110年原住民族四等考試-經建行政-統計學概要詳解

 110年原住民族考試

等別：四等考試
組別：經建行政
科目：統計學概要

解答：

(一) $$斜率b_1={\sum x_iy_i -(\sum x_i\times \sum y_i)/n\over \sum x_i^2-(\sum x_i)^2/n} ={2385-50\cdot 66/20\over 1975-50^2/20} =\bbox[red,2pt]{1.2}$$ (二) $$\cases{\bar x=\sum x_i/n=50/20=5/2\\ \bar y=\sum y_i/n=66/20=33/10} \Rightarrow 迴歸直線方程式:y=b_1(x-\bar x)+\bar ={6\over 5}(x-{5\over 2})+{33\over 10} \\ x=0代入迴歸直線，可得截距b_0=-{6\over 5}\times {5\over 2}+{33\over 10} =\bbox[red,2pt]{0.3}$$ (三) $$相關係數r={\sum x_iy_i -(\sum x_i\times \sum y_i)/n \over \sqrt{\sum x_i^2-(\sum x_i)^2/n}\cdot \sqrt{\sum y_i^2-(\sum y_i)^2/n}} \\ ={2385-50\cdot 66/20\over \sqrt{1975-50^2/20}\cdot \sqrt{3278-66^2/20}} \approx\bbox[red,2pt]{0.933}$$ (四) $$SSR={\left(\sum x_iy_i-(\sum x_i)(\sum y_i)/n \right)^2 \over \sum x_i^2- (\sum x_i)^2/n} ={(2385-50\cdot 66/20)^2\over 1975-50^2/20} = 2664 \Rightarrow MSR=SSR/1=2664\\ SST= \sum y_i^2-(\sum y_i)^2/n= 3278-66^2/20=3060.2\\ SSE= SST-SSR = 3060.2-2664=396.2 \Rightarrow MSE=SSE/(n-2)=396.2/18=22.01\\ \Rightarrow 檢定統計量F^*=MSR/MSE=2664/22.01\approx \bbox[red,2pt]{121.03}$$

解答： $$(一)\cases{上班薪資X\sim N(35570,60^2) \\ 打工薪資Y\sim N(14430,(20\sqrt 7)^2)} \Rightarrow X+Y \sim N(35570+ 14430,60^2+ (20\sqrt 7)^2)\\ \qquad \Rightarrow 總薪資平均值=35570+ 14430= \bbox[red,2pt]{50000}元\\(二)總薪資標準差= \sqrt{60^2+ (20\sqrt 7)^2} =\sqrt{6400} =\bbox[red, 2pt]{80}元\\(三)P(X+Y \gt 50100) = P\left(Z\gt {50100-50000\over 80} \right) =P(Z\gt 1.25)= 0.5-0.3944=\bbox[red,2pt]{ 0.1056}\\(四)P(49900 \lt X+Y \lt 50200) = P\left({49900-50000\over 80} \lt Z\lt {50200-50000\over 80}\right) =P(-1.25\lt Z\lt 2.5) \\\qquad = P(0\le Z\lt 2.5) +P(0\le Z\lt 1.25) = 0.4938+ 0.3944 = \bbox[red,2pt]{0.8882}$$

解答：

(一) $$信賴區間為(20,30) \Rightarrow 樣本平均數\bar x={20+30\over 2}=\bbox[red, 2pt]{25}$$ (二) $$信賴區間為(20,30) =(\bar x-{z_{\alpha/2}\cdot {\sigma\over \sqrt n},\bar x+z_{\alpha/2}{\sigma\over \sqrt n}}) \Rightarrow  z_{\alpha/2}\cdot {\sigma\over \sqrt n} =5\\ 由於\cases{z_{0.1/2}=1.645\\ z_{0.05/2} = 1.96} \Rightarrow 新的信賴區間=(\bar x\pm {z_{0.1/2}\over z_{0.05/2}}\cdot 5) =(25\pm {1.645\over 1.96}\cdot 5)= \bbox[red,2pt]{(24.16,25.84)}$$ (三) $$檢定統計量z={22-25\over \sigma/\sqrt n} ={-3\over 5/1.96} =\bbox[red,2pt]{-1.176}\\ 相對應的p值為1-z_{-1.176/2}=1-z_{-0.588}= 0.5-0.22 =\bbox[red,2pt]{0.28}$$ (四) $$n_1E_1^2 = n_2E_2^2 \Rightarrow 49\times 5^2= n_2\times 1.5^2 \Rightarrow n_2 \ge {49\times 5^2\over 1.5^2} =544.44 \Rightarrow n=\bbox[red,2pt]{545}$$

解答：

(一) $$\cases{P(Y=-1)=0.3\\ P(Y=0)=0.5\\ P(Y=2)=0.2} \Rightarrow \cases{E(Y)=-1\cdot 0.3+0\cdot 0.5+ 2\cdot 0.2 =0.1\\ E(Y^2)= (-1)^2\cdot 0.3+0^2\cdot 0.5+ 2^2\cdot 0.2 =1.1}\\ \Rightarrow Var(Y)= E(Y^2)-(E(Y))^2= 1.1-0.1^2= \bbox[red,2pt]{1.09}$$ (二) $$\cases{P(X=-2)=0.2\\ P(X=1)=0.4\\ P(X=3)=0.4} \Rightarrow  E(X)=-2\cdot 0.2+1\cdot 0.4+ 3\cdot 0.4 =  \bbox[red,2pt]{1.2}$$ (三) $$\cases{P(X=-2)=0.2\\ P(X=1)=0.4\\ P(X=3)=0.4} \Rightarrow  \cases{E(X)=-2\cdot 0.2+1\cdot 0.4+ 3\cdot 0.4 =1.2\\E(X^2)=(-2)^2\cdot 0.2+1^2\cdot 0.4+3^2\cdot 0.4=4.8} \\ \Rightarrow Var(X)= E(X^2)-(E(X))^2 = 4.8-1.2^2=3.36\\ 因此Var(X-2Y) = Var(X)+(-2)^2 Var(Y) =3.36+4\cdot 1.09= \bbox[red, 2pt]{7.72}$$ (四) $$E(2X+3Y)= 2E(X)+3E(Y)=2\times 1.2+3\times 0.1=\bbox[red,2pt]{2.7}$$

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