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2021年12月4日 星期六

109年初等考試-統計學大意詳解

109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:$${0.1\over 0.1+0.4}={1\over 5}=0.2,故選\bbox[red,2pt]{(C)}$$
解答:$$2X+Y=4 \Rightarrow 斜率為負值\Rightarrow 相關係數為-1,故選\bbox[red,2pt]{(D)}$$
解答:$$樣本平均數的標準誤與母體平均數無關,與標準差、樣本數有關,故選\bbox[red,2pt]{(A)}$$
解答:$$本題 \bbox[blue,2pt]{送分}$$
解答:$$\cases{X_1\sim N(\mu,\sigma^2)\\ X_2\sim N(\mu,\sigma^2)} \Rightarrow \mu\left({2X_1-X_2\over 2}\right) =\mu(X_1)-{1\over 2}\mu(X_2)=\mu-{1\over 2}\mu={1\over 2}\mu \ne \mu,故選\bbox[red,2pt]{(D)}$$
解答:$$(A)\bigcirc:P(900\lt X\lt 1500)= P({900-1200\over 100}\lt z\lt {1500-1200\over 100}) =P(-3\lt z\lt 3)\approx 1\gt 89\% \\(B)\times: P(1100\lt X\lt 1300)=P(-1\lt z\lt 1)=0.6826 \not \gt 75\%\\(C)\times: P(-Z\lt z\lt Z)=0.95 \Rightarrow Z=1.96 \Rightarrow P(1004\lt X\lt 1396)=0.95\\\qquad \Rightarrow P(1000\lt X\lt 1400) \gt 95\% \\(D)\times: 由(A)知: P(900\lt X\lt 1500)\gt 99.7\%\\,故選\bbox[red,2pt]{(A)}$$
解答:$$信賴區間[75,85] =80\pm 5 \Rightarrow 5= z_{0.025}\times \sigma/\sqrt n =1.96 \times {20\over \sqrt n} \\ \Rightarrow n=(1.96\times 4)^2=61.5 \Rightarrow n=62,故選\bbox[red,2pt]{(C)}$$
解答:$$$\begin{array}{l|c} &非抽菸者\\\hline 禁止區 &0.44 \\\hline 特定區 & 0.52 \\\hline 沒有限制& 0.04 \end{array}  \Rightarrow \begin{array}{l|c} &非抽菸者\\\hline 禁止區 & p_1={0.44\over 0.44+0.52}=0.4583,n_1=700\times 0.44=308 \\\hline 特定區 & p_2=1-0.4583=0.5417,n_2= 700\times 0.52= 364  \end{array}\\  \Rightarrow 信賴區間=p_2-p_1\pm z_{\alpha/2}\times \sqrt{{p_1(1-p_1) \over n_1}+{p_2(1-p_2) \over n_2}}\\ =0.5417-0.4583\pm 1.96\times \sqrt{{0.4583\times 0.5417 \over 308}+{0.5417\times 0.4583  \over 364}}\\ =0.0834\pm 1.96\times 0.0386 =[0.00779, 0.1590],故選\bbox[red,2pt]{(A)}$$
解答:$$由於無母體元素底冊,僅就同一部落/群集(cluster)抽樣,可節省費用,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{E(\bar X^2)= \mu_{\bar X}^2 +\sigma_{\bar X}^2= \mu^2+{\sigma^2 \over n} \\ E(\sum X_i^2/n^2)= {1\over n^2}E(X_1^2 + \cdots +X_n^2) = {1\over n^2}\cdot n(\mu^2+\sigma^2) ={1\over n}(\mu^2+ \sigma^2)} \\ \Rightarrow E(\bar X^2) \gt E(\sum X_i^2/n^2),故選\bbox[red,2pt]{(D)}$$
解答:$$SK\lt 0 \Rightarrow 負偏(左偏) \Rightarrow 高峰偏右\Rightarrow 眾數\gt 中位數\gt 平均數,故選\bbox[red,2pt]{(B)}$$
解答:$$n=z_{\alpha/2}^2 \cdot {p(1-p)\over E^2} \Rightarrow \cases{n_1=1.96^2\cdot 0.5^2/0.05^2 =384\\ n_2=1.96^2\cdot 0.5^2/0.025^2 =1536} \Rightarrow n_2-n_1=1152,故選\bbox[red,2pt]{(A)}$$
