Econometrics homework
Vincent666
Class6.pptxHypothesis testing
Class 6 for Econometrics 1
Vincent Geloso
Recap
Chapter 6 in full
Forget chapter 7, we will not be using it. However, I added some stuff that is not in the textbook (the last few slides only). Use the slides to complement.
Pages to read
In this unit, we link back up with what we saw in theme 3 (normal distribution and confidence intervals) and we give them more meaning by discussing how we can do inductive statistics.
More precisely, we are interested in how we can test hypotheses.
Before we proceed
We will assume three things:
the population distribution is approximately normal
the data is « interval » data (not nominal or ordinal)
There is no bias in the sampling method
What is a hypothesis?
It is a claim about a population parameter
From theme 3, we know the tools that we could use if we had a sample of phone bills and we wanted to know if μ was close to the sample mean. We could use the z-stat or the t-test (the latter is better though).
Question as reminder from theme 3 if you used t-stat: if you had a sample mean of 50.5$ with 20 phone bills in your sample and a standard deviation of 4$, you would get a standard error of ≈0.92 (SE=s/√(n-1). Since you have 19 degrees of freedom (20-1) and say you want a 95% confidence interval, you get the one-tail area of 2.5% from t-distribution (because the two tails amount to 100%-95%=5%), you get a t-stat of 2.093. Multiply 2.093 by 0.92 and you get ≈ 1.92 (this is SE*t-stat) and you get that your true population mean (μ) will be within the interval of 48.58$ to 52.42$.
Example: The mean monthly cell phone bill of this city is μ = $52
What is a hypothesis?
However, this does not allow us to test something about multiple samples.
Example from textbook is great on this: the Kent and Sussex samples of parishes have relief expenditures for the poor of 20.28 and 26.04 shillings (with standard deviations of 7.64 and 8.04 shillings). The mean in Sussex is 28% greater – what caused the difference?
Chance in the sampling? If this is the case, then just picking another sample would likely eliminate the difference (why? Because of the central limit theorem!)
Systematic and consistent?
The Null Hypothesis, H0
States the assumption (numerical) to be tested
Example: The average number of TV sets in U.S. Homes is equal to three ( )
Is always about a population parameter, not about a sample statistic
The Null Hypothesis, H0
Begin with the assumption that the null hypothesis is true
Similar to the notion of innocent until proven guilty (see next slide for example)
Refers to the status quo
Always contains “=” , “≤” or “” sign
May or may not be rejected
(continued)
An illustration
Goods stolen from a house: A noticed loitering outside the house before the theft
Robbery at drug store: B seen driven away from the store on night of robbery
Break-in at electrical store: C seen in vicinity of store, overheard boasting in a bar he was soon to get a ton of money
Theft at a jewellery store: D caught selling the stolen goods
A, B are acquitted. C and D are found guilty by the jury (majority and unanimity), but C goes away because one juror was uncovinced).
This is like the null hypothesis – the null is the presumption of innocence
The Alternative Hypothesis, H1
Is the opposite of the null hypothesis
e.g., The average number of TV sets in U.S. homes is not equal to 3 ( H1: μ ≠ 3 )
Challenges the status quo
Never contains the “=” , “≤” or “” sign
May or may not be supported
Is generally the hypothesis that the researcher is trying to support
Population
Claim: the
population
mean age is 50.
(Null Hypothesis:
REJECT
Suppose
the sample
mean age
is 20: x = 20
Sample
Null Hypothesis
20
likely if μ = 50?
=
Is
Hypothesis Testing Process
If not likely,
Now select a random sample
H0: μ = 50 )
x
Reason for Rejecting H0
Sampling Distribution of X
μ = 50
If H0 is true
If it is unlikely that we would get a sample mean of this value ...
... then we reject the null hypothesis that μ = 50.
20
... if in fact this were the population mean…
X
Level of Significance,
Defines the unlikely values of the sample statistic if the null hypothesis is true
Defines rejection region of the sampling distribution
Is designated by , (level of significance)
Typical values are 0.01, 0.05, or 0.10
Is selected by the researcher at the beginning
Provides the critical value(s) of the test
This is what you used in the t-stat and z-score tables to get the confidence interval – its what we will use going forward.
