W01 課程介紹

2025-09-10-Wednesday 14:10-17:00 林師模教授 shihmolin@gmail.com (助教Sophia張桂鳳)

W02 Review of Statistics

2025-09-17-Tuesday 14:00-17:00 林師模教授

W03 ANOVA: Analysis of variance

2025-09-24-Tuesday 14:00-17:00 林師模教授

■ Hypothesis testing, ■ANOVA(Analysis of variance) 變異數分析;
ChatGPT說明 The steps of hypothesis
例如: 100學生數學成績，內有分N M S 北中南三區學生
1.what question? is region -> effect of math score.
2.H0: μ1=μ2=μ3 (μN=μM=μS)
H1: μ1≠μ2≠μ3 either one ≠
3.★ select Test Statistic 檢定統計量 -找出critical value. (to know the test statistic)
if the statistic not greater than critical value, fall in acceptance area, means that you can not reject the H0.
- 如何找出test方法 經驗/自己推導-想辦法試驗/如果你不知道-you can develop by yourself or even you can publish/ or use other people’s way, to know how others usely do/.
- 多看別人的,多看文章,多看案例.
- ★ need to know what would be the porbability distribution (with your test)
4. Collect Data and Calculate the Test Statistic
- need dicide alpha: significant value first (0.05 or 0.01, 0.10..)
consulting (F) talbe, then you can find out the coreponding critical value.
★ Alpha value 的意義: TypeI error: Ho is true the probability of rejecting the null only 5% (alpha) 你決定了alpha值，就是你決定可能犯了型1錯誤的機率，在alpha值之內。

5. Find the Critical Value(s) or p-value
   
6. Make a Decision
   
7. State the Conclusion

■ANOVA(Analysis of variance)
if we have three or more normal distribution, and we would like to know whether their means are the same or not?
比如剛剛那來自三個地區的100個學生，是否 H 可以求出一個total mean = μ
對任一個 (Yij-μ) = (Ynj-μj) + (μj-μ) 對所有n點想要總和，需要先square以免被相互抵銷
所以先 sauare (Yij-μ) 全部加總起來

SST (sum square total) = SSE (error sum of square 你自己母體的差) + SSB (between group -sum square) 有時SSE 也叫做 SSW (Within group seim square 的意思)
SST=SSE+SSB
如果各組差異大，那麼SSB會變大。 所以如果兩邊都除以SST 這樣就成了
1= (SSE/SST) + (SSB/SST) 假設 這是 1=0.3 + 0.7 那你怎麼知道 0.7有沒大過critical value?
如果各組差異大，那麼SSB會變大。 所以如果兩邊都除以SST 這樣就成了
1= (SSE/SST) + (SSB/SST) 假設 這是 1=0.3 + 0.7 那你怎麼知道 0.7有沒大過critical value?

問Gemini20250924
SST(sum square total), SSE (error sum of square), SSB (between group -sum square)
設SST=SSE+SSB 所以 1= (SSE/SST) + (SSB/SST)
假如 SSE=0.3 而 SST=0.7 那要怎麼知道 0.7有沒大過critical value?

F Statistical calculation step:
F 統計量的計算公式通常是指變異數分析（ANOVA）中的 F = MSB / MSW，
其中MSB 是組間平均平方（Mean Square Between），
MSW 是組內平均平方（Mean Square Within），
F 值用於檢定多個母體平均數是否完全相同。

