Sampling Methods and the Central Limit Theorem Chapter 8 Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Learning Objectives LO8-1 Explain why populations are sampled and describe four methods to sample a population. LO8-2 Define sampling error. LO8-3 Demonstrate the construction of a sampling distribution of the sample mean. LO8-4 Recite the central limit theorem and define the mean and standard error of the sampling distribution of the sample mean. LO8-5 Apply the central limit theorem to calculate probabilities. 8-*

母體 vs. 樣本 母體：研究某一自然現象或是社會問題，調查研究的全體對象 樣本：從母體中抽取的部分元素的集合

Why Sample a Population? LO8-1 Explain why populations are sampled and describe four methods to sample a population. Why Sample a Population? 收集母體資料非常耗費時間 Contacting the whole population would be time consuming. 收集並研究母體資料的成本極高 The cost of studying all the items in a population may be prohibitive. 可能無法實際審查、收集、計算所有母體資料 The physical impossibility of checking all items in the population 檢定、測試會毀壞受測試的檢體、樣本（因此，檢測母體將會毀壞所有母體，如此一來，對母體檢測，將會得不）償失 The destructive nature of some tests 抽樣的結果（用來推論母體）還算恰當 The sample results are adequate. . 8-*

為何抽樣方法很重要？1/2 若患者等候時間為某醫院的經營效率的評估項目之一，因為等候時間過長，病患會因不耐煩而離去，門診收入會大減，因此，醫院的管理者希望能研究分析病患的等候時間的分配狀況。若母體分配應該如下圖，但醫院中無人能收集齊全所有病患候診的資料，故僅能抽樣分析

為何抽樣方法很重要？2/2 第2組樣本的分配似乎比較接近母體分配，但如果我們不知母體，如何判斷該採用哪一組呢？ 若抽樣50名病患來分析，不論用何種抽樣方法，共抽了2組樣本，請問：哪一組樣本比較能代表母體分配？ 第2組樣本的分配似乎比較接近母體分配，但如果我們不知母體，如何判斷該採用哪一組呢？ 此為第1組樣本： 此為第2組樣本：

抽樣的重要性 統計推論係利用樣本統計量去推論母體的特質，而樣本是否具有代表性，會受到抽樣方法的影響，因此抽樣方法非常重要。 好的抽樣法，能將抽樣誤差降至最低，機率抽樣法純粹依照機率來抽取樣本，又稱隨機抽樣(random sampling)，所得的樣本稱為隨機樣本。 非機率抽樣按個人偏好、判斷來選樣本，所選之樣本又稱非隨機樣本。 機率抽樣比較客觀，而非機率抽樣比較主觀。

樣本（估計）誤差 抽樣誤差(sampling error)：樣本統計量與母體參數間的差異，此差異來自抽樣過程的機遇(chance)、抽樣方法、推論方法之不同所致。 - 若因「機遇」所致的誤差，可藉著增加樣本數來降低此類誤差。 - 不同抽樣法、不同推論法的誤差皆有所不同，必須想法找出使誤差最小的統計方法。 非抽樣誤差(nonsampling error)：主要來自調查時的執行、事後記錄、整理資料時所發生的錯誤。 - 此誤差無法估計，僅能藉著妥善規劃與嚴格審查來降低此類誤差。

抽樣方法 非機率抽樣法： 機率抽樣法： 乃依據一個已知的機率去抽取樣本，可根據此機率來探討樣本特性，並推論其抽樣誤差。 1. 判斷抽樣法：依據研究人員的專長、知識、與目的來選取樣本，又稱為目的抽樣法。如：核廢料處理問題的民間意見調查，可選對此有專業知識者做訪問調查，調查結果對核廢料處理政策的判斷，比對一般民眾訪查更有參考性，然而，此抽樣法相當主觀，受個人好惡（或意識型態）影響，恐易產生偏誤結果。 2. 方便抽樣法：以現有、方便的方式取得樣本，如：站在街頭發問卷、到商場發問卷來作市調，民調。此樣本不具代表性。 ─ 除非特殊原因，非機率抽樣法缺乏客觀、科學根據，不易判斷其代表性，最好不採用。 機率抽樣法： 乃依據一個已知的機率去抽取樣本，可根據此機率來探討樣本特性，並推論其抽樣誤差。

