113年初等考試-統計學大意詳解

113年公務人員初等考試試題

等 別： 初等考試
類 科： 統計
科 目： 統計學大意

解答: $$0.5\times 85+0.5\times 75=80,故選\bbox[red, 2pt]{(B)}$$

解答: $$Var(\cdot)是線性運算，因此Var(X+Y)=Var(X)+ Var(Y),故選\bbox[red, 2pt]{(D)}$$

解答: $$P(\mu-k\sigma \lt X\lt \mu+k\sigma) \ge 1-{1\over k^2} =0.84 \Rightarrow {1\over k^2}=0.16 =0.4^2 \Rightarrow k={1\over 0.4}=2.5\\,故選\bbox[red, 2pt]{(D)}$$

解答:

$$查試題的附表，如上圖，因此2x=0.32 \Rightarrow x=0.16,故選\bbox[red, 2pt]{(D)}$$

解答: $$圖表的寬度(全距)及總共要繪製多少個長條形(組數)，故選\bbox[red, 2pt]{(C)}$$

解答: $$n={(z_{\alpha/2})^2 \sigma^2 \over E^2} \Rightarrow E={z_{\alpha/2} \sigma\over \sqrt n}, \\n變為2倍 \Rightarrow E'={z_{\alpha/2} \sigma\over \sqrt {2n}} ={1\over \sqrt 2}E =0.707E,故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{X:自動沖洗時間\\ Y:人工擦乾時間} \Rightarrow \cases{X\sim N(15,3^2)\\ Y\sim N(10,4^2)} \Rightarrow X+Y \sim N(15+10,3^2+4^2 =N(25,5^2)\\ P(X+Y\gt 30)=P(Z\gt {30-25\over 5}) =P(Z\gt 1) =0.5-0.3413(查表)=0.1587\\ ,故選\bbox[red, 2pt]{(B)}$$

解答: $$當n夠大，二項式分佈近似常態分佈，即B(n=100,p=0.6) \approx N(np,np(1-p)) =N(60,24) \\ \Rightarrow P(X=60=\mu)=0.5,故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{P(Z\gt A)=0.2\\ P(Z\lt B)=0.2} \xrightarrow{查表}\cases{A=0.84\\ B=-0.84} \Rightarrow \cases{X=10\times 0.84+50=58.4\\ Y=10\times (-0.84)+50=41.6} \\ \Rightarrow Y:X=41.6:58.4,故選\bbox[red, 2pt]{(A)}\\ 題目應修正為\bbox[cyan,2pt]{最低與最高},不是最高與最低$$

解答: $$X\sim Exp(\lambda) \Rightarrow E(X)={1\over \lambda }=2 \Rightarrow \lambda={1\over 2}=0.5\\ \Rightarrow P(X\lt 0.5(6個月=0.5年))=1-e^{-\lambda x} =1-e^{-0.5\times 0.5} =1-e^{-0.25},故選\bbox[red, 2pt]{(A)}$$

解答: $$P(72\lt X\lt 84)=0.68 \Rightarrow P({72-78\over \sigma}\lt Z\lt {84-78\over \sigma})=0.68\\ 查表可知：P(-1\lt Z\lt 1)=0.68 \Rightarrow \sigma=6,故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{E(T_1) =E(X_1+2X_2+ 4X_3+2X_4+X_5)/10=10\mu/10=\mu\\ E(T_2) =E(X_1+ 2X_2+ 5X_3+2X_4+X_5)/15 =11\mu/15\\ E(T_3) =E(X_1+X_2+X_3 +X_4+X_5)/5= 5\mu/5=\mu \\ E(T_4) =E(3X_1 + 2X_2 +X_3+2X_4 +3X_5)/11 = 11\mu/11= \mu} \\ \Rightarrow 只有T_1,T_3,T_4的\mu 估計是不偏的， \\ 又\cases{Var(T_1)=(1^2 +2^2+ 4^2+2^2+1^1)\sigma^2=26\sigma^2 \\ Var(T_3)=(1^2+1^2 +1^2 +1^2+1^2) \sigma^2 =5\sigma^2\\ Var(T_4) =(3^2+2^2+ 1^2 +2^2+3^2) \sigma^2= 27\sigma^2} \Rightarrow Var(T_3)最小，故選\bbox[red, 2pt]{(C)}$$