■ wiki F檢定
F 統計量的計算步驟
計算組間平方和(SSB)：: 衡量不同組別之間的平均變異。
計算組內平方和(SSW)：: 衡量每個組別內部數據點與該組平均數之間的變異。
計算自由度(df)：:
組間自由度(dfB) = 組別數量- 1
組內自由度(dfW) = 總樣本數- 組別數量 (但老師在這個例題，用的是n-1 不是n-group number 因為是SST,但注意📌)
計算平均平方(MS)：: 組間平均平方(MSB) = SSB / dfB
組內平均平方(MSW) = SSW / dfW
計算F 值：: F = MSB / MSW。
應用情境
變異數分析(ANOVA)：: 這是F 統計量最常被使用的情境，用於比較三個或三個以上群體平均數的差異。
變異數比率檢定：: F 檢定也可以用於檢定兩個母體變異數是否相等。
如何理解F 值
F 值是組間變異與組內變異的比例。
較大的F 值表示組間的平均差異比組內的平均差異大，這暗示著群體平均數之間存在顯著差異。
📌注意: 正式用的是SSB /SSW 不是SST (SSW/m3 這個m3 就是df 就是 n-組別數量 )

F = (SSB/(J-a)) / (SSW/(n-J)) 老師說:根據經驗 n=100 J=3 時 Falpha .=.3 但如果J=5 可能小於3, 如果J=10 會更小。 不信去查F表。 F表最上面那行是group的df, 左邊直行是n的df。

老師要教。怎樣用excel去做F test data資料 –> data analysis資料分析 單因子變異數分析 –> by row 逐列: @回家用googlesheet 做看看。

ANOVA如果有 2 factor(如男女)但 sample數都一樣，叫做balanced table如果 sample數不一樣就是imbalanced table (如果每一類只有1個sample data)叫做non-repetitions experiment非重複實驗 HomeWork:
老師出一個作業下週三以前sumit report, 題目是: 自己設計question自己收集data 做一個two factor ANOVA data 需要是repetition data.

要回家拜託AI了 (這就是為什麼我要交錢給ChatGPT和Gemini)

W04 Chp02-The Simple Linear Regression Model

2025-10-01-Tuesday 14:00-17:00 林師模教授

• Mastering Econometrics with Joshua Angrist
• 有例題可作 簡單線性迴歸分析
• University of West Georgia Linear_regression_Notes

The Simple Linear Regression Model
Why?
simple: one dependent variable has only one independent variable.
Linear: relationship: the equation show as a line.
regression: (regress vs. progress 進步) use the already data to look back the relationship. observe the relationship of data. useing the data in the past to generate a line to look back the relationship. model: from emperical phenomeno to abstract the idea, make specification of a relationship.

例如: y: expenditure x:income
y=f(x) base on economic model
we might be able to build up a true function of the model by collecting empirical data of observation.

so we may say: y=β1 + β2 x (this is a emperical model)
true value like (xi, yi) may differenct from (xhat, yhat) which is on the regression line. so we have turn the model to: yi= β1 + β2 xi + ei (this is a econometric model) we need to use true data to estimate the β1,β2,ei parameters

yi= β1 + β2 xi + ei
y: dependent, explained被解釋, regressant, response variable
x: independet, explanatory, regressor variable
ei: erro term, residule(after we estimate model),
β1 and β2: regression coefficient, parameter
β1: intercept, constant
β2: slope (=dyi/dxi = deribative 導數 =when detax change one unit, deaty change ratio = marginal effect 邊際效應 = 當你有一個exiting situation 增加一個unit叫做 marginal unit)
這個existing situation很重要，因為可能是個critical point, 越過這個existing situation狀況可能改變，但marginal effect不隨狀況改變。
所以在這裡的modle 這個β1是個constant 和β2 是個(線性的)slope 都不會變，是個marginal effect。
注意：如果是個曲線 那麼slope 會改變，就不能算是marginal effect。
β2=dy/dx 是個marginal effect

econometric model = regression model
for every econometric model we always needs an Assumptions.
there are many Assumptions, we will look at it one by one.

今日的Quick Review:
example: 2 variable: Income=x, Expenditure=y
more income will expend more money, so there is a linear relationship. so we can set up a model:
yi= β1 + β2 xi
the distance from xi to the line we call it erro term (ei) thus come out the econometic model:
yi= β1 + β2 xi + ei
想找出β1,β2就要根據assumption 的定義
1.linear relationship
2.E(ei | xi) =0
3.Cov(xi,ei | xi) =0
4.Cov(ei,ej | xi) =0
5.xi is nonstotastic
6.ei~ N(0,δ^2)

find out the best line: estimation (estimate a value for model ): least square idea is let the total distance from data point to the line should be smallest: the sumation of all erro term shoud be minimized. in case of the summation became 0, we should square it then do the summation.
after you summation of square will became a quadratic line, then we should do firest derivative to find out the tangent line. 得出公式!!!!!!! get regression model.

