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

Info iconThis preview shows pages 76–85. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

Page76 / 100

∙ Answer We can show that if 0 – that is the null...

This preview shows document pages 76 - 85. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online