解答: $$(A)\bigcirc:n夠大就近似常態\\(B)\bigcirc:\cases{\hat p=12/100=0.12\\ \sigma = \sqrt{\hat p(1-\hat p)/ n}=0.0325}  \Rightarrow 95\%信賴區間=(\hat p-2\sigma,\hat p+2\sigma) =(0.055,0.185) \\(C) \bigcirc: 如(B)\sigma 的計算會使用\hat p(1-\hat p) \\(D)\times: n=z_{\alpha/2}{p(1-p) \over E^2} =1.96^2\cdot {0.5\cdot 0.5\over 0.1^2} =96.04\lt 100\\ 故選\bbox[red, 2pt]{(D)}$$

解答: $$信賴區間：\left( \bar x\pm t_{n-1,\alpha/2}{s\over \sqrt n}\right)為t分配，自由度為n-1=20-1=19, 故選\bbox[red, 2pt]{(B)}$$

解答: $$n={(z_{1-\alpha}+z_{1-\beta})^2 \sigma^2 \over (\mu_1-\mu_0)^2} ={(1.645+1.282)^2\times 12^2 \over (115-100)^2} =5.48\\ 若題目\mu_1\bbox[cyan, 2pt]{修正為}105,則n=49.35,故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{H_0:p_1=p_2\\ H_1: p_1\ne p_2} \;且 \;\cases{\hat p_1=35/250= 0.14 \\ \hat p_2=27/300 =0.09\\ n_1=250\\ n_2=300 \\ \hat p=(35+27) / (250+300) \approx 0.113} \\\Rightarrow z=\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{ \hat p (1-\hat p)\left({1\over n_1}+{1\over n_2}\right)}} ={0.14-0.09 \over \sqrt{0.113\cdot 0.887\left({1\over 250}+{1\over 300} \right)}} ={0.05\over 0.027} =1.844 \\ \xrightarrow{查表,雙尾}P值=(0.5-0.4675)\times 2=0.065,故選\bbox[red, 2pt]{(B)}$$

解答: $$已知\cases{k=4\\ b=3} \Rightarrow n=k\times b\times 3(重複3次)=36\\F=MSA/MSE \Rightarrow 4=16/MSE \Rightarrow MSE=4\\ 又MSE = SSE/(n-kb) =SSE/(36-12) \Rightarrow 4=SSE/24 \Rightarrow SSE=96,故選\bbox[red, 2pt]{(B)}$$

解答: $$(A)\times: 非大樣本仍可計算決定係數\\ (B)\times: R^2代表解釋力，只要大於零，X與Y之間皆存在關係 \\(C)\times R^2 \ge 0 \Rightarrow R^2無法判定正負相關\\ 故選\bbox[red, 2pt]{(D)}$$

解答: $$檢定統計量t={4.24\over 3.8} =1.116,查表可得t(df=10-1=9, \alpha=0.05)= 3.250 \\ \Rightarrow 1.116 \lt 3.25 \Rightarrow 不能拒絕H_0,故選\bbox[red, 2pt]{(B)}$$

解答: $$\begin{array}{ccc|rrr} & X:\%& Y:萬戶& X^2 & Y^2 & XY\\\hline & 2& 25& 4& 625& 50\\ & 3& 25& 9 & 625& 75\\ & 5 & 20 & 25& 400 & 100\\ & 1 & 30& 1 & 900 & 30\\ &8 & 16& 64& 256& 128 \\\hdashline \sum & 19 & 116 & 103& 2806& 383\end{array}\\ (A)\bigcirc: 斜率b={n\sum XY-\sum X\sum Y\over n\sum X^2-(\sum X)^2} ={5\cdot 383-19\cdot 116\over 5\cdot 103-19^2} =-{289\over 154} =-1.8766\\ (B)\bigcirc: 截距a={\sum Y\over n}-b{\sum X\over n} ={116\over 5}-(-1.877)\cdot {19\over 5}= 30.33 \\ (C)\times: 估計值＝b\times 5+a=(-1.877)\times 5+30.33=39.715 \Rightarrow 殘差=39.715-20=19.715 \ne 0.95\\ (D)\bigcirc: 最小平方法的目的就是殘差總和為零\\ 故選 \bbox[red, 2pt]{(C)}$$

解答: $$(A)\bigcirc: 截距的估計值為-0.74,年收入的估計值(斜率)=0.39 \\ \qquad \Rightarrow 迴歸直線 \hat y=0.39x-0.74\\ (B)\bigcirc: 斜率為0.39 \Rightarrow x加1則y加0.39\\ (C)\bigcirc: 200萬=20(十萬) \Rightarrow \hat y=0.39\cdot 20-0.74=7.06(萬元)\\ (D)\times: 斜率的P值=0.000 \lt 0.05已達顯著性\\ 故選\bbox[red, 2pt]{(D)}$$