老師展示example: 用Eviews:
food.wfi
先選ubcome再ctrl+foodexp open by group

Quick>estimate equation> [food_exp c income ] method: LS-least squares
OK>就可得到estimate result

W05 Chp02-The Simple Linear Regression Model

2025-10-08-Tuesday 14:00-17:00 林師模教授

(1) y_i = β1 + β2 xi + ei (call population regression line)
This is a regression model and also an econometric model.

we have no population so throug sample we get we estimate parameter β2.
(2) y_i = b1 + b2 xi + e_i^{^} (sample: estimator model)
b1, b2 are estimators of β1, β2
we call e_i^{^} as resedule.
(3) y_i^{^}= b1 + b2 xi (call sample regression line)

acording by (2)(3) we can estimate: e_i^{^} = y_i - ( b₁ + b₂ x_i) = y_i - y_i^{^}
(y_i^{^} is fitted value of y_i, also called pedictied value)

OLS: Ordinary Least Square:
b1 = y^bar - b₂ x^bar
b2 = cov(x,y) / var(x)

圖: a (下面還有公式)

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複習Assumption of Least squaqre (注意-2,3-更正上週說法)
1.linear relationship
2.E(ei | xi) =0 has two implications(兩層意義): E(ei)=0 and Cov(xi,ei|xi)=0
(見📌講義p.9 2.1 )
3.var(ei|xi) =δ^2
4.Cov(ei,ej | xi) =0
5.xi is nonstochastic (非隨機的)
6.ei~ N(0,δ^2) -> ( δs2= Σ(ei-ebar)^2 / n = Σ(ei)^2 /n )
δhat2= Σ(ei^ - e^bar )^2 / (n-1) = Σ(ei^) ^2 /(n-1)

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圖: B相片
2.E(ei | xi) =0 so-> and E(yi | xi) = β1 + β2 xi (=population regression line)
3.var(ei|xi)=0 so-> var(yi|xi)= var(ei|xi) =δ^2
4.Cov(ei,ej | xi) =0 so-> Cov(yi,yj | xi) =0
5.
6.ei~ N(0,δ^2) so-> yi~ N(β1+β2xi,δ^2)

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圖: c 相片 立體圖用 income/expenditure為例 (下面還有公式)
不同的x(比如xi=$1,000元)會有很多y點(expenditure)的分佈
每條x上的y 都會是normal distribution
📌講義p.8 2.2.1 All pairs drawn from the same population are assumed to follow the same joint pdf and are identically distributed i.i.d
(理論上: 每條y的distribution 變異數variance都一樣所以樣子都一樣)
理論如此，但實務上不會那麼剛好，所以可能像藍色線那樣分佈。

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以上是Quick Review for Basic Concept
現在看講義快速過一次, 有10節 但有些節會跳過去(可回家自己慢慢看)

📌講義p.4 2.1 The pdf f(y) describes how expenditures are distributed over the population since Households with an income of $1000 per week would have various food expenditure per person for a variety of reasons.

