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Performing simple regression and interpretating results

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The following problems are to be solved by Excel and/or MegaStat. You need to find the values on the excel regression solutions and interpret them. Part 1 is a simple linear regression as done in the chapter 13 assignment. You should only include Y (fuel consumption) and X1 (temperature) for this analysis. Part 2 is a multiple regression and its solution is just an extension of the linear regression solution. Now you should enter the Y values and both of the X values at the same time. Be sure to include your regression solutions/printouts in your submission using copy/paste.
It was felt by the local utility company that fuel consumption (natural gas) in their small town would be affected by temperature. The following data (ordered by Fuel consumption) was collected for 10 weeks during the winter. FIRST, conduct a linear regression analysis between fuel consumption, Y, and temperature, X1, using Excel or MegaStat. Use alpha = 0.05 and 1.01 for t-tests.
a. What is the Bivariate regression equation using only the first two variables? Interpret it and predict fuel consumption at temperature equal to 50 degrees.
b. Is the slope coefficient in the linear regression model between temperature and fuel consumption statistically significant at alpha=.05 and 0.01? How do you know? Show your test.
d. What is the value of the correlation coefficient, (r)? Interpret. What is the value of the coefficient of determination, R2? Interpret. What is the value of the standard error of the estimate, se?

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The solution gives detailed steps on performing simple regression and interpretating results.

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See the attached file.
a. What is the Bivariate regression equation using only the first two variables? Interpret it and predict fuel consumption at temperature equal to 50 degrees.
First we run the regression analysis using only the first two variables. The excel output is:
Coefficients Standard Error t Stat P-value
Intercept 41.45854 8.023474 5.167156 0.000856
X1 -0.63669 0.194615 -3.27153 0.01133
So the equation is ...

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