Home Appraisal Guide

Home Appraisal sample

Please help with this problem?

A real estate company appraised the market value of 20 homes in a district of san fran. They found that the sample mean and the sample standard deviation were 936,500 and 63,000, respectively/

ANSWER: 90% Resulting Confidence Interval for ‘true mean’: = [912552, 960448]

Why???

SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION

x-bar: Sample mean = 936500
s: Sample standard deviation = 63000
n: Number of samples = 20
df: degrees of freedom = 19

significant digits = 1

Confidence Level = 90
“Look-up” Table (‘t-critical value’) = 1.7
Look-up Table of (‘t critical values’) for confidence and prediction intervals. Central two-sided area = 90% with df = 19. Another Look-up method is to utilize Microsoft Excel function: TINV(probability,degrees_freedom) Returns the inverse of the Student’s t-distribution 90% Resulting Confidence Interval for ‘true mean’: x-bar ± (‘t critical value’) * s/SQRT(n)

= 936500 ± 1.7 * 63000/SQRT(20) = [912552, 960448]

ANSWER: Conclusion: Null Hypothesis H1: μ ≠ μ0 (Null Hypothesis) not true with 99% confidence. Home price not equal to 900,000.

Why??
SINGLE SAMPLE TEST, TWO-TAILED, 6 – Step Procedure for t Distributions, “two-tailed test”

Step 1: State the hypothesis to be tested.
Null Hypothesis H0: μ = μ0
Alternate Hypothesis H1: μ ≠ μ0

Step 2: Determine a planning value for α [level of significance] =
0.01
Step 3: From the sample data determine x-bar, s and n; then compute
Standardized Test Statistic: t = ( x-bar – μ0 )/( s/ SQRT(n) )

x-bar: Est. of the Pop. Mean (statistical mean of the sample) =
936500
n: number of individuals in the sample =
20
s: sample standard deviation =
63000
μ0: Population Mean =
900000
significant digits =
3
Standardized Test Statistic t = ( 936500 – 900000 )/( 63000 / SQRT( 20 )) =
2.591

Step 4: Using Students t distribution, ‘lookup’ the area outside of t = TDIST( 2.591 , 19 , 2 ) using Excel TDIST(x, n-1 degrees_freedom, 2 tails)
using Excel TDIST(x, n-1 degrees_freedom, 2 tails)

Step 5: Area in Step 4 is equal to P value =
0.018
based on n -1 = 19 df (degrees of freedom).

Table look-up value shows area under the 19 df curve outside of t = +/- 2.591 is (approx.)
P value = 0.018 [2 * 0.009] by addition of both ‘tails’ of t distribution.

Step 6: For P ≥ α, fail to reject H0; and for P < α, reject H0 with 99% confidence in the conclusion.

Conclusion: Null Hypothesis H0: μ = μ0 (Null Hypothesis) is true with 99% confidence.

Note: level of significance [α] is the maximum level of risk an experimenter is willing
to take in making a "reject H0" or "conclude H1" conclusion (i.e. it is the maximum
risk in making a Type I error).

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In a perfect world, you have to have willing buyer and a willing seller. Neither is under duress. home appraisal sample Both are in a position to maximize gain and are trying to do this. But in the real world, things are rarely that simple and equally balanced. home appraisal sample Which is why people feel differently about the appraisal value of a house.