115 年國立中正大學企業管理學系碩士班丙組《統計學》考古題與詳解

Question 1

[題組 1~6] 24%(每小題4分)
Salary information for a random sample of male and female employees of a large company is shown below.

	Male	Female
Sample Size	64	36
Sample Mean Salary (in $1,000)	44	41
Sample Variance	128	72

The point estimate of the difference between the means of the two populations is ____.
The standard error for the difference between the two means is ____.
The 95% confidence interval for the difference between the means of the two populations is ____.
If you are interested in testing whether or not the average salary of males is significantly greater than that of females, the test statistic is ____.
The p-value is ____.
At 95% confidence, the conclusion is the ____.

Accepted Answer

本題組主要在檢定兩獨立樣本平均數的差異。

核心觀念：獨立樣本 t 檢定與信賴區間。

已知兩獨立樣本的資料，我們要計算平均數差異的點估計、標準誤、信賴區間，以及進行假說檢定。由於樣本數皆大於 30 (Male: $n_1 = 64$ , Female: $n_2 = 36$ )，我們可以使用 Z 分配來近似 t 分配。

1. 點估計 (Point Estimate)
兩母體平均數差異的點估計為兩樣本平均數的差異。
$\hat{\mu}_1 - \hat{\mu}_2 = \bar{x}_1 - \bar{x}_2 = 44 - 41 = 3$ (千美元)
【答案】3

2. 標準誤 (Standard Error)
兩樣本平均數差異的標準誤計算公式為：
$SE(\bar{x}_1 - \bar{x}_2) = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$
其中， $s_1^2$ 和 $s_2^2$ 分別是樣本變異數， $n_1$ 和 $n_2$ 分別是樣本數。
$SE(\bar{x}_1 - \bar{x}_2) = \sqrt{\frac{128}{64} + \frac{72}{36}} = \sqrt{2 + 2} = \sqrt{4} = 2$ (千美元)
【答案】2

3. 95% 信賴區間 (95% Confidence Interval)
信賴區間的計算公式為：
$(\bar{x}_1 - \bar{x}_2) \pm z_{\alpha/2} \times SE(\bar{x}_1 - \bar{x}_2)$
對於 95% 信賴水準， $\alpha = 0.05$ ， $\alpha/2 = 0.025$ 。

Question 2

[題組 7~9] 12%(每小題4分)
The following information was obtained from matched samples. The daily production rates for a sample of workers before and after a training program are shown below.

Worker	Before	After
1	20	22
2	25	23
3	27	27
4	23	20
5	22	25
6	20	19
7	17	18

The point estimate for the difference between the means of the two populations is ____.

Accepted Answer

本題組針對配對樣本，檢定訓練前後的生產率差異。

核心觀念：配對樣本 t 檢定。

配對樣本的分析是計算樣本差異（配對的兩數值之差），然後對這些差異值進行單一樣本檢定。

7. 點估計 (Point Estimate)
首先，計算每位員工訓練前後的生產率差異 $d = \text{After} - \text{Before}$ ：
Worker 1: $22 - 20 = 2$
Worker 2: $23 - 25 = -2$

Question 3

[題組 7~9] 12%(每小題4分)
The following information was obtained from matched samples. The daily production rates for a sample of workers before and after a training program are shown below.

Worker	Before	After
1	20	22
2	25	23
3	27	27
4	23	20
5	22	25
6	20	19
7	17	18

The point estimate for the difference between the means of the two populations is ____.

Accepted Answer

延續上一題關於配對樣本的分析，此題要求計算檢定統計量。

核心觀念：配對樣本 t 檢定統計量的計算。

我們已經計算出樣本差異的平均數 $\bar{d} = 0$ 。
接下來，我們需要計算樣本差異的標準差 $s_d$ 。
首先，計算樣本差異的變異數 $s_d^2$ ：
$s_d^2 = \frac{1}{n-1} \sum_{i=1}^{n} (d_i - \bar{d})^2$
其中， $n=7$ (樣本數)， $\bar{d}=0$ 。
差異值 $d_i$ 為：2, -2, 0, -3, 3, -1, 1。
$(d_i - \bar{d})^2$ 為： $(2-0)^2=4$ , $(-2-0)^2=4$ , $(0-0)^2=0$ , $(-3-0)^2=9$ , $(3-0)^2=9$ , $(-1-0)^2=1$ , $(1-0)^2=1$ 。

Question 4

[題組 7~9] 12%(每小題4分)
The following information was obtained from matched samples. The daily production rates for a sample of workers before and after a training program are shown below.