較常使用的機率抽樣方法： Commonly Used Sampling Methods LO8-1 較常使用的機率抽樣方法： Commonly Used Sampling Methods 簡單隨機抽樣 Simple Random Sampling 系統隨機抽樣 Systematic Random Sampling 分層隨機抽樣 Stratified Random Sampling 部落抽樣 Cluster Sampling 8-*

1-1 第8章 抽樣 簡單隨機抽樣

Simple Random Sampling LO8-1 Simple Random Sampling SIMPLE RANDOM SAMPLING A sample selected so that each item or person in the population has the same chance of being included. 母體中每個元素被抽中的機率相同 . 範例：（1. 抽籤法） Nitra企業有845個員工，由員工母體中抽出52人，將員工的名字寫在紙條上，將所有紙條放入盒子中，充分攪動混和後，抽出52張紙條。 A population consists of 845 employees of Nitra Industries. A sample of 52 employees will be selected from that population. The name of each employee is written on a small slip of paper and all slips are deposited in a box. After they have been thoroughly mixed, the first selection is made by drawing a slip out of the box without looking at it. This process is repeated until the sample of 52 employees is chosen. 注意：此法乃抽出後不放回，所以每個名字被抽中的機率略有不同，但若母體極大，則其機率的差異很小。此處，差異約為0.001。 Note: This process is sampling without replacement, so the probability of each selection changes: 1/845, 1/844, 1/843, etc. When the population is large, the difference in the probabilities is very small. In this case, the probability for each of the 52 selections is about 0.001. 8-*

抽樣的基本概念 隨機性的條件 母體中的任一元素(element)皆有被抽出的 可能。 任一組樣本被抽出的機率皆為已知的(或可加以計算)。 各組樣本被抽出的過程是獨立的。

LO8-1 簡單隨機抽樣：2. 使用隨機亂數表 p. 250 Simple Random Sampling: Using Table of Random Numbers 由母體Nitra企業的845名員工中抽出52位，而比抽籤法更方便的方法是：將每位員工編號，然後使用附錄B.4的隨機亂數表。由亂數表中，隨便選一行與一列，作為起始值，以下51個數值加上起始值，就是被抽中的52個樣本（此處用前3位數作為被抽中的號碼，若數值大於845，則跳到下一號）。 A population consists of 845 employees of Nitra Industries. A sample of 52 employees will be selected from that population. A more convenient method of selecting a random sample is to use the identification number of each employee and a table of random numbers such as the one in Appendix B.4. 8-*

簡單隨機抽樣：3. 用電腦抽號 p. 250 Simple Random Sampling: Using Excel LO8-1 簡單隨機抽樣：3. 用電腦抽號 p. 250 Simple Random Sampling: Using Excel Jane and Joe Miley 經營位於Tryon, North Carolina的Foxtrot Inn（共有8個房間），下表為此B&B在2013年6月的出租情形，可用Excel選6月中的5個晚上作為樣本。注意：Excel選取樣本時採取抽後放回的方式，因此，5個樣本中可能會抽中同一日。 Jane and Joe Miley operate the Foxtrot Inn, a bed and breakfast in Tryon, North Carolina. There are eight rooms available for rent at this B&B. Listed below is the number of these eight rooms rented each day during June 2013. Use Excel to select a sample of five nights during the month of June. 8-*

Excel: p. 749 函數：rand 或如下一頁所示： Data選data analysis選sampling按ok 在Input range中填入B1:B31（6月共30日），此為母體 因為每欄都有名稱，勾選Labels，然後，點選Random，在樣本數目(nuber of samples)中，填入5 點選 Output Range，然後在你想擺放樣本處點一下，或輸入相對應的欄位，如：D1

