104年普考-經建行政-統計學概要詳解

 104年公務人員普通考試試題

類 科： 經建行政、交通技術
科 目： 統計學概要

解答：$$令\cases{X:台幣薪資\\ Y:日幣薪資} \Rightarrow Y=aX, a為匯率,且a\gt 1\\(一)日幣薪資變異係數={\sigma(Y)\over \mu(Y)} ={\sigma(aX)\over \mu(aX)} ={a\sigma(X)\over a\mu(X)} ={\sigma(X)\over \mu(X)}=台幣薪資變異係數 \Rightarrow \bbox[red,2pt]{不變}\\(二)令Z=年齡 \Rightarrow r(Y,Z)={\sum YZ-\sum Y\sum Z/n\over \sqrt{\sum Y^2-(\sum Y)^2/n} \cdot \sqrt{\sum Z^2-(\sum Z)^2/n}} \\={a\sum XZ-a\sum X\sum Z/n\over \sqrt{a^2\sum X^2-a^2(\sum X)^2/n} \cdot \sqrt{\sum Z^2-(\sum Z)^2/n}}\\ ={\sum XZ-\sum X\sum Z/n\over \sqrt{\sum X^2-(\sum X)^2/n} \cdot \sqrt{\sum Z^2-(\sum Z)^2/n}}=r(X,Z)\\ \Rightarrow R(Y,Z)= R(X,Z) \Rightarrow \bbox[red,2pt]{不變}\\(三)日幣檢定統計量t_Y= {\bar Y-5000a\over \sigma(Y)/\sqrt n}= {a\bar X-5000a\over a\sigma(X)/\sqrt n} = {\bar X-5000\over \sigma(X)/\sqrt n}=t_X \Rightarrow \bbox[red,2pt]{不變}\\(四)日幣迴歸直線斜率b_Y=r(Y,Z)\cdot {\sigma(Y)\over \sigma(Z)} =r(X,Z)\cdot {a\sigma(X)\over \sigma(Z)} =a\cdot b_X \Rightarrow \bbox[red,2pt]{變大}\\(五)假設將員工依教育程度分為k組，每組有n_i人,i=1-k \Rightarrow 總人數n=\sum_{i=1}^k n_i \\ \Rightarrow F_Y={MSTR\over MSE} ={SSTR/(k-1)\over SSE/(n-k)} ={\sum_{i=1}^k n_i(\bar y_i-\bar y)^2/(k-1)\over\sum_{j=1}^k \sum_{i=1}^{n_i}   (y_{i,j}-\bar y_j)^2/(n-k)} \\ ={a^2\sum_{i=1}^k n_i(\bar x_i-\bar x)^2 /(k-1)\over a^2 \sum_{j=1}^k \sum_{i=1}^{n_i}  (x_{i,j}-\bar x_j)^2/(n-k)} =F_X\Rightarrow \bbox[red,2pt]{不變}$$

解答：

(一)$$投擲三個銅板成功(出現剛好兩個正面)的機率p=C^3_2/2^3=3/8，不成功的機率=1-3/8=5/8\\老大獲勝的情形=老大第1次就成功+ 三人第1次都失敗且老大第2次成功\\\qquad + 三人前2次都失敗且老大第3次成功+\cdots\\ 其機率=p+(1-p)^3p+ (1-p)^6p+\cdots =p(1+(1-p)^3+(1-p)^6+\cdots )={p\over 1-(1-p)^3} \\={3/8 \over 1-(5/8)^3} = \bbox[red,2pt]{64\over 129}$$(二)$$老三獲勝的情形=第1次只有老三成功+ 第1次都失敗且第2次只有老三成功\\\qquad + 前2次都失敗且第3次只有老三成功+ \cdots\\ 其機率=(1-p)^2p+ (1-p)^5p + (1-p)^8p =p(1-p)^2(1+(1-p)^3+ (1-p)^6+\cdots )\\= {p(1-p)^2\over 1-(1-p)^3} ={(3/8)(5/8)^2 \over 1-(5/8)^3} = \bbox[red,2pt]{25\over 129}$$