2.2 exogeneity 外生性; 複習為什麼economic model上叫做 marginal effect (看圖d β2=dyi/dxi) will depend on your current situation, always the same. 2.2.4 𝑣𝑎𝑟 𝑒𝑖 𝑥𝑖 = 𝜎2 This is the homoskedasticity (homo=same; skedasticity=variance)

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📌講義p.18 2.2.9 Summarizing the Assumptions 接下來p.19舉例：
想想我們怎樣找出這個yi^ 看p.21 (2.5) (2.6) 如何去fitted regression line

Gauss-Markov Theorem: use 1-5 asumptio OLS.. so that you will find out b1,b2,
which is: the Best Linear Unbiased Estimators. We call this is BLUE estimaors是一個formular, 就是說: 這個formular一旦用OLS方法可以找出b1,b2 我們就稱這個formular是BLUE。
至於為什麼？你可以去看教科書的附錄，那裡有BLUE的公式推導證明。

2025/10/10假日，找了半天沒找到推導。但看到這個利用Transforming找出BLUE estimator的例子:
📘課本 p.375 8.4.1 Transforming the Model: Proportional Heteroskedasticity
看講義p.23有b1,b2的公式

var(b2) = δhats2 / Sumation(xi-xbar)^2 and δhats2 = ehati s2 /n-1

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請參考講義p.29- 2.4.3 Sampling Variation
奇怪，b2不是只有一個數字嗎，怎會有var(b2)？ 這是因為 因為有Repeated sampling的關係 見相片F 有表格 b1 b2 1,2,3….100
會何需要用這想法？你就要回去看立體圖 相片(圖: c) 因為x1…xi 每個x會有很多不同的yi 這樣很多linear line會有不同的var(b2) 看見相片f (跟c很像的立體圖 但有var(b2)公式) 當你作了很多Repeated sampling 後， var(b2)可看出是寬或窄 你的母體怎樣 sampling就會長怎樣 你只要看Repeated sampling的var(b2)就可看出分佈是寬或窄

b2最重要 他就是marginal effect 請記住公式(2.15)

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注意!! 📌講義p.37 2.2.9 average of the squared errors: 公式的δ^2 錯了，應該是沒有hat 因為這是population的關係。

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Exercise food.wf1 重跑Eviews 展示: income +ctl expenditure > open as group > view -ploat sactter ＋ fited line= regression line Quick: estimate equation : expenditure c income > method default to LS -確定 > 解讀result: C INCOME 你可以看到b2 就是income的 SD 他的square就是 var(b2)

• 如果你想產生y^ -> 去做Forcast (他增加一個food_expf 新的變數 focast的意思) income +ctl expenditure +cal food_expf> open as group > view -ploat sactter ＋ fited line= regression line 就可以看到focast 的線條

• income +ctl resid > open as group > view -ploat sactter 可看到residule 分佈

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Residule不用管大或小，但在意是不是constant?

看圖G: 所謂constant就是 在一定範圍內震盪(變化)，而不是越來越大或越變越小。

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舉另外一例 📌講義p.47 2.2.9 Figure 2.14 A Fitted Quadratic Relationship
因為原先x,y的關係是Quadratic Relationship而非Linear Relationship，怎麼辦呢？有兩種方法可以嘗試： 1.是把independent variable SQFT再square一次 設法把曲線變為線性關係
2.另一方法是處理dependent variable y，把y做Logarithmization，即 log y也可以
但兩種方法需要比較，看那個線型比較好? (以能和最多iPair發生關係的最好)

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現在用EXAMPLE 2.6 Baton Rouge House Data 來做練習 (使用資料檔br.wf1)

• sqft + ctl price > open as group > view -graph -sactter >看起來不像learning Eviews有兩種方法可run qudratic or log equation:

1. price c sqft^2 > result > forcast pricef (fitted value) 會出現變數 pricef 這是第一個model

• sqft + ctl price + ctl pricef > open as group > view -graph-sactter

2. log(price) c sqft > result > forcast pricef2 (
   fitted value) 會出現變數 pricef2 這是第2個model

• sqft + ctl price + ctl pricef2 > open as group > view -graph-sactter -ok

將兩個圖疊起來做比較，看誰比較線性！
- sqft + ctl price + ctl pricef2 + ctl pricef > open as group > view -graph -sactter -ok
那一個比較好？請看講義p.52的討論: 2.8.5 Choosing a Functional Form
(在這個case是 (2)比較好，因為他可以掌握最多的資料)

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作業:📘課本 p.93這個練習
2.11.2 Computer Exercises

