Lecture 5

Lecture 5 - Regression with a Single Regressor: Hypothesis...

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Unformatted text preview: Regression with a Single Regressor: Hypothesis Tests and Confidence Intervals (SW Chapter 5) Object of interest: 1 in, Y i = + 1 X i + u i , i = 1,, n 1 = Y / X , for an autonomous change in X ( causal effect ) The Least Squares Assumptions: 1. E ( u | X = x ) = 0. 2. ( X i ,Y i ), i =1,, n , are i.i.d. 3. Large outliers are rare ( E ( X 4 ) < , E ( Y 4 ) < . General setup Null hypothesis and two-sided alternative: H : 1 = 1,0 vs. H 1 : 1 1,0 where 1,0 is the hypothesized value under the null. Null hypothesis and one-sided alternative: H : 1 = 1,0 vs. H 1 : 1 < 1,0 General approach : construct t-statistic, and compute p-value (or compare to N (0,1) critical value) In general: t = estimator - hypothesized value standard error of the estimator where the SE of the estimator is the square root of an estimator of the variance of the estimator. For testing the mean of Y : t = ,0 / Y Y Y s n - For testing 1 , t = 1 1,0 1 ( ) SE - , where SE ( 1 ) = the square root of an estimator of the variance of the sampling distribution of 1 Formula for SE( 1 ) Recall the expression for the variance of 1 (large n ): var( 1 ) = 2 2 var[( ) ] ( ) i x i X X u n - = 2 4 v X n , where v i = ( X i X ) u i . SE ( 1 ) = 1 2 = the standard error of 1 Summary: To test H : 1 = 1,0 v. H 1 : 1 1,0 , Construct the t-statistic t = 1 1,0 1 ( ) SE - = 1 1 1,0 2 - Reject at 5% significance level if |t| > 1.96 The p-value is p = Pr[| t | > | t act |] = probability in tails of normal outside | t act |; you reject at the 5% significance level if the p-value is < 5%. This procedure relies on the large- n approximation; typically n = 50 is large enough for the approximation to be excellent. Example: Test Scores and STR , California data Estimated regression line: test score = 698.9 2.28 STR Regression software reports the standard errors: SE ( ) = 10.4 SE ( 1 ) = 0.52 t-statistic testing 1,0 = 0 = 1 1,0 1 ( ) SE - = 2.28 0 0.52-- = 4.38 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 The p-value based on the large- n standard normal approximation to the t-statistic is 0.00001 (10 5 ) Because the t-statistic for 1 is N(0,1) in large samples, construction of a 95% confidence for 1 is just like the case of the sample mean: 95% confidence interval for 1 = { 1 1.96 SE ( 1 )} Confidence interval example : Test Scores and STR Estimated regression line: testscore= 698.9 2.28 STR SE ( ) = 10.4 SE ( 1 ) = 0.52 95% confidence interval for 1 : { 1 1.96 SE ( 1 )} = {2.28 1.96 0.52} = (3.30, 1.26) A concise (and conventional) way to report regressions:...
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Lecture 5 - Regression with a Single Regressor: Hypothesis...

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