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Model to Predict the Assessed Value of Houses

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Please see the attached file.

You want to develop a model to predict the assessed value of houses, based on heating area. A sample of 15 single-family houses is selected in a city. The assessed value (in thousand of dollars) and the heating area of the houses (in thousands of square feet) are recorded, with the following results, stored in the file HOUSE2.xls(see attached file):
House Assessed Value Heating Area of Dwelling
($000) (Thousands of Square Feet)
1 184.4 2.00
2 177.4 1.71
3 175.7 1.45
4 185.9 1.76
5 179.1 1.93
6 170.4 1.20
7 175.8 1.55
8 185.9 1.93
9 178.5 1.59
10 179.2 1.50
11 186.7 1.90
12 179.3 1.39
13 174.5 1.54
14 183.8 1.89
15 176.8 1.59
(Hint: first,determine which are the independent and dependent variables.)

a) Construct a scatter plot and, assuming a linear relationship, use the least-squares method to compute the regression coefficients B0 and B1.
b) Interpret the meaning of the Y intercept, B0, and the slope, B1, in this problem.
c) Use the prediction line developed in (a) to predict the assessed value for a house whose heating area is 1,750 square feet.
d) Determine the coefficient of determination, r square(r*r), and interpret its meaning in this problem.
e) Perform a residual analysis on your results and determine the adequacy of the fit of the model.

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Solution Summary

Using Excel, we obtain the scatterplot and ANOVA table and determine the regression equation. We can then interpret the meaning of the Y intercept, B0, and the slope in this problem. We can then predict the assessed value for a house whose heating area is 1,750 square feet. We also find the coefficient of determination, the percentage of variability in the assessed values of houses explained by the different heating areas. In Excel, we get the residual output as a table and plot.

Solution provided by:
Changping Wang, MA
Education
  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
  • "Thank you"
  • "Thank you very much for your valuable time and assistance!"
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