2.16 The capital asset pricing model (CAPM)
The capital asset pricing model (CAPM) is an important model in the field of finance. It explains variations in the rate of return on a security as a function of the rate of return on a portfolio consisting of all publicly traded stocks, which is called the market portfolio. Generally, the rate of return on any investment is measured relative to its opportunity cost, which is the return on a risk-free asset. The resulting difference is called the risk premium, since it is the reward or punishment for making a risky investment. The CAPM says that the risk premium on security j is proportional to the risk premium on the market portfolio. That is:

r_j - r_{f = βj ( rm - rf )}

where r_j and r_f are the returns to security j and the risk-free rate, respectively, r_m is the return on the market portfolio, and β_j is the jth security’s “beta” value. A stock’s beta is important to investors since it reveals the stock’s volatility. It measures the sensitivity of security j’s return to variation in the whole stock market. As such, values of beta less than one indicate that the stock is “defensive” since its variation is less than the market’s. A beta greater than one indicates an “aggressive stock.” Investors usually want an estimate of a stock’s beta before purchasing it. The CAPM model shown above is the “economic model” in this case. The “econometric model” is obtained by including an intercept in the model (even though theory says it should be zero) and an error term

r_j − r = α_j + β_j ( r_m − r ) + e_j

a. Explain why the econometric model above is a simple regression model like those discussed in this chapter.
b. In the data file capm5 are data on the monthly returns of six firms (GE, IBM, Ford, Microsoft, Disney, and Exxon-Mobil), the rate of return on the market portfolio (MKT), and the rate of return on the risk-free asset (RISKFREE). The 180 observations cover January 1998 to December 2012. Estimate the CAPM model for each firm, and comment on their estimated beta values. Which firm appears most aggressive? Which firm appears most defensive?
c. Finance theory says that the intercept parameter αj should be zero. Does this seem correct given your estimates? For the Microsoft stock, plot the fitted regression line along with the data scatter.
d. Estimate the model for each firm under the assumption that αj = 0. Do the estimates of the beta values change much?

| 作業 從第9頁開始。|

下週要開始第3章講 Interval Estimation and Hypothesis Testing

W06 Chp03 Interval Estimation and Hypothesis Testing

2025-10-15-Tuesday 14:00-17:00 林師模教授

現在我們要來研究 var(β₂):請看

(圖a)
因為 var(β₂)= σ² / Σ(x_i-x_ibar)²
那首先就要先弄清楚σ² 和Σ(x_i-x_ibar)² 的值。

1.σ²: Sum squared residule的公式是 σ² = Σ ei² / N 但因沒有population只有sample，
所以改用 σ^² = Σ e_i^² / N-2 (少掉b1,b2兩個)
因為Eviews已算出 Sum squared residule =304505.2 這就是Σ ei^²，
而將這個Σ e_i^²開根號(再除以38)就是S.E. of regression = 89.51700 這就是σ^²
(反之: (89.51700)² * 38= 304505.144982 =Sum squared residule)

(表2)
2. Σ(xi-xibar)²:
因為 var(x) = Σ(xi-xibar)² / N-1 在Eviews選變數[income+Ctl food_expd]->Open with group->view->Discriptive Stats->Common Sample就會出現 (表2)可看到:
Std. Dev. of INCOME 也就是 x 的 Std. Dev.= 6.847773
Std. Dev. of FOOD_EXP 也就是 y 的 Std. Dev.= 112.6752
公式 var(x) = Σ(xi - xi bar)² / N-1
所以Σ(xi - xi bar)² = (Std. Dev of x)² * (N-1) = 6.8477736.84777339 =1828.788

Interval Estimation:

b1 and b2 we call estimators, when get value of b1 and b2 so call estimate.
they are called “point estimate” 點估計; empericaly some time we might not just interested just one point, instead, interested in a range. an interval.
那麼range怎樣找出來呢?
idea is “how can we get range” how to establish?
(1) y_i = β₁ + β₂ x_ii + e_i (call population regression line)
(2) y_i = b₁ + b₂ x_i + e_i^ (sample: estimator model)
用Z公式來standardize b₂
Z=b₂ -β₂ /Se(b₂) = (StdE(b₂) 請參考前面的 var(b₂) 公式) StdE(b2)=se=standard error

