= 0 + 1 + 2 age2 + 3 + 4 + ,

lprice is log(house price in $),

age is the age of the house,

age2 is age squared,

rooms is the number of rooms in the house, and

area is the square footage of the house

a. Estimate the model and predict lprice if age=10, rooms=6 and area=2,000

Lprice= 10.47935+-.01049*10+.00004050*10^2+.05877*6+.00030599*2000

Lprice= 11.3431

c. Now construct a prediction interval for a particular property from the population with the characteristics listed in part (a). [Note that a CI for the average person in the subpopulation is not the same as a CI for a particular unit from the population.]

***Answer question c***

\begin{tabular}{|l|l|} \hline Number of Observations Read & 321 \\ \hline Number of Observations Used & 321 \\ \hline \end{tabular} \begin{tabular}{|l|r|r|r|r|r|} \hline \multicolumn{7}{|c|}{ Analysis of Variance } \\ \hline Source & DF & \begin{tabular}{r} Sum of \\ Squares \end{tabular} & \begin{tabular}{r} Mean \\ Square \end{tabular} & F Value & Pr > F \\ \hline Model & 4 & 37.02911 & 9.25728 & 119.84 & <.0001 error corrected total root mse r-square dependent mean adj r-sq coeff var parameter estimates variable label df estimate standard t value pr>tt \\ \hline Intercept & Intercept & 1 & 10.47935 & 0.14115 & 74.24 & <.0001 age of house agesq rooms in area square footage>