2010, ECON 7710 1

Hypothesis Testing 1: Single Coefficient

• Review of hypothesis testing

• Testing single coefficient

• Interval estimation

Objectives

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s.e. (9.3421) (0.8837)R2 = 0.7431, N = 20, SER = 8.5018

ii X3771.63971.103Y

Explaining weight by height (Table 1.1)

Can X really explain Y?

When X=0, what is Y?

If we suspect that the coefficient of X is 5, can we find support from the data?

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1. Hypothesis testing: Revision

The principle of hypothesis testing

The value of the parameter to be tested is assumed in H0. The estimate of this parameter is compared with that assumed value.

If the estimate is far from the assumed value, then H0 is rejected. Otherwise, H0 is not rejected.

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Procedures of Hypothesis Testing

1. Determine null and alternative hypotheses.

2. Specify the test statistic and its distribution as if the null hypothesis were true.

3. Select and determine the rejection region.

4. Calculate the sample value of test statistic.

5. State your conclusions.1. Revision

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2. Testing a Regression Coefficient

PopulationYi = 0 + 1X1i + 2X2i + … + KXKi + i

KiKi22i11oi XˆXˆXˆˆY Sample:

3 types of tests (k = 0, 1, 2, , K):

•Ho: k = c; HA: k c

•Ho: k c; HA: k > c

•Ho: k c; HA: k < c

c is any number meaningful in your study

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Probability Distribution of Least Squares Estimators

kkkˆvar,N~

)1,0(N~ˆvar

ˆZ

k

kk

2. Testing

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Student's t - statistic

t has a Student-t Distribution with N – K – 1 degrees of freedom.

)1,0(N~ˆvar

ˆZ

k

kk

1KN

k

kk t~ˆse

ˆt

2. Testing

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Two-Tail t-test

1. State the null & alternative hypotheses

H0: k = c

HA: k c

2. Compute the estimated t-valuecˆ

ˆSet

k

k

3. Choose a level of significance () and degrees of freedom (N – K – 1). Then find a critical t-value from the t-table (tc = tN-K-1,/2).

2. Testing

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Two-Tail t-test (cont.)

4. State the decision rule.

Version I: If |t| > tc, then reject H0.

Version II: If t > tc or t < -tc, then reject H0.

5. Conclusion Acceptance region

0 tc-tc

rejection regionrejection region

2. Testing

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Example 1: In the following regression results, test whether the estimated coefficient of X1 and X2 are significantly different from zero. ( = 5%)

Y = 14.32 + 0.798 X1 – 0.101 X2

se (6.1361) (0.2535) (0.08333)R2 = 0.2718, N = 30.

Hypotheses: H0: 1 = 0; HA: 1 0.

First test:

2. Testing

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Computed t-value: ....

)ˆ(

ˆ1123

25350

078900

1

1

se

t

Table t-value: For = 0.05 and 30 – 2 – 1 = 27 degrees of freedom, a critical value is t27,0.025 = 2.052.

Decision rule: If |t| > 2.052, then reject H0.

Conclusion: Since |t| = 3.112 > 2.052, Ho can be rejected. The estimated coefficient of X1 is significantly different from zero.

2. Testing

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One-Tail t-test

Step 1: State the null & alternative hypotheses

Right-tail test: Test whether k > c.

H0: k c; HA: k > c.

Left-tail test: Test whether k < c.

H0: k c; HA: k < c.

2. Compute the estimated t-value (same as before)

2. Testing

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3. Choose a level of significance () and degrees of freedom (N – K – 1). Then find a critical t-value from the t-table (tc = tN-K-1,).

4. State the decision rule.

Right-tail test: Reject H0 if t > tc.

Left-tail test: Reject H0 if t < -tc.

One-Tail t-test (cont.)

2. Testing

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0 tc < t

Right-tail

0-tct <

left-tail

2. Testing

One-Tail t-test (cont.)

5. Conclusion

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Example 3: Right-tail test: Test whether 1 is greater than 0.35 at 5% level of significance.

2. Testing

Dependent Variable: YMethod: Least SquaresSample: 1 30Included observations: 30

Coefficient Std. Error t-Statistic Prob.

C 16.42782 5.936234 2.767381 0.0099X 0.7177 0.246886 2.907007 0.0071

R-squared 0.231839 Mean dependent var 32.6Adjusted R-squared 0.204405 S.D. dependent var 12.71871S.E. of regression 11.3446 Akaike info criterion 7.759701Sum squared resid 3603.597 Schwarz criterion 7.853114Log likelihood -114.3955 Hannan-Quinn criter. 7.789585F-statistic 8.450689 Durbin-Watson stat 1.338091Prob(F-statistic) 0.007061

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Example 4: Left-tail test: Test whether 1 in Example 3 is smaller than 1.2. ( = 0.05)

2. Testing

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A Special caseHo: k = 0

HA: k 0

Statistic

k

k

ˆse

ˆt

2. Testing

It is the lowest level of significance at which we could reject the Ho that a parameter is zero.

The p-values Reported by Regression Software

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The t-statistics and P-values

Dependent Variable: YMethod: Least SquaresSample: 1 30Included observations: 30

Coefficient Std. Error t-Statistic Prob.

