110年地方四等-統計學概要詳解

110 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：

(一)$$P(77\le X\le 82)=P({77-77\over 5}\le Z\le {82-77\over 5})=P(0\le Z\le 1)= P(Z\le 1)-P(Z\le 0)\\=1-P(Z\le -1)-P(Z\le 0) =1-0.1587(查表)-0.5(查表)=\bbox[red,2pt]{0.3413}$$(二)$$X\sim N(77,5^2) \Rightarrow \bar X={X_1+X_2+X_3 +X_4\over 4} \sim N(77,({5\over \sqrt 4})^2=({5\over 2})^2) \Rightarrow \bar X\sim N(77,({5\over 2})^2)\\ \Rightarrow P(77\le \bar X\le 82) =P({77-77\over 5/2}\le Z\le {82-77\over 5/2}) = P(0\le Z\le 2)= P( Z\le 2)-P(Z\le 0)\\ =1-P(Z\le -2)-P(Z\le 0)=1-0.0228(查表)-0.5(查表)=\bbox[red,2pt]{0.4772}$$(三)$$查試題附表:P(Z\le -2.81)=0.025 \Rightarrow P(Z\ge 2.81)=0.025 \\\Rightarrow P(Z\ge 2.81)=P(X\ge 2.81\times 5+77)= P(X\ge 91.05) \\ \Rightarrow 至少要消費\bbox[red,2pt]{91.05元}$$(四)$$n= (z_{\alpha}\cdot {\sigma \over E})^2 = z_{0.025}\cdot {5^2\over 2^2} =1.96\times 6.25 =12.25 \Rightarrow n=\bbox[red,2pt]{13}$$

 解答：$$\cases{X=1:(1,1)\\ X=2:(1,2),(2,1),(2,2)\\ X=3:(1,3)(2,3),(3,1), (3,2),(3,3)\\ X=4:(1,4),(2,4),(3,4),(4,1), (4,2),(4.3),(4,4)} \\ \Rightarrow \cases{P(X=1)=1/16\\ P(X=2)=3/16\\ P(X=3)=5/16\\ P(X=4)=7/16} \Rightarrow \cases{E(X)={1\over 16}  (1\cdot 1+2\cdot 3+3\cdot  5+4\cdot 7)=25/8\\ E(X^2)={1\over 16}(1^2\cdot 1+2^2\cdot 3 +3^2\cdot 5 +4^2\cdot 7)=85/8} \\ \Rightarrow Var(X)= E(X^2)-(E(X))^2 = {85\over 8}-({25\over 8})^2={55\over 64}\\ \Rightarrow (一)E(X)=\bbox[red,2pt]{25\over 8};(二)E(X^2)=\bbox[red,2pt]{85\over 8};(三)Var(X)=\bbox[red, 2pt]{55\over 64}$$

 解答：

(一)$$\cases{H_0:男生與女生有相同的意願購買高階手機\\ H_1:男生與女生購買高階手機的意願不同}$$(二)$$\cases{男生購買高階手機意願的比例\hat p_1=280/400=7/10 \\女生購買高階手機意願的比例\hat p_2=200/400=1/2 }\\ \Rightarrow S_{\hat p_1-\hat p_2}^2 ={\hat p_1(1-\hat p_1)\over n_1} + {\hat p_2(1-\hat p_2)\over n_2} ={0.7\times 0.3\over 400} +{0.5\times 0.5\over 400} ={23\over 20000}\\檢定統計量z={(\hat p_1-\hat p_2)-(p_1-p_2)\over S_{\hat p_1-\hat p_2}} ={(0.7-0.5)-0 \over \sqrt{23\over 20000}} = \bbox[red,2pt]{5.8977} \gt 1.96(z_{0.05/2})\\ \Rightarrow 拒絕H_0，即男女購買高階手機意願有明顯的不同，又\hat p_1\gt \hat p_2\\，因此\bbox[red,2pt]{男生購買高階手機意願高於女生}$$(三)$$信賴區間=(\hat p_1-\hat p_2)\pm z_{\alpha/2}S_{\hat p_1-\hat p_2} =0.2\pm 1.96\times \sqrt{23\over 20000} =\bbox[red,2pt]{[0.1334,0.2665]}$$

 解答：

(一)$$依題意，迴歸直線為\hat y=-9.362x_1+7.515x_2+572.824,其中x_1:燃料消耗量,x_2:汽車排氣量\\ 因此\cases{x_1=50\\ x_2=100} \Rightarrow \hat y=-9.362\times 50+7.515\times 100+572.824= 856.224(千元)\\ 殘差=y-\hat y= 549.900-856.224 =\bbox[red,2pt]{ -306.324}(千元)$$(二)$$檢定統計量t={7.515\over 2.641} = \bbox[red,2pt]{2.8455}$$(三)$$\begin{array}{} \hline  & df & SS & MS & F\\\hline \text{Regression} & (1) & (2) & (3) & (4)\\\text{Error} &(5)  & 34,188,066& (6) \\\text{Total} &(7) & 43,506,728\\\hline\end{array} \\\cases{兩個變數\Rightarrow (1)=2\\ (2)=43,506,728- 34,188,066=9,318,662\\ (3)=(2)\div (1)=9318662\div 2=4,659,331\\ (7)=n-1=100-1=99\\ (5)=(7)-(1)=99-2=97\\ (6)=34188066\div (5)= 34188066\div 97 = 352454.2887} \\ \Rightarrow 檢定統計量F=(4)=(3)\div (6)=4659331\div 352454.2887 = \bbox[red,2pt]{13.22}$$(四)$$R^2={SSR\over SST} =1-{SSE\over SST} =1-{34,188,066\over 43,506,728}= \bbox[red,2pt]{0.2142}$$

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