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It Is Possible That the Distance That a City Is =87.6%= 87.6 \% \quad

Question 167

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It is possible that the distance that a city is from the ocean could affect its average January
low temperature. Coast gives an approximate distance of each city from the East Coast or
West Coast (whichever is nearer). Including it in the regression yields the following
regression table: Dependent variable is:JanTemp
R squared =87.6%= 87.6 \% \quad R squared (adjusted) =86.9%= 86.9 \% s=4.878\mathrm { s } = 4.878 with 554=5155 - 4 = 51 degrees of freedom
 Source  Sum of Squares  df  Mean Square  F-ratio  Regression 8611.8632870.62121 Residual 1213.675123.7974\begin{array} { l c c c c } \text { Source } & \text { Sum of Squares } & \text { df } & \text { Mean Square } & \text { F-ratio } \\ \text { Regression } 8611.86 & 3 & 2870.62 & 121 \\ \text { Residual } & 1213.67 & 51 & 23.7974 & \end{array}\\
 Variable  Coefficient  SE(Coeff)  t-ratio  P-value  Intercept 111.8786.16718.10.0001 Lat 2.477220.130719.00.0001 Long 0.2219970.04624.810.0001 Coast 0.6749290.09017.490.0001\begin{array} { l c l l r } \text { Variable } & \text { Coefficient } & \text { SE(Coeff) } & \text { t-ratio } & \text { P-value } \\ \text { Intercept } & 111.878 & 6.167 & 18.1 & \leq 0.0001 \\ \text { Lat } & - 2.47722 & 0.1307 & - 19.0 & \leq 0.0001 \\ \text { Long } & 0.221997 & 0.0462 & 4.81 & \leq 0.0001 \\ \text { Coast } & - 0.674929 & 0.0901 & - 7.49 & \leq 0.0001 \end{array}
And here is a scatterplot of the residuals:  It is possible that the distance that a city is from the ocean could affect its average January low temperature. Coast gives an approximate distance of each city from the East Coast or West Coast (whichever is nearer). Including it in the regression yields the following regression table: Dependent variable is:JanTemp R squared  = 87.6 \% \quad  R squared (adjusted)  = 86.9 \%   \mathrm { s } = 4.878  with  55 - 4 = 51  degrees of freedom  \begin{array} { l c c c c } \text { Source } & \text { Sum of Squares } & \text { df } & \text { Mean Square } & \text { F-ratio } \\ \text { Regression } 8611.86 & 3 & 2870.62 & 121 \\ \text { Residual } & 1213.67 & 51 & 23.7974 & \end{array}\\   \begin{array} { l c l l r } \text { Variable } & \text { Coefficient } & \text { SE(Coeff) } & \text { t-ratio } & \text { P-value } \\ \text { Intercept } & 111.878 & 6.167 & 18.1 & \leq 0.0001 \\ \text { Lat } & - 2.47722 & 0.1307 & - 19.0 & \leq 0.0001 \\ \text { Long } & 0.221997 & 0.0462 & 4.81 & \leq 0.0001 \\ \text { Coast } & - 0.674929 & 0.0901 & - 7.49 & \leq 0.0001 \end{array}   And here is a scatterplot of the residuals:    \text { Write a report on this regression. Interpret the coefficients and } R ^ { 2 } \text {. Are the conditions met? }
 Write a report on this regression. Interpret the coefficients and R2. Are the conditions met? \text { Write a report on this regression. Interpret the coefficients and } R ^ { 2 } \text {. Are the conditions met? }

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