Couple of definitions. Hypothesis tests for one parameter. Some Prob/Stat Review. Likewise. Standard normal distribution. Normal distribution

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1 Couple of defiitios Hypothesis tests for oe parameter Stadard ormal distributio z i ~ N(,1) Normal distributio y i ~ N(μ, σ ) Normal ca always be tured ito a stadard ormal by subtractig mea ad dividig by stadard deviatio z i = (y i - μ)/σ ~ N(,1) 1 Some Prob/Stat Review y i is a ormal radom variable y i ~ N(,σ ) Suppose there are idepedet y i s W 1 yi i1 The W 1 ~ N(,σ ) (Last questio o Problem set #1) Likewise y i ~ N(,σ ) Suppose there are idepedet y i s W biyi i1 Where b i is a costat W N, bi i

2 Some Prob/Stat Review z i is a stadard ormal radom variable z i ~ N(,1) Suppose there are idepedet z i s W 3 z i1 i The W 3 is a Chi-squared radom variable with degrees of freedom 5 [I this graph, K is the degrees of freedom] 6 Suppose W k is a chi-squared distributio with k degrees of freedom W is a chi-squared distributio with degrees of freedom The S=(W k /K)/(W /) is a F distributio with (k,) degrees of freedom Defied over all S> 7 Suppose z is stadard ormal z~n[,1] Suppose W is a chi-squared with degrees of freedom z t W is distributed as a studet t with DOF 8

3 Studet t William Sealy Gosset ( ) Statisticia Employee of Guiess Used statistical models to isolate the highest yieldig varieties of barley Homer is correct Symmetric, ui-modal PDF Defied over all real umbers Shape is a fuctio of the degrees of freedom Sigmoid CDF E[t]= V[t]>1 but approaches 1 as DOF approach PDF shape very similar to stadard ormal with fatter tails Stadard Normal Stadard Normal pdf pdf Studet t with degrees of freedom Studet t with 5 degrees of freedom x x 1 3

4 .4.4 Stadard Normal Stadard Normal pdf Studet t with 1 degrees of freedom pdf Studet t with degrees of freedom x x Plot: Normal ad t() CDF Stadard Normal 1 Normal.8 pdf Studet t with 1 degrees of freedom.6.4. t() x x 16 4

5 Plot: Normal ad t(1) CDF Plot: Normal ad t(1) CDF 1 1 Normal Normal t(1). t(1) x x 18 Normality of ε y i = β + x 1i β 1 + x i β +. x ki β k + ε i There are observatios k+1 parameters to be estimated -k-1 degrees of freedom Assume ε i is ormally distributed. What does that assumptio buy us? 19 ˆ y x i i 1 i ( x x)( y y) ( x x) i i i i i1 i1 1 1 ( xi x) ( xi x) i1 i1 1 i1 w where w i i i i1 ( x x) i ( x x) i 5

6 Note that 1 1 i1 ˆ w ˆ 1 ˆ 1 is a liear estimator, that is i i Note is a liear fuctio of the ε i s A liear fuctio of ormal variables is also ormally distributed Sice the ε i s are assumed to be ormal. 1 the ˆ is ormally distributed 1 ˆ Normal[, V ( )] E[ ˆ ] 1 1 V ( ˆ ) 1 i1 ( x x) i Geeral case : y x x... x the ˆ is ormally distributed ˆ Normal[, V ( )] E[ ˆ ] ( ˆ V ) SST i 1i 1 i ki k i (1 R ) ( i ) i1 where SST x x ad R the R from the regressio of x o all the other x ' s 3 because ˆ is ormally distributed, we could use the std. ormal distributio for test of hypotheses IF WE KNEW ˆ ˆ V ( ˆ ) SST (1 R ) ~ N(,1) 4 6

7 Problem? σ ε is ukow ad must be estimated Ubiased estimate is ˆ i1 ˆ i k 1 ˆ y ˆ x ˆ x ˆ... x ˆ i i 1i 1 i ki k each ˆ is ormally distributed therefore, ˆ a liear combiatio of i ormally distributed var iables 5 6 ˆ i1 ˆ i k 1 The umerator i the estimate looks somethig like a chi squared But because ˆ N. i ~ (, ) ( it does ot have a var. of 1) it is ot exactly i the correct form. Because the observatios are already used to get k+1 parameters, there are oly -k-1 uique estimated errors Therefore, the degrees of freedom of the chisquared distributio are -k-1 Techically k ˆ k,( 1) / ~ ( 1) 7 8 7