Level of Significance and the Rejection Region (or critical region)
H0: μ ≥ 3 H1: μ < 3
0
H0: μ ≤ 3
H1: μ > 3
a
a
Represents
critical value
Lower-tail test
Level of significance =
a
0
Upper-tail test
Two-tail test
Rejection region is shaded
/2
0
a
/2
a
H0: μ = 3 H1: μ ≠ 3
Types of errors
There are two types of errors: type 1 and type 2
Consider the following example of a pedestrian crossing a street. That pedestrian can wait for a safe time to cross (to not be run over). However, if he waits for the maximum safety, he never crosses the street. Thus, if he minimizes the risk, he maximizes the delay. If he minimizes the delay, he maximizes the risks.
The two problems cannot be simultaneously avoided
These are cases of type 1 and type 2 errors because they imply a trade off!
Errors in Making Decisions
Type I Error
Reject a true null hypothesis
Considered a serious type of error (people are more worried about this one)
The probability of Type I Error is
Called level of significance of the test
Set by researcher in advance
Errors in Making Decisions
Type II Error
Fail to reject a false null hypothesis
The probability of Type II Error is β
(continued)
Outcomes and Probabilities
Actual Situation
Decision
Fail to
Reject
H
0
Correct Decision
(1 - )
a
Type II Error
( β )
Reject
H
0
Type I Error
( )
a
Possible Hypothesis Test Outcomes
H0 False
H0 True
Key:
Outcome
(Probability)
Correct
Decision
( 1 - β )
( 1 - β ) is called the power of the test
Outcomes and Probabilities: example from a new drug
Actual Situation
Decision
Fail to
Reject
H
0
Correct Decision
The drug has side effects – do not introduce
Type II Error
The drug has no side effect, not introduced, patients suffer for lack of treatment
Reject
H
0
Possible Hypothesis Test Outcomes
H0 False
H0 True
Key:
Outcome
(Probability)
Correct Decision
The drug can be introduced and benefits patients
A new drug is introduced, the null is « not introduce because it may have adverse effects on patients » The alternative is that there are no side effects and the drug should be introduced
Type I error
The drug has side effects but is introduced, patients suffer
Type I & II Error Relationship
Type I and Type II errors can not happen at
the same time
Type I error can only occur if H0 is true
Type II error can only occur if H0 is false
If Type I error probability ( ) , then
Type II error probability ( β )
Factors Affecting Type II Error
All else equal,
β when
β when σ
β when n
Types of errors relationship
When you increase , you can see that the the critical region gets smaller (see blue space versus green space) – this is a lesser likelihood of a type I error (rejecting the null when it is true)
However, the non-rejection region grew because of that (see space in-between critical regions for two-tails test) which means that we have a greater chance of making a type II error (failing to reject when null is false)
Represents
critical value
Level of significance =
/2
a
/2
Level of significance =
/2
a
/2
a
Zone of no-rejection
WHICH HYPOTHESIS TEST TO USE?
The idea is that we want to take a sample measure (i.e. a sample statistic – e.g. mean, median, variance, correlation coefficient, regression coefficient) and calculate a test statistic that will be subjected to comparison with the sampling distribution (i.e. what we saw under central limit theorem).