請照著講義p.4-5 的說明展開程式，就可找出做interval的正確方法。
先standardize b2在決定confidence value-找出z value 再用公式算出interval

slightly different from Z to t, because we do not know σ only know σ^
注意: 因為沒辦法用 sigama square 需要換成 sigama hat square 所以Z distribution 要換成 t distribution
注意p.7 的 (3.2)公式 的 根號中的σ square 應該改為 σ hate square 才對。

注意t table查表 時須注意正確的degree of freedom. (課本863頁TABLE D.2 Percentiles of the t-distribution查95%-1.685/df38 如果到了∞ 也是1.96和Z一樣)
- p.862-Z表TABLE D.1是查: Cumulative Probabilities for the Standard Normal Distribution (z) = P(Z ≤ z)
注意: p.10 t_c 也寫錯了 應該是 b_k 才對!! 回課本p.116 去看EXAMPLE 3.1 Interval Estimate for Food Expenditure Data
老師用food.wf1跑一次Eviews來解釋interval作法:
- food_exp c food ->Std. Error 你可以用公式自己算，也可以按按Eviews就可以輕鬆算出
- view-> coeffice diagnostics-confidence interval-chose a confidence-或3個都要或改都可以->OK就算出來了 想想這三種confidence的interval的意義。

• 看看TABLE 3.1, 3.2 分幾次 每次只取10個sample去跑 會怎樣?

Hypothesis Testing:

1. y_i = β₁ + β₂ x_i + e_i (call population regression line)
   
2. y_i = b₁ + b₂ x_i + e_i^ (sample: estimator model)
   只要有sample總是可以得到b₁ b₂ (比如food的結果如下)
   yi = 83.41 + 10.2 x_i + e_i^ 但實際上不是那麼可靠！
   但如果 β₂=0 表示 x沒有 effect對於y
   那麼10.2 一直到趨近於0 之間，到那一個點(critical point)這個effect會變得very weak?!
   這就需要做test 看看 β₂=0 是否成立! 看看在這個(critical point)時 是不是significant 說他們是有顯著的關系。
   如果significant 就是
   H0: β₁=0
   H1: β₁≠0

H0: β₂=0
H1: β₂≠0
t = b₂-β₂ /Se(b₂) ~ tα/2, N-2
α=0.05 so α/2=0.025

4.8已進入 rejection area, which is means β₂≠0, it is significantly different from 0.
because t vale is greter than critical value (超出了α 的critical point)
這是針對單一回歸係數的檢定！
這在3.2.2有更詳細的解釋 需要去看
還有3.3.1講的是單尾的假設檢定

下週W07-從p.123的3.4 Examples of Hypothesis Tests開始講

W07 Ch-03 Interval Estimation and Hypothesis Testing

2025-10-22-Tuesday 14:00-17:00 林師模教授

After we have estimator, somtimes we needs to know whethere there is meaning of the parameters: coefficent-b2 (or b1)
whethere we can use the estimat LINE to represent the estimation?
we needs to TEST
set H0: β2=0
H1: β2 != 0

where the normaly distribution of be β2 : we use t-test to see if the average is 0 or not?

p.123 run Eviews [EXAMPLE 3.2 Right-Tail Test of Significance] 用 food.wf1
equation: food_exp c income

(表1)

p.124 [EXAMPLE 3.3 Right-Tail Test of an Economic Hypothesis 經濟假設的右尾檢驗]
這就是用food.wf1 看整個過程的詳細解說

如果是用Eviews(不必手算)，這樣做

Eviews8.1的操作法-省去計算時間
如果是用Eviews(不必手算)，這樣做：
EX3.3-a: 使用food.wf1檔: Quick > Estimae Equation > food_exp c income

EX3.3-b: View > Coefficient Diagnostics > Wald Test-Coefficient Restirctions..