8 Recall that ˆ ˆ V ( ˆ ) SST (1 R ) ~ N(,1) The theoretical var iace for ˆ ( ˆ V ) SST (1 R) ( k1) ˆ ad k ad t z W ~ t( ) ~ ( 1) 9 The estimated var iace is the ˆ ˆ( ˆ ) (1 SST ) R V 3 ˆ V ( ˆ ) N(,1) ~ t ( k1) ˆ ( k 1) ( k 1) k1 k1 Degrees of freedom Stadard ormal Chi-squared 31 ˆ ˆ SST ˆ (1 R ) SST(1 R ) ( k1) ˆ ( k1) ˆ ˆ ( k 1) SST(1 R) k1 ˆ ˆ ~ t ( k1) ˆ SST ˆ Est. Var( ) (1 R) 3 8

9 Istead of workig with ˆ ~ N(,1) SST (1 R ) we use ˆ ˆ SST (1 R ) ~ t ( k1) se( ˆ ) s tadard error of se( ˆ ) ˆ SST (1 R ) ˆ a ~ t ( k1) se( ˆ ) ˆ Testig Hypotheses about a Sigle Parameter: tailed tests Basic model y i = β + x 1i β 1 + x i β +. x ki β k + ε i Ecoomic theory suggests a particular value of the parameter Two-tailed test These are called two tail tests because falsificatio of the ull hypothesis ca be due to either large + or values (i absolute value) Therefore, we use both tails of the uderlyig t-distributio H : β =a H a : β a

10 ˆ ˆ a ~ t ( k1) se( ˆ ) The distributio for Give the hypothesis is true, we ca replace a for β ˆ a tˆ ~ t( k1) se( ˆ ) If the hypothesis is true, the costructed test statistic should be cetered o zero. How far from zero does it have to be to reect the ull? Need to set the cofidece level of the test. Usually 95% Let 1-cofidece level = α With 95% cofidece level, α =.5 α is the probability you reect the ull whe it is i fact true retur to this later t α/ (dof) be the cut-off from a t-distributio with dof degrees of freedom where oly α/ percet of the distributio lies above Give symmetry, α/ percet lies below -t α/ (dof)

11 .45 ˆt If we were to draw at radom, 95% of the time it would be betwee (-t α/, t α/ ).5%.5% 95% ˆt Therefore, if the hypothesized value of β is true, there is a 95% chace ˆt will be betwee (-t α/, t α/ ) -t α/ t α/ 41 4 Decisio Rule ˆ a tˆ ~ t( k1) se( ˆ ) if tˆ t ( k 1) reect ull / if tˆ t ( k 1) caot reect ull / Most basic test y i = β + x 1i β 1 + x i β +. x ki β k + ε i H : β = H a : β Is the parameter estimate statistically distiguishable from zero?

12 Baseball example Regress attedace o wis do wiig teams attract more fas Data o 3 teams, parameters, DOF=-=8 Look at table i the back of the book for critical value of t Vertical axis is DOF Horizotal axis is the value of α 45 Sigificace Level oe-tailed tests two-tailed tests Degrees of freedom P-value for two tailed tests 46 Sigificace Level oe-tailed tests two-tailed tests Degrees of freedom * ru simple regressio. reg attedace wis Source SS df MS Number of obs = F( 1, 8) = 8.89 Model Prob > F =.59 Residual 1.911e R-squared = Ad R-squared =.139 Total.5178e Root MSE = attedace Coef. Std. Err. t P> t [95% Cof. Iterval] wis _cos ˆ a ˆ 31.5 t.98 se( ˆ ) tˆ.98 t ( k1).48 reect ull / 48 1

13 Statistical sigificace Whe we reect the ull hypothesis that H : β =, we say that a variable is statistically sigificat Which is short had for sayig the variable is statistically distiguishable from Statistically isigificat variables are those that we caot reect the ull H : β = College GPA example Data o 141 studets cotiuous variables: HS GPA ACT Score Oe itercept DOF = -k-1 = = 138 O the table, there is o 138, ust fid the closest oe 49 5 Sigificace Level oe-tailed tests two-tailed tests Degrees of Degrees of Freedom ifiity * ru multivariate regressio. reg college_gpa act hs_gpa Source SS df MS Number of obs = F(, 138) = Model Prob > F =. Residual R-squared = Ad R-squared =.1645 Total Root MSE =.343 college_gpa Coef. Std. Err. t P> t [95% Cof. Iterval] act hs_gpa _cos tˆ act.87 tˆ act 1.98 caot reect ull tˆ hsgpa 4.73 tˆ 1.98 reect ull hsgpa 5 13

