Hypothesis Testing

MATH DETAILS
The OLS estimators of the slope and the intercept are
1
1 2
2
1
0 1
( )( )
ˆ (4.7)
( )
ˆ ˆ . (4.8)
n
i i
i XY
n
X
i
i
X X Y Y
s
s
X X
Y X

 


 
 

 


1
β 0
β

MEASURES OF FIT (SW SECTION 4.3)
• Two regression statistics provide complementary measures of how well the regression
line “fits” or explains the data:
• The regression R2 measures the fraction of the variance of Y that is explained by X; it is
unitless and ranges between zero (no fit) and one (perfect fit)
• The standard error of the regression (SER) measures the magnitude of a typical
regression residual in the units of Y.

THE REGRESSION R2 IS THE FRACTION OF
THE SAMPLE VARIANCE OF YI “EXPLAINED”
BY THE REGRESSION
ˆ ˆ OLS prediction OLS residual
ˆ ˆsample var ( ) sample var( ) sample var( )( ?)
total sum of squares “explained” SS “residual” SS
i i i
i i
Y Y u
Y Y u why
   
  
  
2
2 2 1
2
1
ˆ ˆ( )
:
( )



 



n
i
i
n
i
i
Y Y
ESS
Definition of R R
TSS
Y Y
• means ESS = 0
• means ESS = TSS
• For regression with a single X, = the square of the correlation
coefficient between X and Y
2
=0R
2
=1R
2
0 1R •
2
R

DO DISTRICTS WITH SMALLER
CLASSES HAVE HIGHER TEST SCORES?
Scatterplot of test score v. student-teacher ratio
What does this figure show?

WE NEED TO GET SOME NUMERICAL EVIDENCE
ON WHETHER DISTRICTS WITH LOW STRS HAVE
HIGHER TEST SCORES – BUT HOW?
1. Compare average test scores in districts with low STRs to those with high STRs
(“estimation”)
2. Test the “null” hypothesis that the mean test scores in the two types of districts are
the same, against the “alternative” hypothesis that they differ (“hypothesis testing”)
3. Estimate an interval for the difference in the mean test scores, high v. low STR districts
(“confidence interval”)

INITIAL DATA ANALYSIS: COMPARE
DISTRICTS WITH “SMALL” (STR < 20)
AND “LARGE” (STR ≥ 20) CLASS SIZES
Class Size Average score Standard deviation
(s2)
n
Small 657.4 19.4 238
Large 650.0 17.9 182
Y
1. Estimation of Δ = difference between group means
2. Test the hypothesis that Δ = 0
3. Construct a confidence interval for Δ

1. ESTIMATION
largesmall
small large
1 1small large
1 1
657.4 650.0
7.4
nn
i i
i i
Y Y Y Y
n n 
  
 

 
Is this a large difference in a real-world sense?
• Standard deviation across districts = 19.1
• Difference between 60th and 75th percentiles of test score distribution
is 667.6 – 659.4 = 8.2
• This is a big enough difference to be important for school reform
discussions, for parents, or for a school committee?

2. HYPOTHESIS TESTING
2 2
Difference-in-means test: compute the -statistic,
(remember this?)
( )s l
s l
s l s l
s s
s l
n n
t
Y Y Y Y
t
SE Y Y
 
 

2 2
1
where ( ) is the “standard error” of , the subscripts
and/refer to “small” and “large” STR districts, and
1
( ) (etc.)
1
s
s l s l
n
s i s
is
SE Y Y Y Y s
s Y Y
n 
 
 


•

2. HYPOTHESIS TESTING
Compute the difference-of-means t-statistic:
Size s2 n
small 657.4 19.4 238
large 650.0 17.9 182
Y
2 2 2 2
19.4 17.9
238 182
657.4 650.0 7.4
4.05
1.83s l
s l
s l
s s
n n
Y Y
t
 
   

so reject (at the 5% significance level) the null hypothesis that the two
means are the same.
1.96,t

3. CONFIDENCE INTERVAL
A 95% confidence interval for the difference between the means is,
1.96
7.4
( )
1.96 1.83 (3.8, 11.0)
( )S l S lY Y Y YSE  
   
Two equivalent statements:
1. The 95% confidence interval for Δ doesn’t include 0;
2. The hypothesis that Δ = 0 is rejected at the 5% level.

HYPOTHESIS TESTING – OLS REGRESSION

HYPOTHESIS TESTING AND THE STANDARD ERROR OF Β1
The objective is to test a hypothesis, like β1 = 0, using data – to reach a
tentative conclusion whether the (null) hypothesis is correct or incorrect.
General setup
Null hypothesis and two-sided alternative:
H0: β1 = β1,0 vs. H1: β1 ≠ β1,0
where β1,0 is the hypothesized value under the null.
Null hypothesis and one-sided alternative:
H0: β1 = β1,0 vs. H1: β1 < β1,0

GENERAL APPROACH: CONSTRUCT T-
STATISTIC, AND COMPUTE P-VALUE (OR
COMPARE TO THE N(0,1) CRITICAL VALUE)
:• In general
estimator - hypothesized value
standard error of the estimator
t 
where the SE of the estimator is the square root of an
estimator of the variance of the estimator.
,0
:
/
Y
Y
Y
t
S n

• For testing the mean of Y
,• 1For testing 1 1,0
1
ˆ
,
ˆ( )
t
SE
 



1
1
ˆwhere ( ) the square root of an estimator of the variance
ˆof the sampling distribution of
SE 



