The percent of single men have no income and the percent of men with no income are single men.

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Chapter 6, Problem 6.21E

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To Find: The percent of single men have no income and the percent of men with no income are single men.

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2. The SMSA data consisting of 141 observations on 10 variables is fitted by the model below: 1 y = Bo+B1x4 + ẞ2x6 + ẞ3x8 + √1X4X8 + V2X6X8 + €. See Question 2, Tutorial 3 for the meaning of the variables in the above model. The following results are obtained: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.302e+03 4.320e+02 3.015 0.00307 x4 x6 x8 x4:x8 x6:x8 -1.442e+02 2.056e+01 -7.013 1.02e-10 6.340e-01 6.099e+00 0.104 0.91737 -9.455e-02 5.802e-02 -1.630 0.10550 2.882e-02 2.589e-03 11.132 1.673e-03 7.215e-04 2.319 F) x4 1 3486722 3486722 17.9286 4.214e-05 x6 1 14595537 x8 x4:x8 x6:x8 1 132.4836 < 2.2e-16 1045693 194478 5.3769 0.02191 1 1198603043 1198603043 6163.1900 < 2.2e-16 1 25765100 25765100 1045693 Residuals 135 26254490 Estimated variance matrix (Intercept) x4 x6 x8 x4:x8 x6:x8 (Intercept) x4 x6 x8 x4:x8 x6:x8 0.18875694 1.866030e+05 -5.931735e+03 -2.322825e+03 -16.25142055 0.57188953 -5.931735e+03 4.228816e+02 3.160915e+01 0.61621781 -0.03608028 -0.00445013 -2.322825e+03…

In some applications the distribution of a discrete RV, X resembles the Poisson distribution except that 0 is not a possible value of X. Consider such a RV with PMF where 1 > 0 is a parameter, and c is a constant. (a) Find the expression of c in terms of 1. (b) Find E(X). (Hint: You can use the fact that, if Y ~ Poisson(1), the E(Y) = 1.)

Suppose that X ~Bin(n,p). Show that E[(1 - p)] = (1-p²)".