 & 110.4 & \text { CORRELATION COEFFICIENTS } \\ 100.9 & 98.2 & 104.3 & \\ 102.3 & 99.8 & 107.2 & \mathrm { Y } / \mathrm { X } _ { 1 } = .850 \\ 101.5 & 100.5 & 105.8 & \mathrm { Y } / \mathrm { X } _ { 2 } = .742 \\ 101.6 & 103.2 & 107.8 & \\ 101.6 & 107.8 & 103.4 & \\ 99.8 & 96.6 & 102.7 & \text { COEFFICIENT OF DETERMINATION } \\ 100.3 & 88.9 & 104.1 & \\ 97.6 & 75.1 & 99.2 & \mathrm { Y } / \mathrm { X } _ { 1 } = .723 \\ 97.2 & 76.9 & 99.7 & \mathrm { Y } / \mathrm { X } _ { 2 } = .550 \\ 97.3 & 84.6 & 102.0 & \mathrm { Y } / \mathrm { X } _ { 1 } , \mathrm { X } _ { 2 } = .867 \\ 96.0 & 90.6 & 94.3 & \\ 99.2 & 103.1 & 97.7 & \\ 100.3 & 105.1 & 101.1 & \\ 100.3 & 96.4 & 102.3 & \\ 104.1 & 104.4 & 104.4 & \\ 105.3 & 110.7 & 108.5 & \\ 107.6 & 127.1 & 111.3 & \end{array}

\hat { Y } = 52.6 + .462 X _ { 2 }

\hat { Y } = 58.9 + .612 X _ { 1 }

\hat { \mathrm { Y } } = 48.0 + .398 \mathrm { X } _ { 1 } + .228 \mathrm { X } _ { 2 }

\hat { \mathrm { Y } } = 57.6 + .153 \mathrm { X } _ { 1 } + .270 \mathrm { X } _ { 2 }

Find the Best Predicted Value of Y Corresponding to the Given

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