解答:$$H_0:\mu \lt 18\text{ v.s. }H_a: \mu \ge 18,故選\bbox[red,2pt]{(A)}$$
解答:$$\mu(0.6A+0.4B) =0.6\mu(A)+0.4\mu(B)= 0.6\times 0.09+0.4\times 0.13=0.106 \ne 0.206,故選\bbox[red,2pt]{(A)}$$
解答:$$信賴區間為\left[{(n-1)s^2 \over \chi^2_{n-1,\alpha/2}}, {(n-1)s^2 \over \chi^2_{n-1,1-\alpha/2}}\right] =\left[{9\times 16\over \chi^2_{9,0.025}}, {9\times 16\over \chi^2_{9,0.025}}\right] =\left[{144\over 19.0228}, {144\over 2.7004}\right]\\ =[7.57,53.33],故選\bbox[red,2pt]{(A)}$$
解答:$$二獨立的樣本變異數之比呈現一 F-分配,故選\bbox[red,2pt]{(D)}$$
解答:$$依題意\cases{集區個數b=3\\ 處置k=4} \Rightarrow \cases{①=b-1=2\\ ②=k-1=3\\ ③=(k-1)(b-1)=6\\ ④=①+②+③=11}\\由於缺失資料,因此SSE的自由度要減1,即③=6-1=5 \Rightarrow ④=10,故選\bbox[red,2pt]{(C)} $$
解答:$$p-value 小於0.05才達顯著差異,故選\bbox[red,2pt]{(C)}$$
解答:$$計算是否達顯著水準才需要用到\alpha,故選\bbox[red,2pt]{(A)}$$
解答:$$D_3的p-value=0.5001 \not \lt 0.05,未達顯著性,故選\bbox[red,2pt]{(C)}$$
解答:$$並非自變數與應數高度相關,而是自變數間高度相關,故選\bbox[red,2pt]{(C)}$$
解答:$$左低右高的矩形狀並非常態資料圖形,故選\bbox[red,2pt]{(D)}$$
解答:$$依現有數據推論(預估)群體統計為推論統計,故選\bbox[red,2pt]{(C)}$$
解答:$$186\div 300=0.62,故選\bbox[red,2pt]{(B)}$$
解答:$$由於母體分佈未知,以柴比雪夫不等式計算,即P(\mu-k\sigma \lt x\lt \mu+k\sigma)\ge 1-{1\over k^2}\\ 依題意P(58.6\lt X\lt 68.6) \ge 75\%=1-{1\over 2^2},即k=2 \Rightarrow \cases{\mu-2\sigma = 58.6\\ \mu+2\sigma = 68.6} \\ \Rightarrow (\mu,\sigma)= (63.6,2.5),故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{A=59 \Rightarrow z={59-79\over 4.5} =-4.44\\ B=55 \Rightarrow z={55-69\over 3.7}= -3.78} \Rightarrow B科表現較佳\\ 變異係數CV\cases{A班:4.5/79 =0.057 \\ B班:3.7/69=0.054} \Rightarrow B科表現較穩定,故選\bbox[red,2pt]{(D)}$$
解答:$$樣本標準差\cases{s_X =\sqrt{{\sum x^2\over n-1}- {(\sum x)^2\over n(n-1)}} =\sqrt{{1.024861\over 12}- {3.627^2\over 13\cdot 12}} = 0.033\\s_Y =\sqrt{{\sum y^2\over n-1}- {(\sum y)^2\over n(n-1)}} =\sqrt{{1.27306\over 12}- {4.054^2\over 13\cdot 12}} =0.027} \\ \Rightarrow s_X \gt s_Y \Rightarrow B較穩定,標準差為0.027,故選\bbox[red,2pt]{(D)}$$
解答:$$P(B\mid A)={P(B\cap A)\over P(A)}=0.4 \Rightarrow P(B\cap A)=P(A)\times 0.4=0.8\times 0.4=0.32,故選\bbox[red,2pt]{(B)}$$
解答:$$A、B獨立\Rightarrow P(A\cap B)=P(A)P(B) =0.32 \Rightarrow P(B)=0.32\div P(A)=0.32\div 0.8=0.4\\ \Rightarrow (P(B),P(A\cap B)) = (0.4,0.32) ,故選\bbox[red,2pt]{(C)}$$
解答:$$本題與107年地方特考第8題幾乎一樣,但答案選項截然不同,公布的答案是\bbox[red,2pt]{(C)},但107年公布的答案是0.2231$$
解答:$$ 由上題附表可知P(\lambda=1.5,X=0)= e^{-1.5}=0.2231\\ 由於平均10天收3張訂單\Rightarrow 平均5天收到1.5張訂單 \\\Rightarrow P(\lambda=1.5,X=3)={e^{-1.5}\cdot 1.5^3\over 3!} = {0.2231\times 1.5^3 \over 6}= 0.1255,故選\bbox[red,2pt]{(B)}$$