1
ˆ (1of 2)( )Formula for SE 
Recall the expression for the variance of (large n):
2
1 2 2 2 2
var[( ) ]ˆvar( ) , where ( ) .
( ) ( )
i x i v
i i X i
X X
X u
v X u
n n
 
 
 

   
1
2 2
ˆThe estimator of the variance of replaces the unknown population
values of and by estimators constructed from the data:v X

 
1
2
2
2 1
ˆ 22 2
2
1
1
ˆ
estimator of1 1 2
ˆ
(estimator of ) 1
( )
ˆ ˆwhere ( ) .
n
i
v i
n
X
i
i
i i i
v
n
n n
X X
n
v X X u







   
 
 
 
 



1
ˆ (2 of 2)( )Formula for SE 
1
1
2
2 1
ˆ 2
2
1
2
ˆ1 1
1
ˆ
1 2
ˆ ˆ ˆ, where ( ) .
1
( )
ˆ ˆˆ( ) the standard error of
n
i
i
i i i
n
i
i
v
n
v X X u
n
X X
n
SE



  



   
 
 
 
 


This is a bit nasty, but:
• It is less complicated than it seems. The numerator estimates
var(v), the denominator estimates [var(X)]2.
• Why the degrees-of-freedom adjustment n – 2? Because two
coefficients have been estimated (β0 and β1).
1
ˆ( ) is computed by regression softwareSE •
• Your regression software has memorized this formula so you
don’t need to.

SUMMARY:
TO TEST H0: Β1 = Β1,0 V. H1: Β1 ≠
Β1,0
• Construct the t-statistic
1
1 1,0 1 1,0
2
1 ˆ
ˆ ˆ
ˆ( ) ˆ
t
SE

   
 
 
 
• Reject at 5% significance level if
• The p-value is probability in tails of normal
outside |tact|; you reject at the 5% significance level if the p-
value is < 5%.
t 1.96
[ ]Pr act
t tp  
1
ˆThis procedure relies on the large- approximation that is normally
distributed; typically = 50 is large enough for the approximation
to be excellent.
n
n
•

EXAMPLE: TEST SCORES AND STR,
CALIFORNIA DATA (1 OF 2)
Regression software reports the standard errors:
0 1
1 1,0
1,0
1
ˆ ˆ( ) 10.4 ( ) 0.52
ˆ 2.28 0
-statistic testing 0 4.38
ˆ 0.52( )
SE SE
t
SE
 
 


 
  
    
• The 1% 2-sided significance level is 2.58, so we reject the null
at the 1% significance level.
• Alternatively, we can compute the p-value…
Estimated regression line: Test Score = 698.9 – 2.28 x STR

EXAMPLE: TEST SCORES AND STR,
CALIFORNIA DATA (2 OF 2)
The p-value based on the large-n standard normal
approximation to the t-statistic is 0.00001 (10–5)

CONFIDENCE INTERVALS FOR Β1
(SECTION 5.2)
Recall that a 95% confidence is, equivalently:
• The set of points that cannot be rejected at the 5% significance
level;
• A set-valued function of the data (an interval that is a function
of the data) that contains the true parameter value 95% of the
time in repeated samples.
Because the t-statistic for β1 is N(0,1) in large samples, construction of a 95%
confidence for β1 is just like the case of the sample mean:
1 1 1
ˆ ˆ95% confidence interval for { 1.96 ( )}SE    

0 1
1
1 1
ˆ ˆ( ) 10.4 ( ) 0.52
ˆ95% confidence interval for :
ˆ ˆ{ 1.96 ( )} { 2.28 1.96 0.52}
( 3.30, 1.26)
SE SE
SE
 

 
 
     
  
CONFIDENCE INTERVAL EXAMPLE
The following two statements are equivalent (why?)
• The 95% confidence interval does not include zero;
• The hypothesis β1 = 0 is rejected at the 5% level

OLS REGRESSION: READING STATA
OUTPUT
regress testscr str, robust
Regression with robust standard errors Number of obs = 420
F( 1, 418) = 19.26
Prob > F = 0.0000
R-squared = 0.0512
Root MSE = 18.581
-------------------------------------------------------------------------
| Robust
testscr | Coef. Std. Err. t P>|t| [95% Conf. Interval]
--------+----------------------------------------------------------------
str | -2.279808 .5194892 -4.38 0.000 -3.300945 -1.258671
_cons | 698.933 10.36436 67.44 0.000 678.5602 719.3057
-------------------------------------------------------------------------
so:
· 2
1
1
698.9 2.28 , , ,
( ) ( )
( 0) , -value (2-sided)
395% 2-sided conf. .interval
10.4 0.52
.
30for is
0
1.2, 6
5
( )

  
    
  
TestScore STR R SER
t p
1
4.38 0.00
8
0
.6


SUMMARY OF STATISTICAL
INFERENCE ABOUT Β0 AND Β1
Estimation:
0 1
0 1
ˆ ˆOLS estimators and
ˆ ˆand have approximately normal sampling distributions in large samples
 
 
•
•
Testing:
• H0: β1 = β1,0 v. β1 ≠ β1,0 (β1,0 is the value of β1 under H0)
1 1,0 1
ˆ ˆ ˆ( )/ ( )t SE   •
• p-value = area under standard normal outside tact (large n)
Confidence Intervals:
• This is the set of β1 that is not rejected at the 5% level
• The 95% CI contains the true β1 in 95% of all samples.
1 1 1
ˆ ˆ95% confidence interval for is { 1.96 ( )}SE   •

Hypothesis Testing

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