簡單隨機抽樣：3. 用電腦抽號 Simple Random Sampling: Using Excel LO8-1 簡單隨機抽樣：3. 用電腦抽號 Simple Random Sampling: Using Excel 用Excel選5個樣本，列於D欄內。 注意：Excel選取樣本時採取抽後放回的方式，因此，5個樣本中可能會抽中同一日。 Jane and Joe Miley operate the Foxtrot Inn, a bed and breakfast in Tryon, North Carolina. There are eight rooms available for rent at this B&B. Listed is the number of these eight rooms rented each day during June 2012. Use Excel to select a sample of five nights during the month of June. 8-*

系統抽樣 Systematic Random Sampling 系統抽樣法是自母體自然隨機排列的資料中，每隔一定間隔(k)選取一個樣本，直到抽滿 n 個樣本為止。

第8章 抽樣 系統抽樣法

為何使用系統抽樣法？ 簡單隨機抽樣雖好（具代表性），但有時不適用 商場、百貨公司、大賣場、超市等想抽樣調查顧客意見與滿意度，公司沒有完整確定的顧客名單，無法隨機抽樣調查。 例如：某大賣場抽100個顧客調查其留在店中的時間長短，週一到週四（每天抽25人），每天有4個時段（8AM, 11AM, 4PM, 7PM)，可將4天放一盒，4時段放另一盒，抽出哪一天，要在那個時段，做隨機抽樣，若週一抽中4PM做抽樣調查，接著由1到10中隨機選起始值，若為6，則週一4PM開始，第6個顧客是第一個樣本，然後k =10，第16, 26,…個顧客就是樣本，直到抽滿25人為止。（然後派人留意該顧客留在店中的時間長短，做出記錄） 所以，此處用簡單隨機抽樣選出：天、時段、起始點（顧客） 而系統抽樣則被用來抽出實際樣本（顧客）

系統隨機抽樣 Systematic Random Sampling LO8-1 系統隨機抽樣 Systematic Random Sampling SYSTEMATIC RANDOM SAMPLING The items or individuals of the population are arranged in some order. A random starting point is selected and then every kth member of the population is selected for the sample. . 範例：由845為員工中選出52人做為樣本。 1. 先算k值：k = 845/52 = 16.25，四捨五入後，k = 16 2. 先隨機抽出第一個樣本，之後每次往後算，第16個人將被選入樣本中。 A population consists of 845 employees of Nitra Industries. A sample of 52 employees will be selected from that population. First, k is calculated as the population size divided by the sample size. For Nitra Industries, we would select every 16th (845/52) employee list. If k is not a whole number, then round down. Random sampling is used in the selection of the first name. Then, select every 16th name on the list thereafter. 8-*

分層抽樣 Stratified Random Sampling 分層抽樣是將母體依其特性或依與調查目的有關的性質分成幾個類別或組別，母體中的每一個個體或元素都屬於其中的一層，而且是唯一的一層。分層之後再從各層中簡單隨機抽取樣本，各層的抽樣數乃按照各層佔母體總數的比例來抽取。

第8章 抽樣 分層抽樣方法－比例抽樣

分層隨機抽樣 Stratified Random Sampling LO8-1 分層隨機抽樣 Stratified Random Sampling STRATIFIED RANDOM SAMPLING A population is first divided into subgroups, called strata, and a sample is selected from each stratum. This is useful when a population can be clearly divided in groups based on some characteristics. . EXAMPLE: （將在後面說明解釋） Suppose we want to study the advertising expenditures for the 352 largest companies in the United States to determine whether firms with high returns on equity (a measure of profitability) spend more of each sales dollar on advertising than firms with a low return or deficit. We decide to sample a total of 50 companies. To make sure that the sample is a fair representation of the 352 companies, the companies are grouped on percent return on equity and the number to sample in each group is proportional to the relative size of the group. Then, the number of companies is randomly selected from each group. * Number sampled for Stratum 1 is (0.02)(50)=1; Stratum 2 is (0.10)(50)=5, etc. 8-*

為何分層抽樣？ 若樣本容易區分成不同類別 若某些類別（特性）佔總母體比例相當低，則若採用簡單隨機抽樣，某種特性有可能很難被抽中，即使簡單隨機抽樣對母體具代表性，但樣本趨於代表majority，而非全部母體。用分層抽樣則能保證此minority也能被抽到樣本中 見下例 (p. 253)