Worker	Before	After
1	20	22
2	25	23
3	27	27
4	23	20
5	22	25
6	20	19
7	17	18

The point estimate for the difference between the means of the two populations is ____.

Accepted Answer

此題要求根據上一題計算出的檢定統計量，得出統計檢定的結論。

核心觀念：配對樣本 t 檢定結論的判斷。

我們已經計算出檢定統計量 $t = 0$ 。
配對樣本 t 檢定的自由度 (df) 為 $n-1 = 7-1 = 6$ 。

為了得出結論，我們需要設定顯著水準 $\alpha$ 。題目沒有明確給出 $\alpha$ ，但通常在學術考試中，若未指定，則預設為 $\alpha = 0.05$ 。

我們有兩種判斷方法：

臨界值法 (Critical Value Approach):
對於雙尾檢定 (因為 $H_0: \mu_d = 0$ )，在 $\alpha = 0.05$ 且 df = 6 時，查 t 分配臨界值表，臨界值為 $t_{\alpha/2, df} = t_{0.025, 6} \approx 2.447$ 。
檢定區域為 $|t| > 2.447$ 。

Question 5

[題組 10~13] 16%(每小題4分)
It has been suggested that night shift-workers show more variability in their output levels than day workers. Below, you are given the results of two independent random samples.

	Night Shift	Day Shift
Sample Size	9	8
Sample Mean	520	540
Sample Variance	25	23

The null hypothesis is to be tested at the 5% level of significance. Based on the critical value approach, the decision rule for the test is ____.

Accepted Answer

本題組聚焦於比較兩獨立樣本的變異數。

核心觀念：變異數的檢定（F 檢定）。

題目提到「night shift-workers show more variability in their output levels than day workers」，這暗示我們要檢定夜班工人的變異數是否大於日班工人。

虛無假設 ( $H_0$ )：夜班工人的變異數 $\sigma_1^2$ 不大於日班工人的變異數 $\sigma_2^2$ 。
對立假設 ( $H_1$ )：夜班工人的變異數 $\sigma_1^2$ 大於日班工人的變異數 $\sigma_2^2$ 。

由於我們比較的是兩個母體變異數的大小，所以虛無假設通常設定為相等，而對立假設則根據題目敘述來設定。

在此，我們將夜班視為母體 1 ( $n_1=9, s_1^2=25$ )，日班視為母體 2 ( $n_2=8, s_2^2=23$ )。

Question 6

[題組 10~13] 16%(每小題4分)
It has been suggested that night shift-workers show more variability in their output levels than day workers. Below, you are given the results of two independent random samples.

	Night Shift	Day Shift
Sample Size	9	8
Sample Mean	520	540
Sample Variance	25	23

The null and alternative hypotheses to be tested are ____.

Accepted Answer

此題要求計算用來檢定兩個母體變異數是否相等的檢定統計量。

核心觀念：變異數比例的 F 檢定統計量。

檢定兩個獨立樣本變異數的統計量是 F 統計量，其計算公式為兩個樣本變異數的比值：
$F = \frac{s_1^2}{s_2^2}$
其中， $s_1^2$ 是第一個樣本的變異數， $s_2^2$ 是第二個樣本的變異數。

在本題中，我們比較的是「Night Shift」與「Day Shift」的變異數。

Question 7

[題組 10~13] 16%(每小題4分)
It has been suggested that night shift-workers show more variability in their output levels than day workers. Below, you are given the results of two independent random samples.

	Night Shift	Day Shift
Sample Size	9	8
Sample Mean	520	540
Sample Variance	25	23

The null hypothesis is to be tested at the 5% level of significance. Based on the critical value approach, the decision rule for the test is ____.

Accepted Answer

此題要求根據給定的顯著水準和臨界值法，制定變異數檢定的決策規則。

核心觀念：F 檢定的臨界值與決策規則。

我們已經設定了檢定假設：
$H_0: \sigma_1^2 = \sigma_2^2$
$H_1: \sigma_1^2 > \sigma_2^2$ (右尾檢定)

顯著水準 $\alpha = 0.05$ 。
檢定統計量為 $F = s_1^2 / s_2^2$ 。
分子自由度 $df_1 = n_1 - 1 = 9 - 1 = 8$ 。
分母自由度 $df_2 = n_2 - 1 = 8 - 1 = 7$ 。

決策規則是：如果檢定統計量 $F$ 大於臨界值 $F_{\alpha, df_1, df_2}$ ，則拒絕虛無假設 $H_0$ 。

Question 8

[題組 10~13] 16%(每小題4分)
It has been suggested that night shift-workers show more variability in their output levels than day workers. Below, you are given the results of two independent random samples.

