Chapter 5: Hypothesis Tests, Confidence Intervals & Gauss-Markov Result
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1 Chapter 5: Hypothess Tests, Confdence Intervals & Gauss-Markov Result 1-1
2 Outlne 1. The standard error of 2. Hypothess tests concernng β 1 3. Confdence ntervals for β 1 4. Regresson when X s bnary 5. Heteroskedastcty and homoskedastcty 6. Effcency of OLS and the Student t dstrbuton 5-2
3 Object of nterest: β 1 n, Y = β 0 + β 1 X + u, = 1,, n The Samplng Dstrbuton of : To derve ts large-samplng dstrbuton of, we made the followng assumptons: The Least Squares Assumptons: 1. E(u X = x) = (X,Y ), =1,,n, are..d. 3. Large outlers are rare (E(X 4 ) <, E(Y 4 ) <. 5-3
4 The Samplng Dstrbuton of, ctd. Under the Least Squares Assumptons, for n large, s approxmately dstrbuted, 2 σ ~ v N β 1,, where v = (X µ X)u n(σ 2 X ) 2 5-4
5 Hypothess Tests & the Standard Error of Wsh to test a hypothess, e.g. β 1 = 0, usng data General setup Null hypothess and two-sded alternatve: H 0 : β 1 = β 1,0 vs. H 1 : β 1 β 1,0 wth β 1,0 the hypotheszed value under the null. Null hypothess and one-sded alternatve: H 0 : β 1 = β 1,0 vs. H 1 : β 1 < β 1,0 5-5
6 Compute t-statstc & p-/crtcal value t = estmator - hypotheszed value standard error of the estmator where the SE of the estmator s the square root of an estmator of the varance of the estmator. For testng the mean of Y: t = β1 β1,0 For testng β 1, t =, SE( ˆ β ) ˆ Y µ Y,0 s Y / where SE( ) = the square root of an estmator of the varance of the samplng dstrbuton of 1 n 5-6
7 Estmatng var( ) Recall the expresson for the varance of (large n): var[( X var( ) = µ x )u ] =, where v = (X µ X )u. n(σ 2 X ) 2 n(σ 2 X ) 2 The estmator of the varance of replaces the unknown 2 σ v populaton values of and by estmators constructed from the data: 1 = = n estmator of σ 2 v (estmator of σ 2 X ) 2 v where = ( X X ) uˆ. ˆ n n 2 vˆ n = 1 n ( X X ) n = 1 5-7
8 n 1 2 ˆ 2 ˆ = 1 v n 2 = 1 σ, where =. ˆ β v 2 ˆ ( X ˆ X ) u 1 n n 1 2 ( X X ) n = 1 2 SE( ) = σ = the standard error of ˆ ˆ β 1 Note: It s less complcated than t seems. The numerator estmates var(v), the denomnator estmates [var(x)] 2. Why the degrees-of-freedom adjustment n 2? Because two coeffcents have been estmated (β 0 and β 1 ). SE( ) s computed by regresson software Your regresson software computes ths for you 5-8
9 Summary: To test H 0 : β 1 = β 1,0 v. H 1 : β 1 β 1,0, Construct the t-statstc t = = ˆ β β 1 1,0 Reject at 5% sgnfcance level f t > ˆ ˆ β σ 1 The p-value s p = Pr[ t > t act ] = probablty n tals of normal outsde t act ; you reject at the 5% sgnfcance level f the p-value s < 5%. Ths procedure reles on the large-n approxmaton that s normally dstrbuted; typcally n = 50 s large enough for the approxmaton to be excellent. 5-9
10 Example: Test Scores and STR, Calforna data TestScore Estmated regresson lne: = STR Regresson software reports the standard errors: SE( ) = 10.4 SE( ) = 0.52 ˆ β1 β1, SE( ˆ β ) 0.52 t-statstc testng β 1,0 = 0 = = = 4.38 The 1% 2-sded sgnfcance level s 2.58, so we reject the null at the 1% sgnfcance level. Alternatvely, we can compute the p-value
11 The p-value based on the large-n standard normal approxmaton to the t-statstc s (10 5 ) 5-11
12 Confdence Intervals for β 1 Recall that a 95% confdence s, equvalently: The set of ponts that cannot be rejected at the 5% sgnfcance level; A set-valued functon of the data (an nterval that s a functon of the data) that contans the true parameter value 95% of the tme n repeated samples. Because the t-statstc for β 1 s N(0,1) n large samples, constructon of a 95% confdence for β 1 s just lke the case of the sample mean: 95% confdence nterval for β 1 = { ± 1.96 SE( )} 5-12
13 TestScore Example: = STR SE( ) = 10.4 SE( ) = % confdence nterval for : { ± 1.96 SE( )} = { 2.28 ± } = ( 3.30, 1.26) The followng two statements are equvalent: The 95% confdence nterval does not nclude zero; The hypothess β 1 = 0 s rejected at the 5% level 5-13
14 A concse way to report regressons Parenthesze standard errors below estmated coeffcents TestScore = STR, R 2 =.05, SER = 18.6 (10.4) (0.52) Ths s nformatve: The estmated regresson lne s = STR TestScore The standard error of s 10.4 The standard error of s 0.52 The R 2 s.05; the standard error of the regresson s
