slides_12_inferfinite

# ∙ answer we can show that if 0 – that is the null

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Unformatted text preview: ∙ Answer: We can show that, if 0 – that is, the null hypothesis holds – then T has a t n − 1 distribution. Notice that the degrees-of-freedom is the number of observations minus one. ∙ We can easily derive the t distribution using properties of the multivariate normal distribution along with the definition of chi-square and t random variables. 76 ∙ Write T n X ̄ / S / n X ̄ / n − 1 − 1 − 2 ∑ i 1 n X i − X ̄ 2 1/2 ∙ If e n 1,1,...,1 ′ and B n I n − e n e n ′ e n − 1 e n ′ then B n is symmetric and idempotent with rank n − 1. ∙ If X is the n 1 vector of observations, we know X ′ B n X ∑ i 1 n X i − X ̄ 2 . 77 ∙ Now X ′ B n X / 2 n − 1 2 and so S / n − 1 2 / n − 1 ∙ The numerator in T , n X ̄ / Normal 0,1 (under H ). Further, X ̄ and S are independent because B n e n . ∙ The result follows by the definition of a t random variable. 78 ∙ Not surprisingly, T is called a t statistic , in this case for testing the null H : 0 (against any alternative). ∙ For a different null, the t statistic is X ̄ − S / n (estimator − hypothesized value) standard error ∙ The t statistic has a t n − 1 distribution under H : , and we get our critical values from the t distribution. 79 ∙ For example, if the alternative if one-sided (in the positive direction) and df 18, the cv for a 5% size test is about 1.73. This is somewhat above the cv for the standard normal, reflecting the extra estimation uncertainty from not knowing . For the negative alternative, the cv is − 1.73 because the t distribution is symmetric about zero. ∙ The 5% cv for a two-tailed test is about 2.10. ∙ The 1% critical value for df 18 is about 2.55 (or − 2.55) for a one-tailed test. The cv for a two-tailed test is about 2.88. 80 ∙ Even better is if we allow a computer package such as Stata to compute a p-value. It automatically uses the appropriate t distribution. 81 EXAMPLE : Effects of job training grant on worker productivity. Look at change from 1987 to 1988. Only 19 firms were both given a grant and reported a productivity measure in both years. ∙ The response variable is actually the scrap rate – number of items out of 100 that must be scrapped – which falls if productivity increases. 82 . des Contains data from jtrain.dta obs: 157 vars: 4 19 Nov 2009 06:07 size: 2,669 (99.9% of memory free)------------------------------------------------------------------------------- storage display value variable name type format label variable label------------------------------------------------------------------------------- fcode float %9.0g firm code number grant byte %9.0g 1 if received grant clscrap float %9.0g change in log(scrap rate) cscrap float %9.0g change in scrap rate------------------------------------------------------------------------------- Sorted by: fcode 83 . list in 1/20 --------------------------------------- | fcode grant clscrap cscrap | |---------------------------------------| 1. | 410032 . . | 2. | 410440 . . | 3. | 4104953....
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∙ Answer We can show that if 0 – that is the null...

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