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Econometrics-I-8

P this is the probability of a type i error under the

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p This is the probability of a Type I error. Under the null hypothesis, F(3,100) has an F distribution with (3,100) degrees of freedom. Even if the null is true, F will be larger than the 5% critical value of 2.7 about 5% of the time. ™    17/50
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Part 8: Hypothesis Testing Testing Procedures How to determine if the statistic is 'large.' Need a 'null distribution.' If the hypothesis is true, then the statistic will have a certain distribution. This tells you how likely certain values are, and in particular, if the hypothesis is true, then 'large values' will be unlikely. If the observed statistic is too large, conclude that the assumed distribution must be incorrect and the hypothesis should be rejected. For the linear regression model with normally distributed disturbances, the distribution of the relevant statistic is F with J and n-K degrees of freedom. ™    18/50
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Part 8: Hypothesis Testing Distribution Under the Null Density of F[3,100] X .250 .500 .750 .000 1 2 3 4 0 FDENSITY ™    19/50
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Part 8: Hypothesis Testing A Simulation Experiment sample ; 1 - 100 $ matrix ; fvalues=init(1000,1,0)$ proc$ create ; fakelogc = rnn(-.319557,1.54236)$ Coefficients all = 0 regress ; quietly ; lhs = fakelogc Compute regression ; rhs=one,logq,logpl_pf,logpk_pf$ calc ; fstat = (rsqrd/3)/((1-rsqrd)/(n-4))$ Compute F matrix ; fvalues(i)=fstat$ Save 1000 Fs endproc execute ; i= 1,1000 $ 1000 replications histogram ; rhs = fvalues ; title=F Statistic for H0:b2=b3=b4=0$ ™    20/50
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Part 8: Hypothesis Testing Simulation Results 48 outcomes to the right of 2.7 in this run of the experiment. ™    21/50
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Part 8: Hypothesis Testing Testing Fundamentals - II p POWER of a test = the probability that it will correctly reject a “false null” hypothesis p This is 1 – the probability of a Type II error. p The power of a test depends on the specific alternative. ™    22/50
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Part 8: Hypothesis Testing Power of a Test Null: Mean = 0. Reject if observed mean < -1.96 or > +1.96. Prob(Reject null|mean=0) = 0.05 Prob(Reject null|mean=.5)=0.07902 Prob(Reject null|mean=1)=0.170066. Increases as the (alternative) mean rises. B E TA .084 .168 .251 .335 .419 .000 -3 -2 -1 0 1 2 3 4 5 -4 N 2 N 0 N 1 Variable ™    23/50
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Part 8: Hypothesis Testing Test Statistic For the fit measures, use a normalized measure of the loss of fit: ( 29 ( 29 ( 29 ( 29 r r r r r r r 0 since and 0 since - 2 2 u r 2 2 u r 2 u 2 2 u u u yy yy u u u u u u R -R / J F[J,n-K] = R R 1-R / (n-K) Often useful R =1- R =1- S S Insert these in F and it becomes / J F[J,n-K] = / (n-K) e e e e e e e e e e e e e e ™    24/50
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Part 8: Hypothesis Testing An important relationship between t and F 2 Chi Squared[J]/ J F Chi squared[n K]/ (n K) where the two chi-squared variables are independent. If J = 1, i.e., testing a single restriction, Chi Squared[1]/1 F Chi squared[n K]/ (n K) (N[0,1]) Chi s - = - - - - = - - - = - { } 2 2 quared[n K]/ (n K) N[0,1] = t[1] Chi squared[n K]/ (n K) - - = - - - For a single restriction, F[1,n-K] is the square of the t ratio.
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