	Night Shift	Day Shift
Sample Size	9	8
Sample Mean	520	540
Sample Variance	25	23

The null hypothesis is to be tested at the 5% level of significance. Based on the critical value approach, the decision rule for the test is ____.

Accepted Answer

此題要求根據計算出的檢定統計量和決策規則，得出統計檢定的結論。 核心觀念：F 檢定結論的判斷。 我們已經計算出： 檢定統計量 $F = 1.087$。 決策規則為：若 $F 3.59$，則拒絕 $H0$。 現在，我們將計算出的檢定統計量與決策規則進行比較： $1.087 
gtr 3.59$。

Question 9

[題組 14~16] 12%(每小题4分)
An appliance store sells three models of television sets. Average unit selling price, in dollars, and quantities sold, in thousands, in 2022 and 2024 are given below.

Model	\multicolumn{2}{c	}{2022}	\multicolumn{2}{c	}{2024}
	Price	Quant.	Price	Quant.
XX	460	2.1	605	2.0
YY	580	1.7	620	1.6
ZZ	710	0.9	722	0.7

Using 2022 as a base year, the simple unweighted price index for 2024 is ____.

Accepted Answer

本題組涉及物價指數的計算，包括簡單加權、拉斯皮爾、帕許指數。

核心觀念：簡單未加權物價指數 (Simple Unweighted Price Index)。

簡單未加權物價指數的計算方式是將所有商品的當期價格加總，除以基期所有商品價格的加總，然後乘以 100。

\text{Simple Unweighted Price Index} = \frac{\sum P_t}{\sum P_0} \times 100

其中， $P_t$ 是當期 (2024 年) 的價格， $P_0$ 是基期 (2022 年) 的價格。

Question 10

[題組 14~16] 12%(每小题4分)
An appliance store sells three models of television sets. Average unit selling price, in dollars, and quantities sold, in thousands, in 2022 and 2024 are given below.

Model	\multicolumn{2}{c	}{2022}	\multicolumn{2}{c	}{2024}
	Price	Quant.	Price	Quant.
XX	460	2.1	605	2.0
YY	580	1.7	620	1.6
ZZ	710	0.9	722	0.7

Using 2022 as a base year, the simple unweighted price index for 2024 is ____.

Accepted Answer

此題要求計算拉斯皮爾 (Laspeyres) 物價指數。

核心觀念：拉斯皮爾物價指數 (Laspeyres Price Index)。

拉斯皮爾指數使用基期的數量作為權數，計算當期價格與基期價格的加權平均比率。

\text{Laspeyres Index} = \frac{\sum (P_t \times Q_0)}{\sum (P_0 \times Q_0)} \times 100

其中：
$P_t$ = 當期 (2024) 價格
$P_0$ = 基期 (2022) 價格
$Q_0$ = 基期 (2022) 數量

計算分子 $\sum (P_t \times Q_0)$ ：
XX: $605 \times 2.1 = 1270.5$
YY: $620 \times 1.7 = 1054.0$

Question 11

[題組 14~16] 12%(每小题4分)
An appliance store sells three models of television sets. Average unit selling price, in dollars, and quantities sold, in thousands, in 2022 and 2024 are given below.

Model	\multicolumn{2}{c	}{2022}	\multicolumn{2}{c	}{2024}
	Price	Quant.	Price	Quant.
XX	460	2.1	605	2.0
YY	580	1.7	620	1.6
ZZ	710	0.9	722	0.7

Using 2022 as a base year, the simple unweighted price index for 2024 is ____.

Accepted Answer

此題要求計算帕許 (Paasche) 物價指數。

核心觀念：帕許物價指數 (Paasche Price Index)。

帕許指數使用當期的數量作為權數，計算當期價格與基期價格的加權平均比率。

\text{Paasche Index} = \frac{\sum (P_t \times Q_t)}{\sum (P_0 \times Q_t)} \times 100

其中：
$P_t$ = 當期 (2024) 價格
$P_0$ = 基期 (2022) 價格
$Q_t$ = 當期 (2024) 數量

計算分子 $\sum (P_t \times Q_t)$ ：
XX: $605 \times 2.0 = 1210.0$
YY: $620 \times 1.6 = 992.0$

Question 12

[題組 17~21] 10%(每小题2分)
Regression analysis was applied between sales data (in $1,000s) and advertising data (in$ 100s) and the following information was obtained.
$\hat{y} = 12 + 1.8 x$
$n = 17$
$SSR = 225$
$SSE = 75$
$s_{b1} = 0.2683$

Based on the above estimated regression equation, if advertising is $3,000, then the point estimate for sales (in dollars) is ____.