分層隨機抽樣 (p. 253) Stratified Random Sampling LO8-1 分層隨機抽樣 (p. 253) Stratified Random Sampling 範例： 若我們想研究美國352家大企業的廣告支出，以便決定是否股本報酬率高的公司比低報酬或虧損者，花更多錢在廣告上，我們打算抽50家做為樣本。 Suppose we want to study the advertising expenditures for the 352 largest companies in the United States to determine whether firms with high returns on equity (a measure of profitability) spend more of each sales dollar on advertising than firms with a low return or deficit. We decide to sample a total of 50 companies. 為保證樣本能代表這352家企業，我們將它們分成5個類別，按照其股本報酬率來分類，下表列出此5類、家數、以及其相對次數，最後列出每層的抽樣數。每類別的抽樣皆以隨機抽樣方式抽出樣本。 To make sure that the sample is a fair representation of the 352 companies, the companies are grouped on percent return on equity and the number to sample in each group is proportional to the relative size of the group. Then, the number of companies is randomly selected from each group. . *第1組的抽樣數為： (0.02)(50)=1; 第2組的抽樣數為： (0.10)(50)=5, etc. 第4組的抽樣數為： (0.33)(50)=16.5～16 第5組的抽樣數為： (0.01)(50)=0.5～1 8-*

部落抽樣法 Cluster Sampling 若母體自然形成不同群組（如：地理位置），也可將大區分成小區（如：州內區分不同郡縣，又稱 primary units） 部落抽樣是將母體中相鄰的某些群體劃分為 n 個不同的部落(cluster)，母體中的每一個元素均屬於其中的一個部落，且是唯一的一個部落。然後再從這些部落中隨機抽取部落，並對抽出的部落進行普查的抽樣方法（全部放入樣本中），又稱集團抽樣。 或先隨機抽出部落，再由抽中的部落中隨機抽樣

第8章 抽樣 部落抽樣法

部落抽樣法 Cluster Sampling LO8-1 部落抽樣法 Cluster Sampling CLUSTER SAMPLING A population is divided into clusters using naturally occurring geographic or other boundaries. Then, clusters are randomly selected and a sample is collected by randomly selecting from each cluster. 範例： 若想知道大芝加哥區居民對環保署、州環保局的看法如何。 Suppose you want to determine the views of residents in the greater Chicago, Illinois, metropolitan area about state and federal environmental protection policies. 可以將大芝加哥按郡縣分成小區，共可分成12郡，隨機抽選3個郡： La Porte, Cook, and Kenosha，然後再由這3個郡中隨機抽出我們需要的樣本。 You can employ cluster sampling by subdividing the region into small units, perhaps by counties. These are often called primary units. Of the twelve counties, you randomly select three: La Porte, Cook, and Kenosha. Next you select a random sample of residents in each of these counties. 8-*

1-1 第8章 抽樣 母體參數與樣本統計量

1-1 第8章 抽樣 估計誤差

第8章 抽樣 估計誤差

抽樣誤差：Sampling Error 抽樣乃是用來計算樣本統計值，作為母體參數的估計值，因此，樣本統計值與母體參數值的差距，就是抽樣誤差。 LO8-2 Define sampling error. 抽樣誤差：Sampling Error 抽樣乃是用來計算樣本統計值，作為母體參數的估計值，因此，樣本統計值與母體參數值的差距，就是抽樣誤差。 By definition, sampling is used to calculate sample statistics which are estimates of population parameters. So there will always be a difference (usually an unknown difference) between the sample statistic and the population parameter. This difference is called sampling error. Examples: 8-*

抽樣誤差：範例 p. 256 Sampling Error P.250 B&B的例子，2013年6月房間出租數的資料為母體，請計算母體平均數，用Excel抽3組樣本，每組抽出5天，計算每組樣本的平均值，計算抽樣誤差。 解：N=30,n=5，共可抽30!/(5!25!)=142,506組樣本，可計算出142,506個樣本平均值，這些可做成次數分配（機率分配），而E(樣本平均數)＝母體平均值 因為樣本平均是母體平均的不偏估計式 μ= (0+2+3+…+3)/30 = 94/30 = 3.13 1st: 4,7,4,3,1 Mean = 19/5 = 3.8 樣本誤差= 3.8-3.13 = 0.67 2nd : 3,3,2,3,6 Mean = 17/5 = 3.4 樣本誤差= 3.4-3.13 = 0.27 3rd : 0,0,3,3,3 Mean = 1.8 樣本誤差 = -1.33

