113年地方三等經建農業-統計詳解

113年特種考試地方政府公務人員考試

等 別：三等考試
類 科：經建行政、農業行政
科 目：統計學

解答: $$\textbf{(一) }X\sim N(140,3.5^2) \Rightarrow P(X\gt 140.5) =P(Z\gt {140.5-140\over 3.5/\sqrt{25}}) =P(Z\gt 0.71)\\\quad 查考卷附表1,可得P(Z\le 0.71)=0.7611 \Rightarrow P(Z\gt 0.714) =1-0.7611= \bbox[red, 2pt]{0.2389} \\ \textbf{(二) }P(138\lt X\lt 142) =P({138-140\over 5/\sqrt{25}} \lt t\lt {142-140\over 5/\sqrt{25}}) =P(-2\lt t\lt 2) \\\quad 查考卷附表2,在df=24,可得 P(t\gt 2.064)=0.025 \Rightarrow P(|t|\lt 2) =1-2\times 0.025= \bbox[red, 2pt]{0.95}$$

解答: $$\textbf{(一) }f(x_i)= \theta x_i^{\theta-1} \Rightarrow L(\theta) =\prod_{i=1}^n f(x_i)= \theta^n (x_1 \cdot x_2 \cdots x_n)^{\theta-1} \Rightarrow \ln L(\theta) =n\ln \theta+(\theta-1)\ln(x_1\cdot x_2 \cdots x_n) \\ \quad \Rightarrow {d\over d\theta} \ln L(\theta) ={n\over \theta} +\ln(x_1\cdot x_2 \cdots x_n) =0 \Rightarrow \hat \theta=\bbox[red, 2pt]{-{n\over \sum_{i=1}^n \ln x_i}} \\\textbf{(二)) } E(X)= \int_0^1 xf(x)\,dx =\int_0^1 \theta x^\theta\,dx = \left. \left[ {\theta\over \theta+1} x^{\theta+1}\right] \right|_0^1 ={\theta\over \theta+1} =\frac{-{n\over \sum_{i=1}^n \ln x_i}}{1-{n\over \sum_{i=1}^n \ln x_i}} = \bbox[red, 2pt]{n\over n-\sum_{i=1}^n \ln x_i}$$

解答: $$\cases{H_0:使用前與使用後成績無差異\\ H_1:使用後成績高於使用前} \Rightarrow \begin{array} {}& X(使用前)& Y(使用後)& D=Y-X & D^2\\ \hline & 84& 90 & 6& 36\\ & 72&86& 14 & 196\\ & 61& 72& 11 & 121\\ & 78& 80& 2& 4\\ & 83& 85& 2& 4\\ & 86& 88 & 2& 4\\ & 71& 80& 9& 81\\\hdashline \sum & 535& 581& 46 &446\end{array} \\ \Rightarrow   \cases{\bar d=46/7=6.57 \\ \bar x=535/7=76.43\\ \bar y=581/7 = 83}  \Rightarrow s_d =\sqrt{{1\over n-1}\left(\sum D^2-(\sum D)^2/n \right)} = \sqrt{{ 1\over 6}(446-46^2/7)} =4.89 \\ \Rightarrow 檢定統計量t={\bar d\over s_d/\sqrt n} = {6.57 \over 4.89/\sqrt 7}=3.55\\ 查考卷附表(單尾,但此題為雙尾檢定)可得t(\alpha= 0.025,df=6) =2.447 \\ \Rightarrow 3.55\gt 2.447 \Rightarrow 拒絕H_0 \Rightarrow \bbox[red, 2pt]{會提高學習成效}\\ 考量p值檢定，查考卷附表可知:p(df=6,t\gt 3.143) =0.01 \Rightarrow p(|t|\gt 3.143)=0.02 \\ \Rightarrow p(|t|\gt 3.55) \lt 0.02 \lt 0.05(顯著水準)$$

