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Sales ComparisonHARD16.4% of exam

An appraiser is valuing a 3,300-square-foot custom home in a subdivision of one-story tract homes. A regression of 52 subdivision sales, which range from 1,150 to 2,050 square feet, gives sale price = $71,000 + $104 × GLA with an R-squared of 0.84. The appraiser applies the line to the subject and reports $414,200 as the indicated value. What is the main weakness in this analysis?

Correct Answer

C) The subject's size lies far outside the range of the sales

A regression line describes how buyers priced homes within the range of the data. The largest sale had 2,050 square feet, and the subject has 3,300, so applying the line ($71,000 + $104 × 3,300 = $414,200) extrapolates well beyond the evidence; buyers may pay a very different rate per square foot for a much larger custom home. An R-squared of 0.84 and 52 sales are reasonable for a one-variable model, and the intercept belongs in a predicted price.

Answer Options
A
The R-squared of 0.84 is too low to rely on the model
B
The intercept should have been left out of the prediction
C
The subject's size lies far outside the range of the sales
D
Fifty-two sales are too few for a one-variable regression

Why This Is the Correct Answer

Option C is correct because the subject is 1,250 square feet larger than the largest sale in the model. A regression line measures the market's reaction only within the range of the data; projecting it far beyond that range assumes the same price per added square foot holds, which the sales do not show.

Why the Other Options Are Wrong

Option A: The R-squared of 0.84 is too low to rely on the model

Option A says an R-squared of 0.84 is too low. An R-squared of 0.84 means GLA explains about 84% of the price variation in these sales, which is a strong fit for a one-variable model; the problem is where the model is applied, not how well it fits.

Option B: The intercept should have been left out of the prediction

Option B says the intercept should have been left out. A predicted price needs the whole equation, $71,000 + $104 × 3,300; dropping the intercept would understate the prediction and would not cure the extrapolation.

Option D: Fifty-two sales are too few for a one-variable regression

Option D says 52 sales are too few for a one-variable regression. Fifty-two sales give a reasonable sample for a single variable; the data are adequate in number but do not include homes of the subject's size.

Stay inside the data fence

A regression line is fenced in by the smallest and largest sales used to build it. Inside the fence it speaks for the market; outside it is guessing.

How to use: Use this whenever a subject's size, age or other variable is compared with the range of the sales in a model.

Exam Tip

When a regression question gives the range of the data, compare it with the subject first; a subject outside the range points to extrapolation as the flaw.

Common Mistakes to Avoid

  • -Treating a good R-squared as proof the model fits any subject
  • -Ignoring the range of the data when applying a regression
  • -Assuming the price per added square foot is the same at every size

Concept Deep Dive

Analysis

The question asks the candidate to spot extrapolation. The arithmetic is right, but the line was fitted to 1,150 to 2,050 square foot homes, and nothing in the data shows how buyers price a 3,300 square foot custom home.

Background Knowledge

A regression model is reliable for interpolation, meaning predictions within the range of the observed independent variable. Outside that range, the relationship between price and GLA may change, for example as buyers pay less for each added square foot in very large homes, so extrapolated results lack market support.

Real-World Application

An appraiser is asked to value the largest home ever built in a tract. Instead of stretching the tract regression, she searches nearby areas for sales of similar large custom homes and uses them to support the GLA adjustment.

extrapolationregressiondata rangeR-squaredGLA adjustment
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