Which test to use (a summary)
| Value of one mean | One random sample with known population variance | Z-score |
| Value of one mean | One random sample with unknown population variance | T-test |
| Value of one variance | One random sample | Chi-square (see more below) |
| Comparing two means | Two random samples (and independent) with unknown population variance | T-test |
| Comparing two means | One paired sample with unknown variance for population | T-test |
| Comparing two variances | Two random and independent samples drawn from normal population | F-test (see more below and we will discuss this at length in theme 9) |
Hypothesis Tests for the Mean
Known
Unknown
Hypothesis
Tests for
Tests of the Mean of a Normal Distribution (σ Known)
Convert sample result ( ) to a z value
The decision rule is:
σ Known
σ Unknown
Hypothesis
Tests for
Consider the test
(Assume the population is normal)
Do you remember what the name for this is? It’s the standard error we saw in theme 3! (standard deviation over square root of n) --- if for a sample, becomes “s”
Decision Rule
Reject H0
Do not reject H0
a
zα
0
μ0
H0: μ = μ0
H1: μ > μ0
Critical value
Z
Alternate rule:
p-Value
p-value: Probability of obtaining a test statistic more extreme ( ≤ or ) than the observed sample value given H0 is true
Also called observed level of significance
Smallest value of for which H0 can be rejected
p-Value Approach to Testing
Convert sample result (e.g., ) to test statistic (e.g., z statistic )
Obtain the p-value
For an upper
tail test:
Decision rule: compare the p-value to
If p-value < , reject H0
If p-value , do not reject H0
Example: Upper-Tail Z Test for Mean ( Known)
A phone industry manager thinks that customer monthly cell phone bill have increased, and now average over $52 per month. The company wishes to test this claim. (Assume = 10 is known)
H0: μ ≤ 52 the average is not over $52 per month
H1: μ > 52 the average is greater than $52 per month
(i.e., sufficient evidence exists to support the
manager’s claim)
Form hypothesis test:
Suppose that = .10 is chosen for this test
Find the rejection region:
Reject H0
Do not reject H0
= .10
1.28
0
Reject H0
Example: Find Rejection Region
(continued)
Example: Sample Results
Obtain sample and compute the test statistic
Suppose a sample is taken with the following results: n = 64, x = 53.1 ( = 10 was assumed known)
Using the sample results,
(continued)
Example: Decision
Reach a decision and interpret the result:
Reject H0
Do not reject H0
= .10
1.28
0
Reject H0
Do not reject H0 since z = 0.88 < 1.28
i.e.: there is not sufficient evidence that the
mean bill is over $52
z = 0.88
(continued)
Example: p-Value Solution
Calculate the p-value and compare to
(assuming that μ = 52.0)
Reject H0
= .10
Do not reject H0
1.28
0
Reject H0
Z = .88
(continued)
p-value = .1894
Do not reject H0 since p-value = .1894 > = .10
Where to get this? A Z-score table!!!
See here: http://www.z-table.com/
Two-Tail Tests
In some settings, the alternative hypothesis does not specify a unique direction
Do not reject H0
Reject H0
Reject H0
There are two critical values, defining the two regions of rejection
/2
0
H0: μ = 3 H1: μ ¹ 3
/2
Lower critical value
Upper
critical value
3
z
x
-z/2
+z/2
Hypothesis Testing Example
Test the claim that the true mean # of TV sets in US homes is equal to 3.
(Assume σ = 0.8)
State the appropriate null and alternative
hypotheses
H0: μ = 3 , H1: μ ≠ 3 (This is a two tailed test)
Specify the desired level of significance
Suppose that = .05 is chosen for this test
Choose a sample size
Suppose a sample of size n = 100 is selected
Hypothesis Testing Example
Determine the appropriate technique
σ is known so this is a z test
Set up the critical values
For = .05 the critical z values are ±1.96
Collect the data and compute the test statistic
Suppose the sample results are
n = 100, x = 2.84 (σ = 0.8 is assumed known)
So the test statistic is:
(continued)
Hypothesis Testing Example
Is the test statistic in the rejection region?
Reject H0
Do not reject H0
= .05/2
-z = -1.96
0
Reject H0 if z < -1.96 or z > 1.96; otherwise do not reject H0
(continued)
= .05/2
Reject H0
+z = +1.96
Here, z = -2.0 < -1.96, so the test statistic is in the rejection region
Hypothesis Testing Example
Reach a decision and interpret the result
-2.0
Since z = -2.0 < -1.96, we reject the null hypothesis and conclude that there is sufficient evidence that the mean number of TVs in US homes is not equal to 3
(continued)
Reject H0
Do not reject H0
= .05/2
-z = -1.96
0
= .05/2
Reject H0
+z = +1.96
Example: p-Value
Example: How likely is it to see a sample mean of 2.84 (or something further from the mean, in either direction) if the true mean is = 3.0?