解答:$$\cases{P(B\mid A_1)=P(B\cap A_1)/P(A_1)= 0.6 \Rightarrow P(B\cap A_1)=0.6\times 0.3=0.18\\ P(B\mid A_2)=P(B\cap A_2)/P(A_2)= 0.3 \Rightarrow P(B\cap A_2)=0.3\times 0.7=0.21}\\ 由於B= (B\cap A_1) \cup (B\cap A_2) \Rightarrow P(B)=0.18+0.21= 0.39,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{P(B\mid A_1)=P(B\cap A_1)/P(A_1)= 0.6 \Rightarrow P(B\cap A_1)=0.6\times 0.3=0.18\\ P(B\mid A_2)=P(B\cap A_2)/P(A_2)= 0.3 \Rightarrow P(B\cap A_2)=0.3\times 0.7=0.21}\\ 由於B= (B\cap A_1) \cup (B\cap A_2) \Rightarrow P(B)=0.18+0.21= 0.39\\ P(A_1\mid B)={P(A_1 \cap B)\over P(B)} ={0.18\over 0.39}={6\over 13} = 0.46,故選\bbox[red,2pt]{(D)}$$
解答:$$中獎機率p={4\over 350},而至少一張沒中獎機率=1-全中獎機率=1-p^4 \approx 1,故選\bbox[red,2pt]{(D)}$$
解答:$$中獎機率p={4\over 350}\Rightarrow 平均每張獲利:({4\over 350}(10000+ 5000+1500+ 500)-50\times 4)\div 4\\ =-1.42857,,故選\bbox[red,2pt]{(A)}$$
解答:$$X\sim B(p=0.1,n=100) \Rightarrow \cases{\mu=np = 10\\ \sigma = \sqrt{np(1-p)} =3} \\ \Rightarrow P(X\gt 11.5)=P(Z\gt {11.5-10\over 3})= P(Z\gt 0.5) = 1-0.6915=0.3085,故選\bbox[red,2pt]{(B)}\\本題\bbox[blue,2pt]{(送分)},應該是(B),(D)都算對$$
解答:$$100\times 0.3086 \approx 31,故選\bbox[red,2pt]{(D)}$$
解答:$$P越接近0.5,變異越大,因此所需樣本數越大,故選\bbox[red,2pt]{(B)}$$
解答:$$觀察值 \Rightarrow \begin{array}{c|lll} 政黨 & 同意& 不同意 &沒意見\\\hline A & O_{1,1}=50 & O_{1,2}=24 & O_{1,3}=18 \\ B& O_{2,1}=42 & O_{2,2}=20 & O_{2,3}=14 \\ \text{Others }& O_{3,1}=10 & O_{3,2}=16 & O_{3,3}=6 \end{array}\\ \Rightarrow \cases{同意人數=50+42+10=102\\ 不同意人數=24+ 20+16=60 \\ 沒意見人數=18 +14+ 6= 38 \\A黨人數=50+24+18=92 \\B黨人數= 42+20 +14=76 \\  \text{Others}=10 +16+6=32 } \\  期望值\Rightarrow \begin{array}{c|lll} 政黨 & 同意& 不同意 &沒意見\\\hline A & E_{1,1}=92\cdot 102/200 & E_{1,2}=92\cdot 60/200 & E_{1,3}=92\cdot 38/200 \\ B& E_{2,1}=76 \cdot 102/200& E_{2,2}=76 \cdot 60/200 & E_{2,3}=76 \cdot 38/200 \\ \text{Others }& E_{3,1}=32  \cdot 102/200 & E_{3,2}=32  \cdot 60/200 & E_{3,3}=32  \cdot 38/200 \end{array}\\ 卡方檢定統計值\chi^2= \sum_{i,j=1}^3 {(O_{i,j}-E_{i,j})^2 \over E_{i,j}}=8.03 \lt \chi^2_{4,0.05}=9.488\\ \Rightarrow 不能拒絕H_0,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{b=n-1=10-1=9 \\c=1122.1-804.062 = 318.038 }\Rightarrow a=b-1=9-1=8 \\\Rightarrow d=c/a =318.038/8= 39.75475 \Rightarrow e = 804.062/d = 804.062/39.75475= 20.226\\ \Rightarrow (a,b,c,d,e)=(8,9,318.038,39.75475,20.226),故選\bbox[red,2pt]{(C)}$$

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