解答: $$週一至週共請假6+4+ 8+3+9 ＝30 \Rightarrow 平均每日請假{30\over 5}=6\\ \Rightarrow \begin{array}{c|ccccc}  觀察值& 6 & 4& 8& 3& 9\\\hline 期望值& 6& 6& 6& 6& 6\end{array} \\\Rightarrow 檢定統計量\chi^2 ={1\over 6}((6-6)^2+((4-6)^2 +(8-6)^2 +(3-6)^2+ (9-6)^2) \\={26\over 6}=4.33 \Rightarrow 自由度= (5-1)\times(2-1)=4,故選\bbox[red, 2pt]{(A)}$$

解答: $$\begin{array}{c|ccccc}  實際值A_i& 17 & 21& 19& 23&  \\\hline 推估值F_i& F_1=17& F_2& F_3& F_4& F_5\end{array} \\指數平滑法F_{i+1}=F_i+\alpha(A_i-F_i) \Rightarrow F_2=17+0.2(17-17)=17\\ \Rightarrow F_3=17+0.2(21-17)=17.8 \Rightarrow F_4=17.8 +0.2(19-17.8) =18.04 \\ \Rightarrow F_5= 18.04+0.2(23-18.04) =19.032,故選\bbox[red, 2pt]{(B)}$$

解答: $$負偏態\Rightarrow 眾數\gt 中位數\gt 平均數，故選\bbox[red, 2pt]{(B)}$$

解答: $$假設\cases{A:\{-10,10\}\\ B:\{5,6\}} \Rightarrow A全距\gt B全距，但A平均\lt B平均，故選\bbox[red, 2pt]{(D)}$$

解答: $$10\times {4\over 10}=4, 故選\bbox[red, 2pt]{(B)}$$

解答: $$(C) \bigcirc: A,B獨立\Rightarrow P(A\cap B) =P(A)P(B) \ne 0 \Rightarrow 不互斥\\ 故選\bbox[red, 2pt]{(C)}$$

解答: $$依中央極限定理為常態分配，故選\bbox[red, 2pt]{(B)}$$

解答: $$雖然母體為常態分配，但抽樣人數=15為小樣本，故選\bbox[red, 2pt]{(B)}$$

解答: $$X\sim N(47500,2500^2) \Rightarrow P(45000\lt X\lt 50000) =P({45000-47500\over 2500} \lt Z\lt {50000-47500\over 2500}) \\=P(-1\lt Z\lt 1) =  0.3413\times 2=0.6826({查表}),故選\bbox[red, 2pt]{(C)}$$

解答: $${(n-1)s^2 \over \sigma^2} \sim \chi^2(n-1),故選\bbox[red, 2pt]{(C)}$$

解答: $$誤差與樣本數成反比,故選\bbox[red, 2pt]{(A)}$$

解答: $$型I錯誤: H_0正確卻被拒絕, 故選\bbox[red, 2pt]{(A)}$$

解答: $$F({1-\alpha, v_1,v_2}) ={1\over F(\alpha,v_2,v_2)}, 故選\bbox[red, 2pt]{(D)}$$

解答: $$p=0.0102 \lt 0.05 已達顯著,即\sigma_1 \ne \sigma_2,故選\bbox[red, 2pt]{(D)}$$

解答: $$相關係數近乎零代表非正相關也非負相關，故選\bbox[red, 2pt]{(C)}$$

解答: $$斜率b={n\sum XY-\sum X\sum Y\over n\sum X^2-(\sum X)^2} ={10\cdot 2615-90\cdot 210\over 10\cdot 1080-90^2} =2.685,故選\bbox[red, 2pt]{(C)}$$

解答: $$變異數分析功能與適合度檢定不同, 故選\bbox[red, 2pt]{(D)}$$

解答: $$\begin{array}{} SOURCE & SS &df & MS &F\\\hline Regression & 197.2778 & 2-1=1 & 197.2778/1=197.2778 & \\\hline Error & 95.822 & 7-2=5 & 94.822/5=18.9644\\\hline \end{array} \\ \Rightarrow F= 197.2778/18.9644 = 10.4025, 故選\bbox[red, 2pt]{(B)}$$

解答: $$檢定統計量F=MSTR/MSE =241.67/77.33= 3.125\\ 由\cases{v_1=k-1=3-1=2\\ v_2=n-k = 8-3=5 \\ \alpha=0.05} \xrightarrow{查表}F(2,5) =5.79 \gt 3.125 \Rightarrow 沒有顯著差異,故選\bbox[red, 2pt]{(A)}$$

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解題僅供參考，其他歷年試題及詳解