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SUGGESTED_SOLUTIONS_HW4

# SUGGESTED_SOLUTIONS_HW4 - SUGGESTED SOLUTIONS HW4 PAM 305...

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SUGGESTED SOLUTIONS HW4 PAM 305 FALL 2006 Question 1. 4.6 (i) With df = n – 2 = 86, we obtain the 5% critical value from Table G.2 with df = 90. Because each test is two-tailed, the critical value is 1.987. The t statistic for H 0 : 0 β = 0 is about -.89, which is much less than 1.987 in absolute value. Therefore, we fail to reject 0 β = 0. The t statistic for H 0 : 1 β = 1 is (.976 – 1)/.049 -.49, which is even less significant. (Remember, we reject H 0 in favor of H 1 in this case only if | t | > 1.987.) (ii) We use the SSR form of the F statistic. We are testing q = 2 restrictions and the df in the unrestricted model is 86. We are given SSR r = 209,448.99 and SSR ur = 165,644.51. Therefore, (209,448.99 165,644.51) 86 11.37, 165,644.51 2 F = which is a strong rejection of H 0 : from Table G.3c, the 1% critical value with 2 and 90 df is 4.85. (iii) We use the R -squared form of the F statistic. We are testing q = 3 restrictions and there are 88 – 5 = 83 df in the unrestricted model. The F statistic is [(.829 – .820)/(1 – .829)](83/3) 1.46. The 10% critical value (again using 90 denominator df in Table G.3a) is 2.15, so we fail to reject H 0 at even the 10% level. In fact, the p -value is about .23. (iv) If heteroskedasticity were present, Assumption MLR.5 would be violated, and the F statistic would not have an F distribution under the null hypothesis. Therefore, comparing the F statistic against the usual critical values, or obtaining the p -value from the F distribution, would not be especially meaningful.

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