Question 25 - 29 needed answers. 25 )What command would you run to get me the ou

Question

Question 25 - 29 needed answers. 25 )What command would you run to get me the output below, assuming you had all the data loaded already? (I need the complete command, all syntax and variables in the correct order). Source I df MS Number of obs 51 F(5, 45) 46.82 Model 3040.27195 Residual 1 584.41612 5 608.05439 Prob >F 45 12.9870249 R-squared 0.8388 Adj R-squared0.8209 -3.6038 Total I 3624.68807 50 72.4937614 Root MSE MSE ared 0.00 82 dvote l Coef. Std. Err. [95% Conf. Inte rval] dem 1 10.08157 1.081976 9.32 0.000 7.902359 12.26078 .00076780016433 fedfunds 0012055 0002173 5.55 0.000 0.70 0.485 .0005683 .0011793 defense1 .5704592 2.085223-0.27 0.786 4.770313 3.629395 2.608018 7.160717 cons126.27752 2.067772 12.71 0.000 22.11281 30.44223 crime 0003055 0004338 south I 4.884367 1.130205 4.32 0.000 26) According to the table above, what is the consequence to dvote of adding another unit of south? 27) What is the value of 28) of the variables above, which are listed as significant at the 5% level? 29a) You have a of 6 n estimate for 1 of 145, Your rival suspects A-0. Having found a standard deviation , can you persuade him of the significance of your results at a 5% level? 8 for

Explanation / Answer

Q.26 According to the table above, if we add another unit of south then there would be 4.88 unit increase in dvote.

Q.27 Value of o = 10.08157

Q.28 Of all the variables, the variables which have p - value less than 0.05 significance level shall be significant.

WHich are

(a) fedfunds

(b) South

(c) _cons

Q.29(a) Here 1^ = 145

1 = 0

standard deviation for 1 = 68

Test statistic t = 145/68 = 2.1324

at 5% confidence level tcr = 2.014

so here t > tcr so we shall reject the null hypothesis and can say that estiamtor for 1 is significant in nature.

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