108年普考-經建類-統計學概要-詳解

108年公務人員普通考試

類 科 ：經建行政、工業行政、交通技術
科 目：統計學概要

解：
(一)直方圖適合連續性資料，各長條相連，可呈現整體資料分布情形；長條圖主要以分類資料呈現，長條間有間隔，圖型以分類資料呈現。
(二)$$P(X> 60) =P\left({X-76\over 8}> {60-76\over 8}\right)= P(Z>-2)=1-0.05\div 2= \bbox[red, 2pt]{97.5\%}$$

(三)$$\begin{cases}A: Z_A={4.3-5 \over 0.5} =-1.4 \\ B: Z_B={32-46\over 1.5}= -9.3\\ C: Z_C={63-80\over 20}= -0.85 \end{cases} \Rightarrow Z_C> Z_A > Z_B \Rightarrow \bbox[red, 2pt]{C考試}具有最高相對位置$$

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解：

(一)$$0.32\times 0.2= \bbox[red, 2pt]{0.064}$$(二)$$1-(1-0.32)(1-0.2)=1-0.68\times 0.8=1-0.544= \bbox[red, 2pt]{0.456}$$(三)$$(1-0.32)(1-0.2)=0.68\times 0.8= \bbox[red, 2pt]{0.544}$$(四)$$(1-0.32)\times 0.2=0.68\times 0.2 = \bbox[red, 2pt]{0.136}$$(五) $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

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解：$$\begin{array}{|l|c|c|c|}\hline
年齡(X) & 沒有求償 & 有求償 &總和\\\hline
18< X\le 25 & 187 & 66 & 253\\\hline
25< X\le 40 & 228 & 58 & 286\\\hline
40< X\le 55 & 380 & 62 & 442\\\hline
55< X &186 & 33 & 219 \\\hline
總和 & 981 &  219 & 1200\\\hline
\end{array} \\\Rightarrow \begin{array}{|l|l|c|}\hline
 沒有求償 & 有求償 &總和\\\hline
 O_1=187, & O_2=66, & 253\\
E_1=253\times {981\over 1200}=206.83&E_2=253\times {219\over 1200}=46.17&\\\hline
 O_3=228 & O_4=58 & 286\\
E_3=286\times {981\over 1200}=233.81&E_4=286\times {219\over 1200}=52.19&\\\hline
 O_5=380 & O_6=62 & 442\\
E_5=442\times {981\over 1200}=361.34&E_6=442\times {219\over 1200}=80.66&\\\hline
O_7=186 & O_8=33 & 219 \\
E_7=219\times {981\over 1200}=179.03&E_8=219\times {219\over 1200}=39.97&\\\hline
 981 &  219 & 1200\\\hline
\end{array}\\
\begin{cases} H_0:求償與年齡無關 \\H_1:求償與年齡有關\\ \alpha=0.05\end{cases}$$
(一)$$拒絕區域R=\{\chi^2\mid \chi^2 > \chi^2_{df=3,\alpha=0.05}=7.81473(查表)\\ 檢定統計值\chi^2=\sum_{i=1}^8 {(O_i-E_i)^2 \over E_i} = {(187-206.83)^2 \over 206.83}+ {(66-46.17)^2 \over 46.17}+ {(228-233.81)^2 \over 233.81}\\+ {(58-52.19)^2 \over 52.19}+ {(380-361.34)^2 \over 361.34}+ {(62-80.66)^2 \over 80.66}+ {(186-179.03)^2 \over 179.03}+ {(33-39.97)^2 \over 39.97} \\=17.98 \in R\Rightarrow 拒絕H_0 \Rightarrow  \bbox[red, 2pt]{有關}$$(二)$$查表可知P(\chi_{df=3}^2=12.8381)=0.005 \Rightarrow P(\chi_{df=3}^2 = 17.98)<0.005 \Rightarrow 拒絕H_0 \Rightarrow  \bbox[red, 2pt]{有關}$$