Accepted Answer

本題組為迴歸分析相關問題。

核心觀念：迴歸模型的點估計。

題目給出迴歸直線方程式： $\hat{y} = 12 + 1.8x$ 。
其中， $y$ 代表 sales (in $1,000s)，$ x $代表 advertising (in$ 100s)。
題目問當 advertising 是

Question 13

[題組 17~21] 10%(每小题2分)
Regression analysis was applied between sales data (in $1,000s) and advertising data (in$ 100s) and the following information was obtained.
$\hat{y} = 12 + 1.8 x$
$n = 17$
$SSR = 225$
$SSE = 75$
$s_{b1} = 0.2683$

The F statistic computed from the above data is ____.

Accepted Answer

此題要求計算迴歸模型的 F 統計量。

核心觀念：迴歸模型的 F 檢定。

F 統計量用於檢定整個迴歸模型的顯著性，即檢定自變數是否能顯著解釋應變數的變異。
F 統計量的計算公式為：

F = \frac{MSR}{MSE}

其中， $MSR$ 是迴歸均方 (Mean Square Regression)， $MSE$ 是殘差均方 (Mean Square Error)。

Question 14

[題組 17~21] 10%(每小题2分)
Regression analysis was applied between sales data (in $1,000s) and advertising data (in$ 100s) and the following information was obtained.
$\hat{y} = 12 + 1.8 x$
$n = 17$
$SSR = 225$
$SSE = 75$
$s_{b1} = 0.2683$

The critical F value at $\alpha = 0.05$ is ____.

Accepted Answer

此題要求查找迴歸模型 F 檢定的臨界值。

核心觀念：F 分配的臨界值。

F 檢定的臨界值取決於顯著水準 $\alpha$ 以及迴歸模型的自由度。
在此簡單線性迴歸中：
分子自由度 (迴歸自由度) $df_1 = p = 1$ (自變數個數)。
分母自由度 (殘差自由度) $df_2 = n - p - 1 = 17 - 1 - 1 = 15$ 。
顯著水準 $\alpha = 0.05$ 。

Question 15

[題組 17~21] 10%(每小题2分)
Regression analysis was applied between sales data (in $1,000s) and advertising data (in$ 100s) and the following information was obtained.
$\hat{y} = 12 + 1.8 x$
$n = 17$
$SSR = 225$
$SSE = 75$
$s_{b1} = 0.2683$

The t statistic for testing the significance of the slope is ____.

Accepted Answer

此題要求計算檢定迴歸模型斜率顯著性的 t 統計量。

核心觀念：迴歸斜率的 t 檢定。

檢定迴歸模型斜率是否顯著的 t 統計量計算公式為：

t = \frac{b_1 - \beta_{1,0}}{s_{b1}}

其中：
$b_1$ 是樣本迴歸直線的斜率。
$\beta_{1,0}$ 是虛無假設中的母體斜率值，通常檢定斜率是否為零，所以 $\beta_{1,0} = 0$ 。

Question 16

[題組 17~21] 10%(每小题2分)
Regression analysis was applied between sales data (in $1,000s) and advertising data (in$ 100s) and the following information was obtained.
$\hat{y} = 12 + 1.8 x$
$n = 17$
$SSR = 225$
$SSE = 75$
$s_{b1} = 0.2683$

The critical t value for testing the significance of the slope is ____.

Accepted Answer

此題要求查找檢定迴歸模型斜率顯著性的 t 檢定臨界值。

核心觀念：t 分配的臨界值。

檢定迴歸斜率顯著性的 t 檢定，其自由度為 $df = n - p - 1$ ，其中 $n$ 是樣本數， $p$ 是自變數的個數。
在此簡單線性迴歸中， $n=17$ ， $p=1$ 。
所以，自由度 $df = 17 - 1 - 1 = 15$ 。

Question 17

[題組 22~25] 16%(每小題4分)
Employees of MNM Corporation are about to undergo a retraining program. Management is trying to determine which of three programs is the best. They believe that the effectiveness of the programs may be influenced by sex. A factorial experiment was designed. You are given the following information.