第8章 抽樣 樣本平均數的抽樣分配

樣本平均數的抽樣分配 Sampling Distribution of the Sample Mean LO8-3 Demonstrate the construction of a sampling distribution of the sample mean. 樣本平均數的抽樣分配 Sampling Distribution of the Sample Mean 樣本平均數的抽樣分配是機率分配，包含在既定n（樣本數）下（由同一母體中抽樣之樣本），所有可能的樣本平均數值的機率分配。 The sampling distribution of the sample mean is a probability distribution consisting of all possible sample means of a given sample size selected from a population. 樣本平均數的抽樣分配可敘述/描繪抽樣誤差的機率分配（抽樣誤差也是隨機變數） The sampling distribution of the sample mean summarizes the probabilities of sampling error: 8-*

樣本平均數的抽樣分配 LO8-3 Sampling Distribution of the Sample Mean The mean of the distribution of sample means will be exactly equal to the population mean if we are able to select all possible samples of the same size from a given population. 抽樣分配的標準差：The Standard Error of the Mean: 所以，抽樣分配的離散度比母體分配小。 當樣本數目變大時，其標準差變小。 There will be less dispersion in the sampling distribution of the sample mean than in the population. As the sample size increases, the standard error of the mean decreases. . 8-*

樣本平均數的抽樣分配：範例 p.258 Sampling Distribution of the Sample Mean – Example LO8-3 樣本平均數的抽樣分配：範例 p.258 Sampling Distribution of the Sample Mean – Example Tartus企業有7名生產員工（可視為母體），他們的時薪資料如下：N=7 Tartus Industries has seven production employees (considered the population). The hourly earnings of each employee are given in the table below. What is the population mean? What is the population standard deviation? What is the sampling distribution of the sample mean for samples of size 2? n=2 What is the mean of the sampling distribution? What is the standard deviation of the sampling distribution? What observations can be made about the population and the sampling distribution? 8-*

LO8-3 樣本平均數的抽樣分配：範例 p.258-259 Sampling Distribution of the Sample Mean – Example The population standard deviation 8-*

LO8-3 樣本平均數的抽樣分配：範例 p.259 Sampling Distribution of the Sample Mean – Example N＝7、n = 2 抽樣分配共可抽21組樣本，其抽樣結果與樣本平均列於下表： 8-*

LO8-3 樣本平均數的抽樣分配：範例 p.259 Sampling Distribution of the Sample Mean – Example N＝7、n = 2 樣本平均數的抽樣分配，列於下表： 8-*

樣本平均數的抽樣分配：範例 p.259 Sampling Distribution of the Sample Mean – Example LO8-3 樣本平均數的抽樣分配：範例 p.259 Sampling Distribution of the Sample Mean – Example . 8-*

樣本平均數的抽樣分配：範例 p.259 Sampling Distribution of the Sample Mean – Example LO8-3 樣本平均數的抽樣分配：範例 p.259 Sampling Distribution of the Sample Mean – Example . These observations can be made: 抽樣分配的平均值等於母體平均值＝7.71。 The mean of the distribution of the sample mean ($7.71) is equal to the mean of the population. 抽樣分配的離散度低於母體，且當n增大時，離散度變小。（ ） The spread in the distribution of the sample mean is less than the spread in the population values. As size of the sample is increased, the spread of the distribution of the sample mean becomes smaller. ( ) 抽樣分配與母體分配的形狀不同，抽樣分配比較接近常態分配。 The shape of the sampling distribution of the sample mean and the shape of the frequency distribution of the population values are different. The distribution of the sample mean tends to approximate the normal probability distribution. 8-*

