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Chapter 3

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Department
Economics
Course
ECON 3210
Professor
Razvan Sufana
Semester
Fall

Description
CHAPTER 3 Exercise Solutions 54 Chapter 3, Exercise Solutions, Principles of Econometrics, 4e  55 EXERCISE 3.1 (a) The required interval estimator is b1 1 ce( ) . When b183.416, tc  (0.975,38)2.024 and se(b1)  43.410, we get the interval estimate: 83.416  2.024  43.410 = (4.46, 171.30) We estimate that 1 lies between 4.46 and 171.30. In repeated samples, 95% of similarly constructed intervals would contain the true  1. (b) test H 0 10 against H 1 1  we compute the t-value b1 1 83.416 0 t1    1.92 se(b1) 43.410 Stiece t = 1.92 value does not exceed the 5% critical value tt  c (0.975,38)2.024 , we do not reject H 0. The data does 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 p/2  0.0622 2 0.0311. The level of significance, , is given by the sum of the areas under the PDF for ||t|t,c so the area under the curve for tt is /2 .025 and likewise for tt  . Therefore not rejecting the null hypothesis c c because /2 p/2, or  p, is the same as not rejecting the null hypothesis because tc c . From Figure xr3.1(c) we can see that having a p-value > 0.05 is equivalent to having ttt . c c Figure xr3.1(c) Critical and observed t values   Chapter 3, Exercise Solutions, Principles of Econometrics, 4e  56 Exercise 3.1 (continued) (d) Testing H 0 10 against H,1 10  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, ct  (0.95,38).686 . Since t = 1.92 > tc = 1.69, H 0 is rejected, the alternative is accepted, and we conc lude that the intercept is positive. In this case p-value = P(t > 1.92) = 0.0311. We see from Figure xr3.1(d) that having the p-value < 0.05 is equivalent to having t > 1.69. 0.4 Rejection Region 0.3 PDF 0.2 0.1 1.92 0.0 -3 -2 -1 0 1 t 2 3 c T Figure xr3.1(d) Rejection region and observed t value (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 yi elding an interval that includes the true parameter. When carrying out a two-tailed test of the form H c:  versus H c:,  0 k 1 k non-rejection of H implies c lies within the confidence interval, and vice versa, 0 providing the level of significance is equal to one minus the level of confidence. (f) False. The test in (d) uses the level of significance 5%, which is the probability of a Type I error. That is, in repeated samples we ha ve a 5% chance of rejecting the null hypothesis when it is true. The 5% significance is a probability statement about a procedure not a probability statement about  . It is careless and dangerous to equate 5% level of 1 significance with 95% confidence, which relate s to interval estimation procedures, not hypothesis tests. Chapter 3, Exercise Solutions, Principles of Econometrics, 4e  57 EXERCISE 3.2 (a) The coefficient of EXPER indicates that, on average, a technical artist's quality rating goes up by 0.076 for every additional year of experience. 3.9 3.8 3.7 3.6 3.5 RA3.4G 3.3 3.2 3.1 0 1 2 3 4 5 6 7 8 EXPER Figure xr3.2(a) Estimated regression function (b) Using the value tt   2.074 , the 95% confidence interval for  is given by c (0.975,2
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