Factor A: Program	Male	Female
Program A	320	380
Program B	240	300
Program C	290	380

The partial ANOVA table is shown below:

Source of Variation	Sum of Squares	df	Mean Square	F
Factor A		2	18,075	12.76
Factor B	16,133.333	1		11.39
Interaction			758.334	.54
Error	8,500
Total	62,300

Is there a significant interaction effect between program and sex at $\alpha = 0.05$ ? Why?

Accepted Answer

核心觀念 本題考查二因子變異數分析（two-way ANOVA）中，平方和（Sum of Squares）、自由度（df）與均方（Mean Square）之間的關係： $$ MSA=\frac{SSA}{dfA} $$ 因此可變形為： $$ SSA=dfA	imes MSA $$ 其中： - $SSA$：Factor A（訓練方案）的平方和 - $dfA$：Factor A 的自由度 - $MSA$：Factor A 的均方 解題方法 由題目中的 ANOVA 表可得： $$ dfA=2,\qquad MSA=18{,}075 $$ 代入公式：

Question 18

[題組 22~25] 16%(每小題4分)
Employees of MNM Corporation are about to undergo a retraining program. Management is trying to determine which of three programs is the best. They believe that the effectiveness of the programs may be influenced by sex. A factorial experiment was designed. You are given the following information.

Factor A: Program	Male	Female
Program A	320	380
Program B	240	300
Program C	290	380

The partial ANOVA table is shown below:

Source of Variation	Sum of Squares	df	Mean Square	F
Factor A		2	18,075	12.76
Factor B	16,133.333	1		11.39
Interaction			758.334	.54
Error	8,500
Total	62,300

Is there a significant interaction effect between program and sex at $\alpha = 0.05$ ? Why?

Accepted Answer

此題要求計算 ANOVA 表中的 Mean Square Error (MSE)。

核心觀念：Mean Square Error (MSE) 的計算。

MSE 是誤差平方和 (SSE) 除以誤差自由度 (df_Error)。

MSE = \frac{SSE}{df_{Error}}

我們已知 SSE = 8500。
我們需要計算 $df_{Error}$ 。

從 ANOVA 表的結構，我們知道：
$df_{Total} = df_A + df_B + df_{AB} + df_{Error}$

我們已知：
$df_A = 2$ (Factor A has 3 levels)
$df_B = 1$ (Factor B has 2 levels)
$df_{AB} = df_A \times df_B = 2 \times 1 = 2$

因此， $df_{Total} = df_A + df_B + df_{AB} + df_{Error} = 2 + 1 + 2 + df_{Error} = 5 + df_{Error}$ 。
$df_{Total}$ 通常等於總樣本數 N 減 1。

讓我們嘗試從 F 值和 MS 值來反推 MSE。
我們有三個 F 值：

$F_A = \frac{MS_A}{MSE} = 12.76$
$F_B = \frac{MS_B}{MSE} = 11.39$
$F_{Interaction} = \frac{MS_{AB}}{MSE} = 0.54$

我們也知道：
$MS_A = 18075$
$MS_{AB} = 758.334$
$SS_B = 16133.333$ , $df_B = 1 \Rightarrow MS_B = 16133.333 / 1 = 16133.333$

利用 $F_A$ ：
$MSE = \frac{MS_A}{F_A} = \frac{18075}{12.76} \approx 1416.536$

Question 19

[題組 22~25] 16%(每小題4分)
Employees of MNM Corporation are about to undergo a retraining program. Management is trying to determine which of three programs is the best. They believe that the effectiveness of the programs may be influenced by sex. A factorial experiment was designed. You are given the following information.

Factor A: Program	Male	Female
Program A	320	380
Program B	240	300
Program C	290	380

The partial ANOVA table is shown below:

Source of Variation	Sum of Squares	df	Mean Square	F
Factor A		2	18,075	12.76
Factor B	16,133.333	1		11.39
Interaction			758.334	.54
Error	8,500
Total	62,300

Is there a significant interaction effect between program and sex at $\alpha = 0.05$ ? Why?