14 Cofidece itervals Cofidece itervals The CI represet the 95% most likely values of the parameter β If the hypothesized value a (H : β =a) is ot part of the cofidece iterval, it is ot a likely value ad we reect the ull If iterval cotais a we caot reect ull The t-test ad CI should provide the same decisio if ot, you did somethig wrog 53 if the ull is true, the ˆ a t ( k1) t ( k1) / / se( ˆ ) which meas that ˆ se( ˆ ) t ( k 1) a / ˆ se( ˆ ) t ( k 1) / 54 Cofidece iterval. * ru simple regressio. reg attedace wis Source SS df MS Number of obs = F( 1, 8) = 8.89 Model Prob > F =.59 Residual 1.911e R-squared = Ad R-squared =.139 Total.5178e Root MSE = ˆ se( ˆ ) t ( k 1) / 55 attedace Coef. Std. Err. t P> t [95% Cof. Iterval] wis _cos cofidece it. ˆ ˆ t /( k1) se( ) (13.98) [97.4,53.5] reect ull 56 14

15 Sigificace Level oe-tailed tests two-tailed tests Degrees of freedom * ru multivariate regressio. reg college_gpa act hs_gpa Source SS df MS Number of obs = F(, 138) = Model Prob > F =. Residual R-squared = Ad R-squared =.1645 Total Root MSE =.343 college_gpa Coef. Std. Err. t P> t [95% Cof. Iterval] act hs_gpa _cos cofidece it. ˆ ˆ t /( k1) se( ) (.178) caot reect ull [.119,.37] 58 Regress l(q) o l(p) Problem Set 3 Test whether the cigarette demad elasticity is a elastic respose, that is ζ d =-1 years worth of data, 51 states = 1 DOF = - = Sigificace Level oe-tailed tests two-tailed tests Degrees of freedom ifiity

16 Just lookig at cofidece iterval, we ca reect the ull Source SS df MS Number of obs = F( 1, 118) = Model Prob > F =. Residual R-squared = Ad R-squared =.461 Total Root MSE =.316 l_q Coef. Std. Err. t P> t [95% Cof. Iterval] l_p _cos l(q i ) = β + l(p i )β 1 + ε i H o : β 1 = -1 ˆ a tˆ 7.3 se( ˆ ) t /( dof) t.5(118) 1.96 tˆ 1.96 reect ull 61 P-value Alterative way of characterizig the data cotaied i the t-test Give that the ull is true, the p-value is probability of obtaiig a result at least as extreme as the oe that was actually observed Calculate ˆt I the two tailed test, the p-value is the p value Pr( t tˆ ) Pr( t tˆ ) 6. * ru simple regressio. reg attedace wis Source SS df MS Number of obs = F( 1, 8) = 8.89 Model Prob > F =.59 Residual 1.911e R-squared = Ad R-squared =.139 Total.5178e Root MSE = P-value=(.3)=.6 attedace Coef. Std. Err. t P> t [95% Cof. Iterval] wis _cos %.3% ˆt 16

17 . * ru multivariate regressio. reg college_gpa act hs_gpa Source SS df MS Number of obs = F(, 138) = Model Prob > F =. Residual R-squared = Ad R-squared =.1645 Total Root MSE =.343 college_gpa Coef. Std. Err. t P> t [95% Cof. Iterval] act hs_gpa _cos P-value=(.1915)= % ˆt Note A small p-value gives you cofidece that you ca reect the ull hypothesis you would ot get a value this large (i absolute value) at radom, therefore, the H o must be false Usig p-value, t-test, cofidece iterval are three ways to get the same results The decisio rule (reect or ot reect) should ot vary across test methods Good check o your work

18 Errors i Predictio Statistical tests ca be used as tests of theoretical hypothesis Do demad curves slope dow? Do wages icrease w/more educatio? These are oly statistical tests they ask, i a probabilistic sese, what is the likely state of the world Cosider H o : β = Suppose the t-test is small, so you caot reect the ull hypothesis. There is always a chace that you are wrog. Example 1: New Drug Reduces deaths from stroke by 1%. However a cliical trial caot reect the ull hypothesis that there is o effect t-statistic o the active igrediet is 1.1 Caot reect ull that β = 69 7 Two possible situatios Drug does ot work ad your test is correct Type I ad II Errors Decisio Drug does work, but the statistical model did ot have eough power to detect a statistically sigificat impact 71 True State H o true H o false Caot reecth o Correct decisio Type II error Accept false hypothesis Reect H o Type I error Reect true hypothesis Correct decisio 7 18

