MIT14_30s09_lec23

# MIT14_30s09_lec23 - MIT OpenCourseWare http/ocw.mit.edu...

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 14.30 Introduction to Statistical Methods in Economics Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . 1 14.30 Introduction to Statistical Methods in Economics Lecture Notes 23 Konrad Menzel May 12, 2009 Examples Example 1 Assume that babies’ weights (in pounds) at birth are distributed according to X ∼ N (7 , 1) . Now suppose that if an obstetrician gave expecting mothers poor advice on diet, this would cause babies to be on average 1 pound lighter (but have same variance). For a sample of 10 live births, we observe ¯ X 10 = 6 . 2 . • How do we construct a 5% test of the null that the obstetrician is not giving bad advice against the alternative that he is? We have H : µ = 7 against H A : µ = 6 We showed that for the normal distribution, it is optimal to base this simple test only on the sample ¯ ¯ ¯ mean, X 10 , so that T ( x ) = x ¯ 10 . Under H , X 10 ∼ N (7 , . 1) and under H A , X 10 ∼ N (6 , . 1) . The test rejects H if X ¯ 10 < k . We therefore have to pick k in a way that makes sure that the test has size 5%, i.e. . 05 = P ( X ¯ 10 < k | µ = 7) = Φ k √ − . 1 7 where Φ( ) is the standard normal c.d.f.. Therefore, we can obtain k by inverting this equation · 1 . 645 k = 7 + √ . 01Φ − 1 (0 . 05) ≈ 7 − √ 10 ≈ 6 . 48 Therefore, we reject, since X ¯ 10 = 6 . 2 < 6 . 48 = k . • What is the power of this test? ¯ P ( X 10 < 6 . 48 µ = 6) = Φ 6 . 48 − 6 ≈ Φ(1 . 518) ≈ 93 . 55% | √ . 1 • Suppose we wanted a test with power of at least 99%, what would be the minimum number n of newborn babies we’d have to observe? The only thing that changes with n is the variance of the sample mean, so from the first part of this example, the critical value is k n = 1 . 645 7 − √ n , whereas the ¯ power of a test based on X n and critical value k n is ¯ 1 − β = P ( X n < k n | µ = 6) = Φ √ n − 1 . 645 1 Setting 1 − β ≥ . 99 , we get the condition √ n − 1 . 645 ≥ Φ − 1 (0 . 99) = 2 . 326 n ≥ 3 . 971 2 ≈ 15 . 77 ⇔ This type of power calculations is frequently done when planning a statistical experiment or survey - e.g. in order to determine how many patients to include in a drug test in order to be able to detect an effect of a certain size. Often it is very costly to treat or survey a large number of individuals, so we’d like to know beforehand how large the experiment should be so that we will be able to detect any meaningful change with suﬃciently high probability. Example 2 Suppose we are still in the same setting as in the previous example, but didn’t know the variance. Instead, we have an estimate S 2 = 1 . 5 . How would you perform a test? As we argued earlier, the statistic ¯ X n − µ T := S/ √ n ∼ t n − 1 is student-t distributed with n − 1 degrees of freedom if the true mean is in fact µ . Therefore we reject H if ¯ X n − 7 T = S/ √ 10 < t 9 (5%)...
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MIT14_30s09_lec23 - MIT OpenCourseWare http/ocw.mit.edu...

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