.0228
/2 = .025
-1.96
0
-2.0
Z
1.96
2.0
x = 2.84 is translated to a z score of z = -2.0
p-value
= .0228 + .0228 = .0456
.0228
/2 = .025
Example: p-Value
Compare the p-value with
If p-value < , reject H0
If p-value , do not reject H0
Here: p-value = .0456
= .05
Since .0456 < .05, we reject the null hypothesis
(continued)
.0228
/2 = .025
-1.96
0
-2.0
Z
1.96
2.0
.0228
/2 = .025
Tests of the Mean of a Normal Population (σ Unknown)
Convert sample result ( ) to a t test statistic
σ Known
σ Unknown
Hypothesis
Tests for
The decision rule is:
Consider the test
(Assume the population is normal)
9.3
For a two-tailed test:
The decision rule is:
Consider the test
(Assume the population is normal, and the population variance is unknown)
(continued)
Tests of the Mean of a Normal Population (σ Unknown)
Still the standard error!
Example: Two-Tail Test ( Unknown)
The average cost of a hotel room in Chicago is said to be $168 per night. A random sample of 25 hotels resulted in x = $172.50 and
s = $15.40. Test at the
= 0.05 level.
(Assume the population distribution is normal)
H0: μ = 168 H1: μ ¹ 168
Example Solution: Two-Tail Test
a = 0.05
n = 25
is unknown, so
use a t statistic
Critical Value:
t24 , .025 = ± 2.064
Do not reject H0: not sufficient evidence that true mean cost is different than $168
Reject H0
Reject H0
a/2=.025
-t n-1,α/2
Do not reject H0
0
a/2=.025
-2.064
2.064
1.46
H0: μ = 168 H1: μ ¹ 168
t n-1,α/2
Two-tails test between two means
| Kent | Sussex | |
| Sample size | 24 | 42 |
| Mean of relief expenditures per capita (shillings) | 20.28 | 26.04 |
| St.Dev | 7.641 | 8.041 |
| Standard error of the mean (st.dev/√n) | 1.56 | 1.241 |
H0: mean Sussex = mean Kent H1: mean Sussex ¹ mean Kent
How do we check in such a case?
Two-tails test between two means
First : calculate the standard error of the difference between the means (if the population variance can be assumed equal – we’ll leave this for later, let’s just assume for now)
To do this, you calculate the pooled variance as follows
Second, you can calculate the standard error of the difference between the two sample means
Pooled variance
Sample size
Variance of sample
Numbers for Kent and Sussex above
Two-tails test between two means
| Kent | Sussex | |
| Sample size | 24 | 42 |
| Mean of relief expenditures per capita (shillings) | 20.28 | 26.04 |
| St.Dev | 7.641 | 8.041 |
| Standard error of the mean (st.dev/√n) | 1.56 | 1.241 |
H0: mean Sussex = mean Kent H1: mean Sussex ¹ mean Kent
We have 66 observations total and two lost degrees of freedom so we have only 64 degrees of freedom. If we picked a=0.05 in a two-tails test (see table 5.1 in textbook), the critical value is t=2.000 Thus, our calculated t-stat is well above the critical value. We can thus reject the null and point out that the difference between Kent and Sussex means is statistically significant (i.e. it is a meaningful difference).
HYPOTHESIS TESTING FOR SIMPLE LINEAR REGRESSION COEFFICIENTS
Remember this from theme 5?