EX3.3-c: 輸入c(2)=5.5 > OK

EX3.3-d: 幫你算好t-statistic=2.249904 及p-value=0.0303 (因為要求是0.01 所以fail to reject)

Wald-test的 c(2)其實就是 c去幫你減掉c(2)這個值 的意思。

這兩個案例很重要，詳細做過，就了解了!!

會影響準確度有兩個因素 1.是confidence level; 2.是se(b₂)-這可去看interval(分散程度)
這是簡單的例子，但其中有很多需要深思的意義。

p.125 [EXAMPLE 3.4 Left-Tail Test of an Economic Hypothesis 經濟假設的左尾檢驗]
看看他是怎樣算出結果 和如何reject null 的

3.5 The p-Value

p.126 [EXAMPLE 3.6 Two-Tail Test of Significance 雙尾顯著性檢定]

p.129 [3.6 Linear Combinations of Parameters]

t=(b1+2b2-5)/Se(b1+2b2)

Var(c1b1+c2b2) = var(λ)
covariance matrix 就是

view coe diag wal-test > hypo null 公式: c(1)+2*c(2)=5
你可以把Se 39.476 平方就是variance
var(b1)= 188 var(b2)=4.38

p.130 [EXAMPLE 3.7 Estimating Expected Food Expenditure]
這個例子就是 [3.6 Linear Combinations of Parameters] 這個沒有做什麼test

下個例子，是做出 interval estimate
p.131 [EXAMPLE 3.8 An Interval Estimate of Expected Food Expenditure] 這個計算需要先找出 se(c1b1+c2b2)

view -> Cofeficcient Dignostics -> Wald-test: hypo null 公式: c(1)+2c(2)=5
c(1)+20c(2)=(任何number: 因為只是想找出se 就是 14.178

p.132 [EXAMPLE 3.9 Testing Expected Food Expenditure] “I expect that a household with $2,000 weekly income will spend, on average, more than $250 a week on food.” How can we use econometrics to test this conjecture?

也是view -> Cofeficcient Dignostics -> Wald-test: c(1)+20*c(2)=250

下週講Chp04 Prediction, Goodness-of-Fit, and Modeling Issues
老師會跳過Prediction因為有其他工具可用(有興趣可自學)，只講後面的Goodness-of-Fit, and Modeling Issues
就是講義4.2-4.6

W08 Chp04 Prediction, Goodness-of-Fit, and Modeling Issues

2025-10-29-Tuesday 14:00-17:00 林師模教授 (換到402A教室)

p.153 [4.1 Least Squares Prediction]

p.156 [4.2 Measuring Goodness-of-Fit 配適度測量(R²) 測量擬合優度]

p.157 [4.2.1 Correlation Analysis]

p.158 [4.2.2 Correlation Analysis and R²]

p.160 [4.3 Modeling Issues]

p.160 [4.3.1 The Effects of Scaling the Data 數據規模的影響]

p.161 [4.3.2 Choosing a Functional Form]

p.163 [4.3.3 A Linear-Log Food Expenditure Model]

p.165 [4.3.4 Using Diagnostic Residual Plots]

p.167 [4.3.5 Are the Regression Errors Normally Distributed?]

p.169 [4.3.6 Identifying Influential Observations]

p.171 [4.4 Polynomial Models]

p.171 [4.4.1 Quadratic and Cubic Equations]

p.173 [4.5 Log-Linear Models]

p.175 [4.5.1 Prediction in the Log-Linear Model]

p.176 [4.5.2 A Generalized R² Measure]

p.177 [4.5.3 Prediction Intervals in the Log-Linear Model]

p.177 [4.6 Log-Log Models]

p.179 [4.7 Exercises]

p.192 [Appendix]

W09 --期中考--

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W18 --期末報告--

2026-01-07-Tuesday 14:00-17:00 林師模教授