15 OLS regresson: readng STATA output regress testscr str, robust Regresson wth robust standard errors Number of obs = 420 F( 1, 418) = Prob > F = R-squared = Root MSE = Robust testscr Coef. Std. Err. t P> t [95% Conf. Interval] str _cons so: TestScore = STR,, R 2 =.05, SER = 18.6 (10.4) (0.52) t (β 1 = 0) = 4.38, p-value = (2-sded) 95% 2-sded conf. nterval for β 1 s ( 3.30, 1.26) 5-15
16 Summary of statstcal nference about β 0 and β 1 Estmaton: OLS estmators and and s samplng dstrbutons are approxmately normal n large samples Testng: H 0 : β 1 = β 1,0 v. β 1 β 1,0 (β 1,0 s the value of β 1 under H 0 ) t = ( β 1,0 )/SE( ) p-value = area under standard normal outsde t act (large n) Confdence Intervals: 95% confdence nterval for β 1 s { ± 1.96 SE( )} Ths s the set of β 1 that s not rejected at the 5% level The 95% CI contans the true β 1 n 95% of all samples. 5-16
17 One-sded versons Testng: H 0 : β 1 = β 1,0 v. β 1 <β 1,0 (β 1,0 s the value of β 1 under H 0 ) Reject at 5% sgn.level. f t < H 0 : β 1 = β 1,0 v. β 1 >β 1,0 (β 1,0 s the value of β 1 under H 0 ) Reject at 5% sgn.level. f t >
18 Regresson when X s Bnary Sometmes a regressor s bnary: X = 1 f small class sze, = 0 f not X = 1 f female, = 0 f male X = 1 f treated (expermental drug), = 0 f not Bnary regressors are sometmes called dummy varables. Recall, β 1 s slope, any further nterpretatons here? 5-18
19 Interpretng regressons wth a bnary regressor Y = β 0 + β 1 X + u, where X s bnary (X = 0 or 1): When X = 0, Y = β 0 + u the mean of Y s β 0 that s, E(Y X =0) = β 0 When X = 1, Y = β 0 + β 1 + u the mean of Y s β 0 + β 1 that s, E(Y X =1) = β 0 + β 1 so: β 1 = E(Y X =1) E(Y X =0) = populaton dfference n group means 5-19
20 Example: Let D = 1 f STR 20 0 f STR > 20 OLS regresson: TestScore = D (1.3) (1.8) Tabulaton of group means: Class Sze Average score ( ) Std. dev. (s Y ) N Small (STR > 20) Large (STR 20) Dfference n means: Y small Y large = = 7.4 Match! 95% CI for dfference n means 7.4 ± 1.96*1.8. Alternatve ch3 5-20
21 Summary: regresson when X s bnary (0/1) Y = β 0 + β 1 X + u β 0 = mean of Y when X = 0 β 0 +β 1 = mean of Y when X = 1 β 1 = dfference n group means, X =1 mnus X = 0 SE( ) has the usual nterpretaton t-statstcs, confdence ntervals constructed as usual Ths s another way (an easy way) to do dfference-n-means analyss The regresson formulaton s especally useful when we have addtonal regressors (as we wll see soon) 5-21
22 Heteroskedastcty X Homoskedastcty 1. Skedastcty = randomness 2. Consequences of homoskedastcty 3. Implcaton for computng standard errors Defnton Errors are homoskedastc f var(u X=x ) s ndependent of datum (hence of X) Summarzed by a constant σ 2 u: var(u X=x ) = σ 2 u Otherwse, errors are sad to be heteroskedastc. 5-22
23 Heteroskedastcty n a pcture: E(u X=x) = 0 (u satsfes Least Squares Assumpton #1) The varance of u does depend on x 5-23
24 A real-data example from labor economcs: average hourly earnngs vs. years of educaton (data source: CPS) Heteroskedastc! 5-24
25 What f the errors are n fact homoskedastc? Sps. OLS assn s homoskedastcty var(u X =x)=σ 2 u. Then the formulas for s varance & standard error smplfy var( β ) = σ 2 u 1 2 ( X X ) 2 su 2 1 SE( β 1) = where s : 2 u = u ( X ) n 2 X Gst of proof: 2 β ( X X ) u X X β = = w u where w : = ( X X ) ( X X ) 1 1 ch u 1 ndep. hom osked. u u 2 ( X X ) var( β ) = var( w u ) = w var( u ) = w σ = σ w = σ 5-25
26 Got two formulas for standard errors of Homoskedastcty-only standard errors these are vald f the errors are homoskedastc. Alleged advantage: Computatonal smplcty. (Wth today s computers, ths s moot.) Real dsadvantage: Unrelablty wthout homoskedastcty Formula n ch.4 s correct, whchever the skedastcty. That formula also known as heteroskedastcty robust standard error. 5-26
27 Usng software Homoskedastcty-only standard errors are the default settng n software sometmes the only settng (e.g. Excel). For heteroskedastcty-robust SE, do overrde the default. If you neglect to overrde the default and there s n fact heteroskedastcty, your standard errors (and t-statstcs and confdence ntervals) wll be wrong. (Typcally, homoskedastcty-only SEs are too small.) 5-27