C 16.42782 5.936234 2.767381 0.0099X 0.7177 0.246886 2.907007 0.0071

R-squared 0.231839 Mean dependent var 32.6Adjusted R-squared 0.204405 S.D. dependent var 12.71871S.E. of regression 11.3446 Akaike info criterion 7.759701Sum squared resid 3603.597 Schwarz criterion 7.853114Log likelihood -114.3955 Hannan-Quinn criter. 7.789585F-statistic 8.450689 Durbin-Watson stat 1.338091Prob(F-statistic) 0.007061

2. Testing

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The p-value of 1-hat for a two-sided test

t0

f(t)

-2.91 2.91

p/2 = 0.00355

red area = p-value = 0.0071

p/2 = 0.00355

2.048-2.048

critical values

2. Testing

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3. Confidence Intervals for Regression Coefficients

Yi = 0 + 1Xi + ui (i = 1,n)

The OLS estimators for 0 and 1 are point estimators.The OLS estimates are likely to be different from the theoretical values

We have no idea of how close the OLS estimates to the theoretical values

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Interval estimation:

.1 0 where, - 1 ) ˆ -ˆP(

such that 0 a findcan then we

known, is ˆ ofon distributi theIf estimator. its

be ˆ andparameter population a be Let

k

k

kk

k

k

We know the chance of including the population parameter (k )in the intervals constructed from repeated samples.

3. Confidence Interval

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) ˆ , - ˆ( :estimator Interval kk

Confidence coefficient : 1 -

Level of significance :

Interval estimate : (k* - , k* + )

Population parameter: k

Estimator of k :

Estimate of k : k*

kβ

Confidence limits

3. Confidence Interval

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Constructing Confidence Interval for k

Actual estimated k could be fallen intothese regions

fk

k E k k

3. Confidence Interval

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fk

k E k k

a

b( ) )(

interval a interval b

Constructing Confidence Interval for k

3. Confidence Interval

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( )ta )(

tb

interval ta interval tb

f(t)

ˆSe

ˆt

k

kk

0

Constructing Confidence Interval for k

3. Confidence Interval

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Probability statements

P(-tc < t < tc ) = 1

P( t < -tc ) = P( t > tc ) =

1)ˆ(ˆ)ˆ(ˆ

1)ˆ(

kckkkck

c

k

kkc

setsetP

tse

tP

3. Confidence Interval

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A 95% confidence interval means that, using the interval estimator and drawing samples from the population, 95% of the interval estimates would include the population value .

The probability that a particular interval estimate contains this population value is either 0 or 1.

3. Confidence Interval

The (1-)100% CI for k is

)ˆ(setˆ),ˆ(setˆk/2 ,1KNkk/2 ,1KNk

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Example 5: Regressing WEIGHT on HEIGHT, 2005

3. Confidence Interval

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95% confidence limits for 1:

95% confidence interval for 0:

..,.

...ˆˆ.,

7811037650

10110000325788010250681

set

3. Confidence Interval

The 95% confidence interval for 1 is (0.3765, 0.7811).

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4. Applied Examples

Example 6: Restaurant location (Section 3.2)

Suppose you have been hired to determine a location for the next Woody’s. Woody’s is a moderately priced, 24-hour, family restaurant chain. Two choices are:

Location A: NN = 4.4, PP = 104, II = 20.6

Location B: NN = 2, PP = 50, II = 20

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YYi = 0 + NNi + PPPi + IIIi + i. -ve +ve ?

YY: Number of customers served in thousand

N: Number of direct market competitors

PP: Population in thousand within a 3-mile radius

II: Average household income in thousand

Example 6: Restaurant location (Cont’d)

4. Examples

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Woody’s: Null and Alternative Hypotheses

1. Ho: N 0; HA: N < 0

2. Ho: P 0; HA: P > 0

3. Ho: I = 0; HA: I 0

YY = 102.19*** – 9.07***N + 0.35***PP + 1.29**II se (2.0527) (0.07268) (0.5433)R2 = 0.618, R2 = 0.579, N = 33.

^

_

4. Examples

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If i is normally distributed, then

• hat is normally distributed with mean k and variance var( ).

•Z = has standard normal distribution.

kk

k

kk

ˆvar

ˆ

k kˆrav •Var( ) is unobservable. is used inst

ead and is denoted by se.

•t = has t distribution with N – K – 1

degrees of freedom. k

kk

ˆse

ˆ

4. Examples

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Woody’s two-sided test

Hypotheses: Ho: I = 0; HA: I 0

Statistics:

.freedom of degrees 294-33 with

ondistributi t a has ˆse

0ˆt

N

N

Decision rule: Let = 0.05. From the table the critical values are tc = t29,0.025 = 2.045. Reject Ho if |t| > 2.045.

.37.25433.0

0288.1ˆse

0ˆt

I

I

Computed t-value:Decision: Since t = 2.37 > 2.045, reject Ho. Thus I hat is significantly different from zero.

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Woody’s one-sided tests

Hypotheses: Ho: N 0; HA: N < 0

Decision rule: Let = 0.05. From the table the critical value is tc = -t29,0.05 = -1.699. Reject Ho if t < -1.699.

.42.4053,2

0075,9ˆse

0ˆt

N

N

Computed t-

value:

Decision: Since t = - 4.42 < -1.699, reject Ho. Thus N hat is significantly smaller than zero.

4. Examples

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Example 7: Sales of Hamburger

TRi = 0 + pPi + AAi + i. ? +

Data: Weekly observations for a hypothetical hamburger chain

TR : Weekly revenue in $1,000

P : Price in $

A : Advertising expenditure in $1,000

4. Examples

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TR = 113.83*** – 10.26***P + 2.68***A se (1.6007) (0.1189)R2 = 0.8739, N = 78.

^Regression results:

a. Is the demand significantly elastic or inelastic in price?

b. Is the increase in total revenue stimulated by more advertisements significantly greater than the corresponding increased cost of advertising?

Let = 0.01. Answer the following two questions statistically.