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解：
$$\begin{array} {|r|r|}\hline i & x_i(女) & y_i(男) & x_i^2 & y_i^2 & x_iy_i  \\ \hline 1&35&65&1225&4225&2275\\\hline
2&51&204&2601&41616 &10404\\\hline
3&29&101&841&10201&2929\\\hline
4&5&12&25&144 &60\\\hline
5&12&43&144&1849&516\\\hline
6&4&6&16&36&24\\\hline
7&10&32&100&1024&320\\\hline
\sum&146&463&4952&59095&16528\\\hline
  \end{array}\\
\Rightarrow \begin{cases} s_x^2=\sum x_i^2-{(\sum x_i)^2 \over n}=4952-146^2/7=1906.86\\ s_y^2=\sum y_i^2-{(\sum y_i)^2 \over n}=59095-463^2/7=28470.86 \\ s_{xy} =\sum x_iy_i-{(\sum x_i)( \sum y_i) \over n} =16528-146\times 463/7= 6871.14\end{cases}
$$(一)$$迴歸方程式\hat y=b_0+b_1x,其中b_1={s_{xy}\over s_x^2} = {6871.14 \over 1906.86}= 3.6; \\ 經直線經過 (\bar x,\bar y)=\left({146\over 7},{463\over 7} \right) \Rightarrow b_0= {463\over 7}-3.6\times {146\over 7}= -8.94 \\ \Rightarrow \hat y=-8.94+3.6x \Rightarrow \begin{array}{|c|r|r|r|r|r|} \hline i& x_i & y_i & \hat y_i & y_i-\hat y_i & (y_i-\hat y_i)^2\\\hline1 &35 &65 &117.06 &-52.06 &2710.24 \\\hline
2 &51 &204 &174.66 &29.34 &860.84 \\\hline
3 &29 &101 &95.46 &5.54 &30.69 \\\hline
4 &5 &12 &9.06 &2.94 &8.64 \\\hline
5 &12 &43 &34.26 &8.74 &76.39 \\\hline
6 &4 &6 &5.46 &0.54 &0.29 \\\hline
7 &10 &32 &27.06 &4.94 &24.40 \\\hline
\sum &146 &463 &463.02 &-0.02 &3711.50 \\\hline
\end{array} \\\Rightarrow S_{b1}= \sqrt{\sum(y_i-\hat y_i)^2 \over n-2}/\sqrt{\sum x_i^2-(\sum x_i)^2/n}= \sqrt{3711.5\over 5}/\sqrt{4952-146^2/7}= 27.25/43.67\\=0.624\\
 \begin{cases} H_0:b_1=0\\ H_1:b_1\ne 0\\ \alpha=0.05\end{cases} \Rightarrow 拒絕區域R=\{ t\mid |t|>t_{\alpha/2,n-2}=t_{0.025,5}=2.571(查表)\}\\ 檢定統計量t={b_1\over S_{b1}}= {3.6\over 0.624}=5.77 \in R \Rightarrow 拒絕H_0 \Rightarrow 
女性員工與男性員工人數\bbox[red, 2pt]{有顯著相關}$$(二)$$x_0=21代入迴歸直線\hat y=-8.94+3.6x =-8.94+3.6\times 21=66.66\\
其1-\alpha=95\%的預測區間為\pm t_{\alpha/2,n-2}\times\sqrt{MSE}\times \sqrt{1+{1\over n}+{(x_0-\bar x)^2\over s_x^2}}\\ =\pm 2.571\times\sqrt{3711.5\over 5}\times \sqrt{1+{1\over 7}+{(21-146/7)^2\over 1906.86}} =\pm 2.571\times \sqrt{742.3}\times \sqrt{1.143}\\ =\pm 74.89 \Rightarrow 預測區間為[66.66-74.89,66.66+74.89]= \bbox[red, 2pt]{[-8.23,141.55]}$$

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