解答: $$\begin{array} {} &A & & B & & C\\\hdashline &x_A & x_A^2 & x_B& x_B^2 & x_C& x_C^2 \\\hline &20 & 400  &27 &729 & 21& 441\\ &27 &729& 24&576& 17& 289 \\ &26& 676& 21& 441& 20& 400\\ &24& 576& 25&625& 22& 484 \\\hline \sum & 97& 2381& 97 & 2371& 80& 1614\end{array} \Rightarrow \cases{\overline{x_A} =97/4=24.25\\ \overline{x_B} =97/4=24.25\\ \overline{x_C} =80/4=20\\ \bar x=(97+97+80)/12=22.83} \\\Rightarrow \cases{SS_B=4(\overline{x_A}-\bar x)^2+ 4(\overline{x_B}-\bar x)^2+ 4(\overline{x_C}-\bar x)^2= 48.17\\ SS_W= \sum_{i=1}^4(x_{Ai}-\overline{x_A})^2 +\sum_{i=1}^4(x_{Bi}-\overline{x_B})^2  +\sum_{i=1}^4(x_{Ci}-\overline{x_C})^2 \\\qquad \;=28.75 +18.75+14=61.5} \\ \Rightarrow \begin{array} {} Source & SS& df & MS & F\\\hline SS_B& 48.17& 3-1=2 & 48.17/2=24.09& 24.09/6.83= 3.52\\ SS_W & 61.5 & 11-2=9 & 61.5/9= 6.83\\ \hdashline total & 109.67 & 4\cdot 3-1=11\\\hline\end{array}\\ \Rightarrow F=3.52 \gt F(2,9)_{0.1}=3 \Rightarrow 達顯著性,即三種肥料對農作物的平均生長高度\bbox[red, 2pt]{不同}$$

解答: $$\textbf{(一) }\begin{array} {} & X & Y & X^2 &Y^2 & XY\\\hline & 2& 20& 4& 400 & 40\\ &3& 20 & 9& 400& 60\\ & 4& 60& 16& 3600& 240\\ & 2 & 25 & 4& 625& 50\\ & 5& 75& 25& 5625& 375\\ & 1& 10 & 1& 100 & 10\\ & 3& 40& 9& 1600& 120\\ & 4 & 60& 16& 3600& 240 \\\hdashline \sum & 24& 310& 84& 15950& 1135\end{array} \Rightarrow \cases{\bar x=24/8=3\\ \bar y=310/8=155/4 } \\ \Rightarrow b={ n\sum xy-\sum x\sum y \over n\sum x^2-(\sum x)^2} ={8\cdot 1135-24\cdot 310\over 8\cdot 84-24^2} ={205\over 12} \Rightarrow a=\bar y-b\bar x ={155\over 4}-{205\over 12}\cdot 3=-{25\over 2} \\ \Rightarrow y=bx+a \Rightarrow \bbox[red, 2pt]{y={205\over 12}x-{25\over 2}}\\\textbf{(二) }檢定統計量F={MSR\over MSE} ={SSR\over SSE/(8-2)}={b_1^2SS_{xx} \over (SS_{yy}-b_1^2 SS_{xx})/6}\\= {6b_1^2 \left( \sum x^2 -(\sum x)^2/n\right) \over   \sum y^2-(\sum y)^2/n -b_1^2 \left( \sum x^2 -(\sum x)^2/n \right)} ={6(205/12)^2 (84-24^2/8) \over 15950-310^2/8-(205/12)^2 (84-24^2/8) } =48.26\\ F_\alpha(1,n-2)=F_{0.05}(1,6) =(t_{0.025}(6))^2 =2.447^2 =5.99 (考卷沒有F表，改用t表)\\ \Rightarrow F=48 \gt F_{0.05}(1,6) \Rightarrow \bbox[red, 2pt]{有顯著影響} \\\textbf{(三) }R^2= {SSR\over SST}={b_1^2SS_{xx} \over  SS_{yy} } =  {b_1^2 (\sum x^2-(\sum x)^2/n) \over \sum y^2-(\sum y)^2/n} ={(205/12)^2 \left( 84-24^2/8 \right) \over 15950-310^2/8} = \bbox[red, 2pt]{0.889}$$

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解題僅供參考，高普考歷年試題及詳解