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CHAPTER3 Exercise Solutions 31 Chapter 3 Exercise Solutions Principles of Econometrics 3e32 EXERCISE 31 a The required interval estimator is sebtb When 83416b2024tt11c1097538cand se43410b we get the interval estimate 1202443410446 17130 83416 We estimate thatlies between 446 and 17130 In repeated samples 95 of similarly 1 constructed intervals would contain the true 1 0H against 0H we compute the tvalue b Totest 0111 b83416011 192t1se43410b1 2024tt we doSincethe t192 value does not exceed the 5 critical value 097538cnot reject The data do not reject the zerointercept hypothesis H0 c The pvalue 00622 represents the sum of the areas under the t distribution to the left of t192 and to the right of t192Since the t distribution is symmetric each of the tail areas that make up the pvalue areThe level of significance200622200311ptt so the area under the curve for is given by the sum of the areas under the PDF for c isand likewise forTherefore not rejecting the null hypothesis tt2025ttcc or is the same as not rejecting the null hypothesis because because 22pp From Figure xr31a we can see that having a pvalue005 is equivalent to tttccttt having ccFigure xr31a Critical and observed t values for Exercise 31c
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