28 Heteroskedastcty-robust standard errors n STATA regress testscr str, robust Regresson wth robust standard errors Number of obs = 420 F( 1, 418) = Prob > F = R-squared = Root MSE = Robust testscr Coef. Std. Err. t P> t [95% Conf. Interval] qqqqstr qq_cons Wth the, robust opton, STATA computes vald SE Wthout, computes unrelable homoskedastcty-only standard errors 5-28
29 Strengths of OLS estmator Learned of OLS estmator: unbased, consstent; got formulas for SE, confdence ntervals & test statstcs. All good reasons to use OLS. Another: Result: OLS s lnear n Y 1,, Y n (gven X 1,, X n ) β 1 = = ( X X ) ( X X ) ( X X ) ( X X )(Y Y ) ( X X ) Y = w Y where w : = X X ( X X ) Lnearty makes OLS easy to compute, even wth large data. 2 Y 0 = ( X X ) 5-29
30 Any estmators better than OLS? Seen OLS estmator s unbased & lnear n Y-data. Consder all unbased & lnear n Y-data estmators β*: E( β*) = β β* = wy for some weghts w = w ( X,..., X ) 1 n Mght one of these be better than OLS n that ts varance var(β*) s smaller than OLS s varance var(β hat )? That s, whose spread around true β s tghter than OLS s spread? To answer ths, need extra assumptons
31 Extended OLS Assumptons Add one assumpton to the 3 so far: 1. E(u X=x) = (X,Y ), =1,,n, are..d 3. Large outlers are rare (E(Y 4 ), E(X 4 ) are fnte) 4. u s homoskedastc Assumptons should be checked for any data set for whch results to follow are nvoked. 5-31
32 Effcency of OLS - part I Gauss-Markov theorem Suppose OLS assumptons 1-4. Then the OLS estmator has the smallest varance among all unbased Y-lnear estmators. For proof, see p.177 Its varance beng smallest, t s best. Summary: OLS s BLUE - Best Lnear Unbased Estmator 5-32
33 Jab at OLS BLUE Gauss-Markov theorem Suppose OLS assumptons 1-4. Then the OLS estmator has the smallest varance among all unbased Y-lnear estmators. Jab: BLUE says OLS s best n a small competton (lnear estmators). What f there s an unbased nonlnear estmator wth smaller varance? Lnearty s no excuse to hold on to OLS, when cheap, powerful computers can handle nonlnearty. More wllng now (vs. 30 years ago) to yeld lnearty for smaller varance. 5-33
34 OLS counter: Best n great competton Add yet another assumpton: 1. E(u X=x) = (X,Y ), =1,,n, are..d 3. Large outlers are rare (E(Y 4 ), E(X 4 ) are fnte) 4. u s homoskedastc 5. u s normal Lehmann-Scheffe result Suppose OLS assumptons 1-5. Then the OLS estmator has the smallest varance among all unbased estmators, lnear and nonlnear. Jab II: Assn 4,5 clash wth economc data (e.g. hourly earnngs). OLS unrelable wthout 3. Gvng up 1 allows lttlebased estmators wth even smaller varance. 5-34
35 What about small samples - nference? Our confdence ntervals, statstcal tests assumed large samples. What about such nferences wth small samples? Result Sps. OLS assumptons 1-5. Then for any sample sze n: - and are normally dstrbuted, -the t-statstc has a t-dstrbuton wth df = n 2 Gst of proof: β ( X X ) u β = = ( X X ) A weghted sum of normals (as the u are under assn 5) s normal. wu 5-35
36 In addton, under assumptons 1 5, under the null hypothess the t statstc has a Student t dstrbuton wth n 2 degrees of freedom Why n 2? Has to do wth 2 parameters, β 0 and β 1 For n < 30, t s crtcal values notceably larger than N(0,1) s For n > 50 or so, t n 2 & N(0,1) dstrbutons dffer neglgbly degrees of freedom 5% t-dstrbuton crtcal value
37 Rule of thumb If n < 50 and you beleve that, n your applcaton, u s homoskedastc and normally dstrbuted, then use the t n 2 nstead of the N(0,1) crtcal values for hypothess tests and confdence ntervals. In most econometrc applcatons, there s no reason to beleve that u s homoskedastc and normal usually, there are good reasons to beleve nether. Modern data sets usually have n > 50, so OK to rely on the large-n results for hypothess tests and confdence ntervals. 5-37
38 Summary and Assessment The ntal polcy queston: Suppose new teachers are hred so the student-teacher rato falls by one student per class. Is the expected effect of ths polcy nterventon ( treatment ) on test scores really +2.28? Not really dstrcts wth low STR tend to enjoy other resources and hgher ncomes for better learnng opportuntes outsde school suggestng cov(u, STR ) > 0, volatng assn1. We have omtted varables (such as dstrct s ncomes) from our analyss, basng the estmates... Ch 6 tres to fx ths. 5-38
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