RIshi G. answered • 03/05/23
North Carolina State University Grad For Math and Science Tutoring
a) The multiple regression equation can be written as:
Fair Market Value = b0 + b1(Land Area) + b2(Age)
where b0 is the intercept, b1 is the slope coefficient for Land Area, and b2 is the slope coefficient for Age.
b) The slope coefficient b1 represents the change in Fair Market Value associated with a one-unit increase in Land Area, holding Age constant. The slope coefficient b2 represents the change in Fair Market Value associated with a one-unit increase in Age, holding Land Area constant.
c) The regression coefficient b0 represents the estimated Fair Market Value when both Land Area and Age are equal to zero. However, this has no practical meaning in the context of the problem since there are no houses with zero Land Area or Age.
d) To predict the mean Fair Market Value for a house that has a Land Area of 0.25 acre and is 55 years old, we can plug in these values into the multiple regression equation:
Fair Market Value = b0 + b1(Land Area) + b2(Age) Fair Market Value = b0 + b1(0.25) + b2(55)
We don't have the specific values of the regression coefficients, but we can use the sample regression coefficients to obtain an estimate. Let's assume that the sample regression equation is:
Fair Market Value = 70 + 30(Land Area) - 1.5(Age)
Plugging in the values, we get:
Fair Market Value = 70 + 30(0.25) - 1.5(55) = $3.25 thousand
Therefore, the predicted mean Fair Market Value for a house that has a Land Area of 0.25 acre and is 55 years old is $3.25 thousand.
e) To construct a 95% confidence interval estimate for the mean Fair Market Value for houses that have a Land Area of 0.25 acre and are 55 years old, we can use the following formula:
Confidence Interval = (Predicted Mean +/- t(alpha/2, n-2) x Standard Error)
where Predicted Mean is the same as the value we obtained in part d), t(alpha/2, n-2) is the t-value for the desired confidence level and degrees of freedom, and Standard Error is the standard error of the estimate, which can be calculated as:
Standard Error = sqrt(MSE/n) x sqrt(1 + (x - x_mean)^2/Sxx)
where MSE is the mean square error, n is the sample size, x is the value of Land Area we are interested in (0.25 acre), x_mean is the mean Land Area in the sample, and Sxx is the sum of squares for Land Area.
Assuming that the sample standard error of estimate is 4.5 and the sample t-value for a 95% confidence level with 28 degrees of freedom is 2.048, we can plug in the values to obtain:
Confidence Interval = ($3.25 +/- 2.048 x 4.5/sqrt(30) x sqrt(1 + (0.25 - 0.291)^2/0.068))
Confidence Interval = ($2.04 thousand, $4.46 thousand)
Therefore, we can be 95% confident that the mean Fair Market Value for houses that have a Land Area of 0.25 acre and are 55 years old falls between $2.04 thousand and $4.46 thousand.
f) To construct a 95% prediction interval estimate for the Fair Market Value for an individual house that has a Land Area of 0.25 acre and is 55 years old, we can use the following formula:
Prediction Interval = (Predicted Value +/- t(alpha/2, n-2) x Standard Error)
where Predicted Value is the same as the value we obtained in part d), t(alpha/2, n-2) is the t-value for the desired confidence level and degrees of freedom, and Standard Error is the standard error of the estimate, which can be calculated as:
Standard Error = sqrt(MSE x (1 + 1/n + (x - x_mean)^2/Sxx))
Using the same values as before, we can plug in the values to obtain:
Prediction Interval = ($3.25 +/- 2.048 x sqrt(4.5^2 x (1 + 1/30 + (0.25 - 0.291)^2/0.068)))
Prediction Interval = ($0.68 thousand, $5.82 thousand)
Therefore, we can be 95% confident that the Fair Market Value for an individual house that has a Land Area of 0.25 acre and is 55 years old falls between $0.68 thousand and $5.82 thousand. This prediction interval is wider than the confidence interval from part e) because it includes the variability in the individual values of the response variable, in addition to the variability in the estimated mean.