Total variation is made up of two parts:
Total Sum of Squares
Regression Sum of Squares
Error (residual) Sum of Squares
where:
= Average value of the dependent variable
yi = Observed values of the dependent variable
i = Predicted value of y for the given xi value
Estimation of Model Error Variance
An estimator for the variance of the population model error is
Division by n – 2 instead of n – 1 is because the simple regression model uses two estimated parameters, b0 and b1, instead of one
is called the standard error of the estimate
Excel Output
| Regression Statistics | ||||||
| Multiple R | 0.76211 | |||||
| R Square | 0.58082 | |||||
| Adjusted R Square | 0.52842 | |||||
| Standard Error | 41.33032 | |||||
| Observations | 10 | |||||
| ANOVA | df | SS | MS | F | Significance F | |
| Regression | 1 | 18934.9348 | 18934.9348 | 11.0848 | 0.01039 | |
| Residual | 8 | 13665.5652 | 1708.1957 | |||
| Total | 9 | 32600.5000 | ||||
| Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
| Intercept | 98.24833 | 58.03348 | 1.69296 | 0.12892 | -35.57720 | 232.07386 |
| Square Feet | 0.10977 | 0.03297 | 3.32938 | 0.01039 | 0.03374 | 0.18580 |
Comparing Standard Errors
Y
Y
X
X
se is a measure of the variation of observed y values from the regression line
The magnitude of se should always be judged relative to the size of the y values in the sample data
i.e., se = $41.33K is moderately small relative to house prices in the $200 - $300K range
Statistical Inference: Hypothesis Tests and Confidence Intervals
The variance of the regression slope coefficient (b1) is estimated by
where:
= Estimate of the standard error of the least squares slope
= Standard error of the estimate
Excel Output
| Regression Statistics | ||||||
| Multiple R | 0.76211 | |||||
| R Square | 0.58082 | |||||
| Adjusted R Square | 0.52842 | |||||
| Standard Error | 41.33032 | |||||
| Observations | 10 | |||||
| ANOVA | df | SS | MS | F | Significance F | |
| Regression | 1 | 18934.9348 | 18934.9348 | 11.0848 | 0.01039 | |
| Residual | 8 | 13665.5652 | 1708.1957 | |||
| Total | 9 | 32600.5000 | ||||
| Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
| Intercept | 98.24833 | 58.03348 | 1.69296 | 0.12892 | -35.57720 | 232.07386 |
| Square Feet | 0.10977 | 0.03297 | 3.32938 | 0.01039 | 0.03374 | 0.18580 |
Comparing Standard Errors of the Slope
Y
X
Y
X
is a measure of the variation in the slope of regression lines from different possible samples (the practice question below will help you see this).
Inference about the Slope: t Test
t test for a population slope
Is there a linear relationship between X and Y?
Null and alternative hypotheses
H0: β1 = 0 (no linear relationship)
H1: β1 0(linear relationship does exist)
Test statistic
where:
b1 = regression slope
coefficient
β1 = hypothesized slope
sb1 = standard
error of the slope
Inference about the Slope: t Test
| House Price in $1000s (y) | Square Feet (x) |
| 245 | 1400 |
| 312 | 1600 |
| 279 | 1700 |
| 308 | 1875 |
| 199 | 1100 |
| 219 | 1550 |
| 405 | 2350 |
| 324 | 2450 |
| 319 | 1425 |
| 255 | 1700 |
Estimated Regression Equation:
The slope of this model is 0.1098
Does square footage of the house significantly affect its sales price?
(continued)
Inferences about the Slope: t Test Example
H0: β1 = 0
H1: β1 0
From Excel output:
| Coefficients | Standard Error | t Stat | P-value | |
| Intercept | 98.24833 | 58.03348 | 1.69296 | 0.12892 |
| Square Feet | 0.10977 | 0.03297 | 3.32938 | 0.01039 |
t
b1
Inferences about the Slope: t Test Example
H0: β1 = 0
H1: β1 0
Test Statistic: t = 3.329
There is sufficient evidence that square footage affects house price
From Excel output:
Reject H0
| Coefficients | Standard Error | t Stat | P-value | |
| Intercept | 98.24833 | 58.03348 | 1.69296 | 0.12892 |
| Square Feet | 0.10977 | 0.03297 | 3.32938 | 0.01039 |
t
b1
Decision:
Conclusion:
Reject H0
Reject H0
a/2=.025
-tn-2,α/2
Do not reject H0
0
a/2=.025
-2.3060
2.3060
3.329
d.f. = 10-2 = 8
t8,.025 = 2.3060
(continued)
tn-2,α/2
Inferences about the Slope: t Test Example
H0: β1 = 0
H1: β1 0
P-value = 0.01039
There is sufficient evidence that square footage affects house price
From Excel output:
Reject H0
| Coefficients | Standard Error | t Stat | P-value | |
| Intercept | 98.24833 | 58.03348 | 1.69296 | 0.12892 |
| Square Feet | 0.10977 | 0.03297 | 3.32938 | 0.01039 |
P-value
Decision: P-value < α so
Conclusion:
(continued)
This is a two-tail test, so the p-value is
P(t > 3.329)+P(t < -3.329) = 0.01039
(for 8 d.f.)