Accepted Answer

此題要求判斷 Factor B (Sex) 是否對結果有顯著影響。

核心觀念：二因子變異數分析中的因子效應檢定。

我們要檢定的是 Factor B (Sex) 的主效應。
在 ANOVA 表中，Factor B 的 F 值是 11.39。
顯著水準 $\alpha = 0.05$ 。

首先，我們需要找出 Factor B 的臨界 F 值。
Factor B 的自由度是 $df_B = 1$ 。
誤差自由度 $df_{Error}$ 。根據我們之前的推算，如果 MSE 約為 1416.54，則 $df_{Error} \approx 6$ 。
我們假設 $df_{Error} = 6$ 。
那麼，Factor B 的臨界 F 值為 $F_{\alpha, df_B, df_{Error}} = F_{0.05, 1, 6}$ 。

Question 20

[題組 22~25] 16%(每小題4分)
Employees of MNM Corporation are about to undergo a retraining program. Management is trying to determine which of three programs is the best. They believe that the effectiveness of the programs may be influenced by sex. A factorial experiment was designed. You are given the following information.

Factor A: Program	Male	Female
Program A	320	380
Program B	240	300
Program C	290	380

The partial ANOVA table is shown below:

Source of Variation	Sum of Squares	df	Mean Square	F
Factor A		2	18,075	12.76
Factor B	16,133.333	1		11.39
Interaction			758.334	.54
Error	8,500
Total	62,300

Is there a significant interaction effect between program and sex at $\alpha = 0.05$ ? Why?

Accepted Answer

此題要求判斷 Program 和 Sex 之間的交互作用是否顯著。

核心觀念：二因子變異數分析中的交互作用檢定。

我們要檢定的是 Program (Factor A) 和 Sex (Factor B) 之間的交互作用效應。
在 ANOVA 表中，Interaction 的 F 值是 0.54。
顯著水準 $\alpha = 0.05$ 。

首先，我們需要找出交互作用的臨界 F 值。
交互作用的自由度為 $df_{AB} = df_A \times df_B = 2 \times 1 = 2$ 。
誤差自由度 $df_{Error}$ 。根據我們之前的推算，如果 MSE 約為 1416.54，則 $df_{Error} \approx 6$ 。
我們假設 $df_{Error} = 6$ 。
那麼，交互作用的臨界 F 值為 $F_{\alpha, df_{AB}, df_{Error}} = F_{0.05, 2, 6}$ 。

Question 21

[題組 26~27] 10%(每小題5分)
In a random sample of UTC students 50% indicated they are business majors, 40% engineering majors, and 10% other majors. Of the business majors, 60% were females; whereas, 30% of engineering majors were females. Finally, 20% of the other majors were female.

What percentage of students in this sample was female?

Accepted Answer

本題組為機率問題，運用了條件機率和全機率定理。

核心觀念：全機率定理 (Law of Total Probability)。

題目給出了各主修的比例，以及在各主修中女性的比例。要求計算樣本中女性的總比例。
我們可以用全機率定理來解決這個問題。

設事件：
B = 學生主修商學 (Business Major)
E = 學生主修工程 (Engineering Major)
O = 學生主修其他 (Other Major)
F = 學生是女性 (Female)

已知機率：
$P(B) = 0.50$
$P(E) = 0.40$
$P(O) = 0.10$
(檢查： $0.50 + 0.40 + 0.10 = 1.00$ ，主修是互斥且窮盡的)

Question 22

[題組 26~27] 10%(每小題5分)
In a random sample of UTC students 50% indicated they are business majors, 40% engineering majors, and 10% other majors. Of the business majors, 60% were females; whereas, 30% of engineering majors were females. Finally, 20% of the other majors were female.

Given that a person is female, what is the probability that she is an engineering major?

Accepted Answer

此題要求計算在已知某學生是女性的條件下，該學生是工程主修的機率。

核心觀念：貝氏定理 (Bayes' Theorem)。

我們需要計算 $P(E|F)$ 。
根據貝氏定理：

P(E|F) = \frac{P(F|E) P(E)}{P(F)}

我們已經知道：
$P(F|E) = 0.30$ (工程主修的學生中，30% 是女性)
$P(E) = 0.40$ (工程主修的學生比例)
$P(F) = 0.44$ (從上一題計算得出，樣本中女性的總比例)

115 年國立中正大學企業管理學系碩士班丙組《統計學》

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115 年中正大學的其他科目

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其他學校的企管與商學考古題

115 年 國立中正大學企業管理學系碩士班丙組《統計學》

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其他考古題

115 年中正大學的其他科目

中正大學《統計學》其他年度

其他學校的企管與商學考古題

115 年國立中正大學企業管理學系碩士班丙組《統計學》