113年地方四等經建行政-統計學概要詳解

 113年特種考試地方政府公務人員考試

等 別：四等考試
類 科：經建行政
科 目：統計學概要

解答:$$\textbf{(一) }\cases{p(X=1) = 84/300\\ p(X=2)=150/300\\ p(X=3)=66/300} \Rightarrow E(X) =\sum_{k=1}^3kp(X=k) ={84\over 300}+2\cdot {150\over 300}+3\cdot {66\over 300}={97\over 50} \\\quad \Rightarrow E(X^2) =\sum_{k=1}^3k^2p(X=k) ={84\over 300}+2^2 \cdot {150\over 300}+3^2\cdot {66\over 300}= {213\over 50} \\ \quad \Rightarrow Var(X)=E(X^2)-(E(X))^2 ={213\over 50}-\left( {97\over 50}\right)^2 ={1241\over 2500} \Rightarrow \bbox[red, 2pt]{\cases{E(X)=97/50\\ Var(X)=1241/2500}} \\\textbf{(二) }\cases{p(Y=1) = 78/300\\ p(Y=2)= 117/300\\ p(Y=3) =105/300} \Rightarrow E(Y) =\sum_{k=1}^3kp(Y=k) ={78 \over 300}+2\cdot {117\over 300}+3\cdot {105\over 300}= {209\over 100} \\\quad \Rightarrow E(Y^2) =\sum_{k=1}^3k^2p(Y=k) ={78 \over 300}+2^2 \cdot {117 \over 300}+3^2\cdot {105\over 300}= {497\over 100} \\ \quad \Rightarrow Var(Y)=E(Y^2)-(E(Y))^2 ={497\over 100}-\left( { 209 \over 100}\right)^2 ={6019\over 10000} \Rightarrow \bbox[red, 2pt]{\cases{E(Y)= 209/100\\ Var(Y)=6019/10000}}\\ \textbf{(三) }Var(X+Y)=Var(X)+Var(Y)+2 Cov(X,Y) =1.6691 \\\quad \Rightarrow Cov(X,Y)={1\over 2}\left( 1.6691-{1241\over 2500}-{6019\over 10000}\right) =0.2854\\\quad  \Rightarrow Cor(X,Y)={Cov(X,Y) \over \sigma_X\cdot \sigma_Y} ={0.2854 \over \sqrt{1241\over 2500}\cdot \sqrt{6019\over 10000}} \approx \bbox[red, 2pt]{0.5221}$$

解答:$$\cases{n=8000\\ p=1/1000}\Rightarrow 二項分配近似卜瓦松分配 \Rightarrow \lambda=np=8 \Rightarrow P(x=k)=e^{-\lambda}\cdot {\lambda^k\over k!} \\ \Rightarrow 不良品少於10件的機率=\sum_{k=0}^9 P(x=k)，查考卷的附表(如下圖)可得\\ \sum_{k=0}^9 P(x=k) =0.0003+0.0027+ \cdots +0.1241= \bbox[red, 2pt]{0.7166}$$

解答:$$\cases{n=200\\ p=0.15} \Rightarrow \cases{np=30\\ np(1-p) =25.5} \Rightarrow 可用常態分布近似二項分配\\ \Rightarrow P(X\ge35) \approx P(Z\ge{34.5-30\over \sqrt{25.5}}) =P(Z\ge  0.89) =1-P(Z\le 0.89) =1-0.8133 \\=\bbox[red, 2pt]{0.1867}\quad (查考卷附表可得P(Z\le 0.89)=0.8133)$$


解答:$$\bar x={1\over 2}(244.051+ 255.949)= \bbox[red, 2pt]{250} \Rightarrow \bar x+ z_{\alpha/2}\cdot {\sigma \over \sqrt n} =255.949 \\ \Rightarrow 2.576\cdot {12 \over \sqrt n}=5.949 \Rightarrow n=\left( 2.576 \cdot 12 \over 5.949\right)^2 = \bbox[red, 2pt]{27}$$

解答:$$\textbf{(一)}兩獨立母體,母體變異數未知,且為小樣本\\\quad \Rightarrow df=\left \lfloor {(s_1^2/n_1+s_2^2/n_2)^2 \over {(s_1^2/n_1)^2 \over n_1-1}+{ (s_2^2/n_2)^2 \over n_2-1}} \right \rfloor= \left\lfloor { (230^2/10 +250^2/12)^2 \over {(230^2/10)^2 \over 9} +{(250^2/12)^2 \over 11}} \right \rfloor= \lfloor 19.76\rfloor =19 \\ \quad \Rightarrow t_{df,\alpha/2} =t_{19,0.025} =2.093(查考卷附表)\\ \Rightarrow S_{\bar x_1-\bar x_2} =\sqrt{(s_1^2/n_1) +(s_2^2/n_2)} =\sqrt{230^2/10 +250^2/12} = 102.46\\ \Rightarrow 起薪差異的95\%信賴區間=(30500-30000)\pm 2.093\cdot 102.46 =\bbox[red, 2pt]{(285.55,714.45)} \\\textbf{(二)}\cases{H_0: \sigma_1^2=\sigma_2^2\\ H_1:\sigma_1^2 \ne \sigma_2^2} \Rightarrow  檢定統計量F={s_2^2/ \sigma_2^2 \over s_1^1/\sigma_1^2} ={250^2\over 230^2} =1.18 \lt F_{0.05}(11,9) =3.103 (查考卷附表)\\\quad \Rightarrow 無法拒絕H_0   \Rightarrow 不存在顯著差異,即\bbox[red, 2pt]{\cases{虛無假設H_0: \sigma_1^2 = \sigma_2^2\\ 對立假設H_1: \sigma_1^2 \ne \sigma_2^2 \\檢定統計量F=1.18\\ 拒絕域R=\{F\mid F\gt F_{0.05}(11,9)=1.18\}\\ 檢定結果:無法拒絕虛無假設\\ 結論:兩系畢業生起薪變異數不存在顯著差異}} \\\textbf{(三)}\cases{H_0: \mu_1 =\mu_2 \\ H_1:\sigma_1 \ne \sigma_2} \Rightarrow  檢定統計量t={\bar x_1-\bar x_2-(\mu_1-\mu_2) \over S_{\bar x_1-\bar x_2}} ={30000-30500-(0)\over 102.46} =-4.88\\\quad \Rightarrow |t|\gt  t_{19,0.05}=1.729 \Rightarrow 拒絕虛無假設\\\quad \Rightarrow  \bbox[red, 2pt]{\cases{虛無假設H_0: \mu_1 =\mu_2\\ 對立假設H_1: \mu_1 \ne \mu_2 \\檢定統計量t=-4.88\\ 拒絕域R=\{t\mid |t|\gt t_{19,0.05} =1.729\}\\ 檢定結果:拒絕虛無假設\\ 結論:兩系畢業生平均起薪有顯著差異}}$$

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解題僅供參考，高普考歷年試題及詳解