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Here Are Data About the Average January Low Temperature in Cities

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Here are data about the average January low temperature in cities in the United States, and factors that might allow us to
predict temperature. The data, available for 55 cities, include: $Here are data about the average January low temperature in cities in the United States, and factors that might allow us to predict temperature. The data, available for 55 cities, include: We will attempt to make a regression model to help account for mean January temperature and to understand the effects of the various predictors. At each step of the analysis you may assume that things learned earlier in the process are known. Units Note: The degrees of temperature, given here on the Fahrenheit scale, have only coincidental language relationship to the degrees of longitude and latitude. The geographic degrees are based on modeling the Earth as a sphere and dividing it up into 360 degrees for a full circle. Thus 180 degrees of longitude is halfway around the world from Greenwich, England (0°) and Latitude increases from 0 degrees at the Equator to 90 degrees of (North) latitude at the North Pole. -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 \mathrm { R } squared (adjusted) = 86.9 \% s = 4.878 with 55 - 4 = 51 degrees of freedom \begin{array}{lllcc} \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}{lclll} \text { Variable } & \text { Coefficient } & \text { SE }(\text { 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? }$ We will attempt to make a regression model to help account for mean January temperature and to understand the effects of
the various predictors.
At each step of the analysis you may assume that things learned earlier in the process are known.
Units Note: The "degrees" of temperature, given here on the Fahrenheit scale, have only coincidental language relationship to
the "degrees" of longitude and latitude. The geographic "degrees" are based on modeling the Earth as a sphere and dividing it
up into 360 degrees for a full circle. Thus 180 degrees of longitude is halfway around the world from Greenwich, England
(0°) and Latitude increases from 0 degrees at the Equator to 90 degrees of (North) latitude at the North Pole. $Here are data about the average January low temperature in cities in the United States, and factors that might allow us to predict temperature. The data, available for 55 cities, include: We will attempt to make a regression model to help account for mean January temperature and to understand the effects of the various predictors. At each step of the analysis you may assume that things learned earlier in the process are known. Units Note: The degrees of temperature, given here on the Fahrenheit scale, have only coincidental language relationship to the degrees of longitude and latitude. The geographic degrees are based on modeling the Earth as a sphere and dividing it up into 360 degrees for a full circle. Thus 180 degrees of longitude is halfway around the world from Greenwich, England (0°) and Latitude increases from 0 degrees at the Equator to 90 degrees of (North) latitude at the North Pole. -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 \mathrm { R } squared (adjusted) = 86.9 \% s = 4.878 with 55 - 4 = 51 degrees of freedom \begin{array}{lllcc} \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}{lclll} \text { Variable } & \text { Coefficient } & \text { SE }(\text { 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? }$
-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 $= 87.6 \% \quad \mathrm { R }$ squared (adjusted) $= 86.9 \%$ $s = 4.878$ with $55 - 4 = 51$ degrees of freedom
$\begin{array}{lllcc}\text { Source } & \text { Sum of Squares } & \text { df } & \text { Mean Square } & \text { F-ratio } \\\text { Regression } 8611.86 & 3 & 2870.62 & 121 \\\text { Residual } & 1213.67 & 51 & 23.7974 &\end{array}$

$\begin{array}{lclll}\text { Variable } & \text { Coefficient } & \text { SE }(\text { Coeff }) & \text { t-ratio } & \text { P-value } \\\text { Intercept } & 111.878 & 6.167 & 18.1 & \leq 0.0001 \\\text { Lat } & -2.47722 & 0.1307 & -19.0 & \leq 0.0001 \\\text { Long } & 0.221997 & 0.0462 & 4.81 & \leq 0.0001 \\\text { Coast } & -0.674929 & 0.0901 & -7.49 & \leq 0.0001\end{array}$
And here is a scatterplot of the residuals: $Here are data about the average January low temperature in cities in the United States, and factors that might allow us to predict temperature. The data, available for 55 cities, include: We will attempt to make a regression model to help account for mean January temperature and to understand the effects of the various predictors. At each step of the analysis you may assume that things learned earlier in the process are known. Units Note: The degrees of temperature, given here on the Fahrenheit scale, have only coincidental language relationship to the degrees of longitude and latitude. The geographic degrees are based on modeling the Earth as a sphere and dividing it up into 360 degrees for a full circle. Thus 180 degrees of longitude is halfway around the world from Greenwich, England (0°) and Latitude increases from 0 degrees at the Equator to 90 degrees of (North) latitude at the North Pole. -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 \mathrm { R } squared (adjusted) = 86.9 \% s = 4.878 with 55 - 4 = 51 degrees of freedom \begin{array}{lllcc} \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}{lclll} \text { Variable } & \text { Coefficient } & \text { SE }(\text { 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? }$
$\text { Write a report on this regression. Interpret the coefficients and } R ^ { 2 } \text {. Are the conditions met? }$

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