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vn[C(k 2 ) - c(k,)l/4 where kl and k2 are the closest integers to (n + 1)/ 2 vn. Our argument consists of deriving two approximate 95%-confidence intervals for the median of the error distribution, both based on normal approximations, and equating their lengths. Recall from the preceding subsection that T in LAD regression is analogous to (T in least-squares regression. The standard deviation of an LS regression estimate has the form (TC, whereas the standard deviation of the corresponding LAD regression estimate is approximately TC for the same quantity c. In particular, consider the special case of "regression" when p = 0, that is, when there are no explanatory variables and we simply have a random sample of n independent observations from the same population, such as e 1, e 2 , , ell. The LS "regression" estimate is the sample mean e, because a = e is the value of a that gives the least sum of squared deviations Dei - a)2. Its standard deviation is (T / vn. The LAD "regression" estimate is the sample median e, because a = e is the value of a that gives the least sum of absolute deviations Lie, - al. Its standard deviation is approximately T/vn.
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The central limit theorem says that, for large sample sizes n, the distribution of the sample mean is approximately normal. It turns out that for large n the distribution of the sample median is also approximately normal. Also, the expectation of the sample median is approximately equal to the population median. Letting II denote the median of the population of errors, we can state that, for large n, e has approximately a normal distribution with mean II and standard deviation T / vn. Hence, if 7 is an estimate of T, we can construct an approximate 95%-confidence interval for II to be e 27/ vn. We are not actually interested in II and are not interested in the confidence interval itself but only in its length, 47/ vn. Next, it can be shown that the interval from e(k,) to e(k 2 ) is also an approximate 95%-confidence interval for II. It turns out that the two intervals are similar, at least for large sample sizes. In particular, their lengths are similar: 47/ vn ::::: e(k 2) - e(k,). Choosing 7 to make this an exact equality, replacing e by C, and replacing n by m = n - 2, we obtain formula (4.5). It has been found that using only the m nonzero residuals improves the performance of the test for small samples. The LAD test procedure assumes that, when t has a t distribution with n - 2 degrees of freedom. This is not exactly true but theorems have been proved which imply that when n is large and f3 = 0, then t has approximately a standard normal distribution. This justifies the LAD test procedure when n is large, because a t distribution with many degrees of freedom is very similar to the standard normal distribution. When n is small, to be on the safe side so that we do not The Distribution of t.
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The Birth Rate Data. With a sample size as small as n = 14, we should be cautious about the validity of the LAD test described in this section, but as an illustration, let us apply the test to the birth rate data. In Table 4.2 the ordered nonzero residuals are listed. Since m = n - 2 = 12 and (m + 1)/2 - ..;m = 13/2 - Ifi = 3.04, we have k, = 3. Similarly, k2 = 10. So T = lfi[e(lO) - e(3)l/4. In Table 4.2 we see e(3) = -1.203 and e(lO) = 7.733. Hence T = 7.739. Next we calculate L(x i - j=)Z = 3151. So est.SD(~) = 7.739/ V3151 = 0.1379, and It I = 1-0.53781/0.1379 = 3.900. To calculate the p-value we use the t distribution with n - 2 = 12 degrees of freedom. From the t table in the Appendix, we find that the p-value is between 0.001 and 0.01. We conclude that the slope of the regression line is truly negative. As a check on the LAD test, let us perform the LS test. In Figure 1.1 note the position of the Trinidad/Tobago data point. We saw in Section 1.6 that this point has a disproportionate influence on the slope of the LS regression line. Note that its influence is in the direction of pulling the slope toward O. So in the presence of such influence, if the LS test concludes f3 0, we can believe it. When the LS test is applied to these data, it yields a p-value of 0.018. This stilI indicates f3 0, but not as strongly as the LAD test does.
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