解答：$$\begin{array}{r|cc | ccc} 抽中三球& S & T & S^2 &T^2 & S\times T\\\hline (1,2,3) & 2 & 5 & 4 & 25 & 10\\ (1,2,4) & 3 & 6 & 9 & 36 & 18 \\ (1,3,4) & 3 & 7 & 9 & 49 & 21 \\ (2,3,4) & 2 & 7 & 4 & 49 & 14\\\hline \sum & 10 & 25 & 26 & 159 & 63\end{array}\\ \Rightarrow S,T相關係數r_{ST}= {\sum ST -\sum S\sum T/n\over \sqrt{\sum S^2-(\sum S)^2/n} \cdot \sqrt{\sum T^2- (\sum T)^2/n}}\\ ={63-10\cdot 25/4 \over \sqrt{26-10^2/4} \cdot \sqrt{159-25^2/4}}  ={0.5\over 1\cdot 1.658} = \bbox[red,2pt]{0.3015}$$(二)$$T=7只有兩種情況: (S,T)=(3,7),(2,7) \Rightarrow \cases{P(S=3\mid T=7)=0.5 \\P(S=2\mid T=7)=0.5} \\ \Rightarrow \cases{ E(S\mid T=7)= 3\times 0.5+2\times 0.5=2.5 \\ E(S^2\mid T=7) =3^2\times 0.5 + 2^2\times 0.5=6.5} \\ \Rightarrow Var(S\mid T=7)= 6.5-2.5^2= \bbox[red,2pt]{0.25}$$

解答：

(一)$$T=F(X) \Rightarrow F_T(t)= P(T\le t) =P(F(X)\le t)= P(X\le F^{-1}(t)) =F(F^{-1}(t)) =t\\ \Rightarrow f_T(t)= F'_T(t)=1,0\le t\le 1 \Rightarrow \bbox[red,2pt]{T\sim U(0,1)}$$(二)$$與(一)剛好相反，因此\bbox[red, 2pt]{f_Y(y)=\cases{4y^2e^{-2y},0\lt y\lt \infty\\ 0,其他}}$$

解答：

(一)$$\begin{array}{}X& Y &X^2 & Y^2 & XY \\ \hline 0 & 2 & 0 & 4 & 0\\4 & 4 & 16 & 16 & 16\\2 & 3 & 4 & 9 & 6\\1 & 1 & 1 & 1 & 1\\3 & 5 & 9 & 25 & 15\end{array} \Rightarrow \cases{\sum X =10 \\ \sum Y=15\\ \sum X^2=30\\ \sum Y^2=55 \\ \sum XY=38}\\ \Rightarrow 相關係數r={\sum XY-\sum X\sum Y/n\over \sqrt{\sum X^2-(\sum X)^2/n}\cdot \sqrt{\sum Y^2-(\sum Y)^2/n}}\\ ={38-10\cdot 15/5 \over \sqrt{30-10^2/5}\cdot \sqrt{55-15^2/5}} ={8\over 10}\\ \cases{H_0:相關係數=0\\ H_1:相關係數\ne 0\\ 拒絕域C=\{t\mid t\gt |t_{\alpha/2}(n-2)|= |t_{0.025}(3)|= 3.1824\}} \\ 檢定統計量t=r\cdot \sqrt{n-2\over 1-r^2} ={8\over 10}\cdot \sqrt{3\over 36/100}={4\over \sqrt 3}=2.309 \not \in C \Rightarrow 不能拒絕H_0 \\ \Rightarrow \bbox[red,2pt]{沒有證據顯示相關係數不為0}$$(二)$$Y_i= \beta X_i+\varepsilon_i \Rightarrow \hat Y_i=\beta X_i \Rightarrow f(\beta)=SSE = \sum (Y_i-\hat Y_i)^2 = \sum(Y_i-\beta X_i)^2\\ 令f'=0 \Rightarrow 2\sum(Y_i-\beta X_i) (-X_i)=0 \Rightarrow \sum (Y_i-\beta X_i) X_i=0 \\ \Rightarrow \sum X_iY_i-\beta\sum X_i^2=0 \Rightarrow \beta ={\sum X_iY_i\over \sum X_i^2} = {38\over 30} = \bbox[red,2pt]{1.2667}$$

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