19 Type I false positive Type II false egative I regressio, H : β = Type I you reect that β = whe it equals Type II you caot reect β = whe β What is the probability you will make a wrog decisio Type I error reect ull whe it is i fact true H o : β = Get large t-statistic so reect ull There is a chace that, by accidet, you will get a large t-stat What is that chace? 1 - cofidece level = α so α is the probabilty Type II errors: Do ot reect ull whe it is i fact false H o : β = Get small t-statistic so do ot reect ull What is the probability this will happe? The type II error rate (false egative) labeled β 1-β called the power of the test Factors that icrease power Icrease sample size Icrease variatio i X s Depedig o the problem, eed to balace the probabilities of Type I ad II errors If cocered about Type I errors, so you icrease the size of the cofidece iterval Icrease the chace of Type II error

20 Example: Crimial Court Example: Mammography Cosider crimial court: H o : ot guilty Job of ury decide guilty or ot guilty Type I error reect true hypothesis covict a iocet ma Type II error accept a false hypothesis let guilty ma go free Decisio rule: guilt beyod a reasoable doubt Requires low p-value, high cofidece level (99.99% cofidece iterval) to covict miiimze Type I 77 Low level radiatio exam to detect breast exam H o : o breast cacer Type I error False positive fid a cacer growth but it does ot exist Type II error False egative fails to detect a growth What do you miimize? 78 Cosider the doctor s liability Suppose a Type II error happes failed to fid tumor -- patiet dies gets sued for malpractice Suppose a Type I error detect tumor, perform surgery whe oe was eeded For the doctor, what type of error has more dowside risk? Chagig cofidece level 95% CI is idustry stadard Oly 5% error rate But, maybe wat to decrease Type I error rate Decrease false positives Icrease cofidece level to 99% Maybe you really eed to be sure somethig causes cacer before you ba the substace 79 8

21 I STATA I cotrast, you might wat to decrease chace of Type II error Reduce the size of the cofidece iterval Maybe do ot require as defiitive evidece before you let o the market a ew drug to treat i ucurable disease reg y x1 x x3, level(#) The # is a umber from 1 to the top umber has a low Type I (.1%) very high Type II error rare 81 8 Test score data from CA 4 schools 6 graders give math/readig exams Outcome is average score o both exams Four covariates Studet/teacher ratio Average family icome (i thousads of $) % ESL % o free ad reduced luchs. desc average_score studet_teacher avg_icome esl_pct meal_pct storage display value variable ame type format label variable label - average_score float %9.g average score (math+read) std test studet_teacher float %9.g studet/teacher ratio avg_icome float %9.g average family icome esl_pct float %9.g pct studet with eglish secod laguage meal_pct float %9.g % kids o free/reduced prices meals. sum average_score studet_teacher avg_icome esl_pct meal_pct Variable Obs Mea Std. Dev. Mi Max average_sc~e studet_te~r avg_icome esl_pct meal_pct

22 . * ru mv regressio wth 4 x's. reg average_score studet_teacher avg_icome esl_pct meal_pct Source SS df MS Number of obs = F( 4, 415) = 49.1 Model Prob > F =. Residual R-squared = Ad R-squared =.834 Total Root MSE = average_sc~e Coef. Std. Err. t P> t [95% Cof. Iterval] studet_te~r avg_icome esl_pct meal_pct _cos Notice the t-statistic o Studet/teacher ratio is.45 At 95% level, ca reect ull β =. Notice the P-value is.15, so a 99% CI would ot reect ull 85 Sigificace Level oe-tailed tests two-tailed tests Degrees of freedom ifiity To chage CI, use this optio. * ru the same regressio but ask for a 99% cofidece level. reg average_score studet_teacher avg_icome esl_pct meal_pct, level(99) Source SS df MS Number of obs = F( 4, 415) = 49.1 Model Prob > F =. Residual R-squared = Ad R-squared =.834 Total Root MSE = average_sc~e Coef. Std. Err. t P> t [99% Cof. Iterval] studet_te~r avg_icome esl_pct meal_pct _cos cofidece it. ˆ ˆ t /( k1) se( ) (.86) [ 1.15,.31] 87