Confidence Interval Estimate for the Slope
Confidence Interval Estimate of the Slope:
Excel Printout for House Prices:
At 95% level of confidence, the confidence interval for the slope is (0.0337, 0.1858)
| Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
| Intercept | 98.24833 | 58.03348 | 1.69296 | 0.12892 | -35.57720 | 232.07386 |
| Square Feet | 0.10977 | 0.03297 | 3.32938 | 0.01039 | 0.03374 | 0.18580 |
d.f. = n - 2
Why « 2 » instead of « 1 » as
often said above? Because there are two lost degrees of freedom with b-zero and b-1
Confidence Interval Estimate for the Slope
Since the units of the house price variable is $1000s, we are 95% confident that the average impact on sales price is between $33.70 and $185.80 per square foot of house size
| Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
| Intercept | 98.24833 | 58.03348 | 1.69296 | 0.12892 | -35.57720 | 232.07386 |
| Square Feet | 0.10977 | 0.03297 | 3.32938 | 0.01039 | 0.03374 | 0.18580 |
This 95% confidence interval does not include 0.
Conclusion: There is a significant relationship between house price and square feet at the .05 level of significance
(continued)
PRACTICE AT HOME
These are the same prices as above (in the first two columns) but I just changed the last value of Square Feet in the last two columns.
What happens to the estimate of the standard error of the slope? What happens to the coefficient? What happens to the 95% boundaries?
| House Price in $1000s | Square Feet | House Price in $1000s | Square Feet |
| (y) | (x) | (y) | (x) |
| 245 | 1400 | 245 | 1400 |
| 312 | 1600 | 312 | 1600 |
| 279 | 1700 | 279 | 1700 |
| 308 | 1875 | 308 | 1875 |
| 199 | 1100 | 199 | 1100 |
| 219 | 1550 | 219 | 1550 |
| 405 | 2350 | 405 | 2350 |
| 324 | 2450 | 324 | 2450 |
| 319 | 1425 | 319 | 1425 |
| 255 | 1700 | 255 | 3000 |
CHI-SQUARE AND F-TESTS
Chi-Square and F-Test
What happens if we want to compare variances?
Chi-square )* or F-test
The chi-square will be useful in this theme while the F-test will be introduced later when we deal with theme 9 (the classical regression model)
has a chi-square distribution with (n – 1) degrees of freedom
*this is why chi-square tests are often presented in intro classes with categorical data and expected values (like dice rolls, gender distributions etc.) – because we know that coin tosses should arrive at 0.5:0.5 etc and we can thus test to see if a sample of coin tosses produces statistically meaningful differences with expected (e.g. like detecting a tricked coin)
Goal: Test hypotheses about the population variance, σ2 (e.g., H0: σ2 = σ02) (you need to know the variance of population or assume it)*
If the population is normally distributed,
Tests of the Variance of a Normal Distribution
The test statistic for hypothesis tests about one population variance is
(continued)
Tests of the Variance of a Normal Distribution
Decision Rules: Variance
Population variance
Lower-tail test:
H0: σ2 σ02
H1: σ2 < σ02
Upper-tail test:
H0: σ2 ≤ σ02
H1: σ2 > σ02
Two-tail test:
H0: σ2 = σ02
H1: σ2 ≠ σ02
a
a/2
a/2
a
Reject H0 if
Reject H0 if
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0
0
t
n
s
μx
t if H Reject
00
μμ:H
01
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μμ:H
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1.46
25
15.40
168172.50
n
s
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t
1n
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SSR
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å
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2
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2
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y
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n
1
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2
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2
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2
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s
=
41.33032
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small
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0.03297
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(sq.ft.)
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98.25
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3.32938
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