solutions chapter 3 - 31 CHAPTER 3 Exercise Solutions...

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Unformatted text preview: 31 CHAPTER 3 Exercise Solutions Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 32 EXERCISE 3.1 (a) The required interval estimator is 1 1 se( ) c b t b ± . When 1 83.416, b = (0.975,38) 2.024 c t t = = and 1 se( ) 43.410, b = we get the interval estimate: 83.416 ± 2.024 × 43.410 = ( − 4.46, 171.30) We estimate that 1 β lies between − 4.46 and 171.30. In repeated samples, 95% of similarly constructed intervals would contain the true 1 β . (b) To test 1 : H β = against 1 1 : H β ≠ we compute the t-value 1 1 1 1 83.416 0 1.92 se( ) 43.410 b t b −β − = = = Since the t = 1.92 value does not exceed the 5% critical value (0.975,38) 2.024 c t t = = , we do not reject H . The data do not reject the zero-intercept hypothesis. (c) The p-value 0.0622 represents the sum of the areas under the t distribution to the left of t = − 1.92 and to the right of t = 1.92. Since the t distribution is symmetric, each of the tail areas that make up the p-value are / 2 0.0622 2 0.0311. p = = The level of significance, , α is given by the sum of the areas under the PDF for | | | |, c t t > so the area under the curve for c t t > is / 2 .025 α = and likewise for c t t < − . Therefore not rejecting the null hypothesis because / 2 / 2, p α < or , p α < is the same as not rejecting the null hypothesis because . c c t t t − < < From Figure xr3.1(a) we can see that having a p-value > 0.05 is equivalent to having . c c t t t − < < Figure xr3.1(a) Critical and observed t values for Exercise 3.1(c) Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 33 Exercise 3.1 (continued) (d) Testing 1 : H β = against 1 1 : , H β > uses the same t-value as in part (b), t = 1.92. Because it is a one-tailed test, the critical value is chosen such that there is a probability of 0.05 in the right tail. That is, (0.95,38) 1.686 c t t = = . Since t = 1.92 > c t = 1.69, H is rejected, the alternative is accepted, and we conclude that the intercept is positive. In this case p-value = P ( t > 1.92) = 0.0311. We see from Figure xr3.1(b) that having the p-value < 0.05 is equivalent to having t > 1.69. 0.0 0.1 0.2 0.3 0.4-3-2-1 1 2 3 T PDF 1.92 c t Rejection Region Figure xr3.1(b) Rejection region and observed t value for Exercise 3.1(d) (e) The term "level of significance" is used to describe the probability of rejecting a true null hypothesis when carrying out a hypothesis test. The term "level of confidence" refers to the probability of an interval estimator yielding an interval that includes the true parameter. When carrying out a two-tailed test of the form : k H c β = versus 1 : , k H c β ≠ non-rejection of H implies c lies within the confidence interval, and vice versa, providing the level of significance is equal to one minus the level of confidence....
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This note was uploaded on 11/23/2011 for the course ECON MPBE taught by Professor Jansveceny during the Spring '11 term at Metropolitní Univerzita Praha.

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solutions chapter 3 - 31 CHAPTER 3 Exercise Solutions...

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