中央極限定理 Central Limit Theorem

中央極限定理 Central Limit Theorem LO8-4 Recite the central limit theorem and define the mean and standard error of the sampling distribution of the sample mean. 中央極限定理 Central Limit Theorem CENTRAL LIMIT THEOREM 樣本平均數的抽樣分配接近常態分配（若在固定n之下的所有可能樣本組合都被抽出，做成抽樣分配），n越大越接近常態分配。 If all samples of a particular size are selected from any population, the sampling distribution of the sample mean is approximately a normal distribution. This approximation improves with larger samples. 若母體為常態分配，則不管n多小，樣本平均數的抽樣分配也是常態分配。 If the population follows a normal probability distribution, then for any sample size the sampling distribution of the sample mean will also be normal. 若母體為對稱分配（非常態分配），若n≧10，則樣本平均數的抽樣分配即接近常態分配。 If the population distribution is symmetrical (but not normal), the normal shape of the distribution of the sample mean emerges with samples as small as 10. 若母體為偏態分配，若n≧30，則樣本平均數的抽樣分配即接近常態分配。 If a distribution is skewed or has thick tails, it may require samples of 30 or more to observe the normality feature. 抽樣分配的平均值為μ，變異數為σ2/n，標準差為 The mean of the sampling distribution is equal to μ. The variance is equal to σ2/n and the standard deviation is equal to . 8-*

LO8-4 8-*

中央極限定理：範例 p.263-264 Central Limit Theorem LO8-4 中央極限定理：範例 p.263-264 Central Limit Theorem Spence Sprockets公司雇用40名員工，必須為他們買健保，在選購健保方案前，由員工中選出5位代表成立委員會，委員會必須謹慎評估研究員工的健康需求，然後推薦何種方案最適合員工需求，老闆Ed發現新員工與老員工的需求不同，若他隨機抽樣，平均年資＝？樣本平均年資與母體平均年資是否相等？40名員工的年資（四捨五入）列於下表。 Spence Sprockets, Inc. employs 40 people and faces some major decisions regarding health care for these employees. Before making a final decision on what health care plan to purchase, Ed decides to form a committee of five representative employees. The committee will be asked to study the health care issue carefully and make a recommendation as to what plan best fits the employees’ needs. Ed feels the views of newer employees toward health care may differ from those of more experienced employees. If Ed randomly selects this committee, what can he expect in terms of the mean years with Spence Sprockets for those on the committee? How does the shape of the distribution of years of experience of all employees (the population) compare with the shape of the sampling distribution of the mean? The lengths of service (rounded to the nearest year) of the 40 employees currently on the Spence Sprockets, Inc., payroll are as follows. 8-*

中央極限定理：範例 p.264-267 Central Limit Theorem LO8-4 中央極限定理：範例 p.264-267 Central Limit Theorem 25 Samples of 5 Employees 25 Samples of 20 Employees 8-*

中央極限定理：範例 p.264-267 Central Limit Theorem 全體員工的年資分配右偏少數員工的年資較高 母體平均值 μ =4.8 40C5= 658,008 採抽出放回的方式，他抽取25組內含5(n=5)名員工的樣本， 並求得各樣本平均值 Table 8-5 n=5的時候，樣本平均值的抽樣分配 Chart 8-5右偏狀況較不明顯 若進一步，抽取25組內含20名員工的樣本， 並求得各樣本平均值 Table 8-6 樣本觀察點數= 20，樣本平均值的抽樣分配 Chart 8-6趨近常態分配 = (3.95+3.25+5.95+…+5.05)/25=4.676 （已經相當接近4.8了）

1-1 樣本平均數抽樣分配的形狀 應用中央極限定理的注意事項

採用樣本平均數的抽樣分配 Using the Sampling Distribution of the Sample Mean LO8-5 Apply the central limit theorem to calculate probabilities. 採用樣本平均數的抽樣分配 Using the Sampling Distribution of the Sample Mean 若母體為常態分配，則樣本平均數的抽樣分配也是常態分配 If a population follows the normal distribution, the sampling distribution of the sample mean will also follow the normal distribution 若為非常態分配，但樣本數大於30，中央極限定理保證樣本平均的抽樣分配為常態分配 If the shape is known to be non-normal but the sample contains at least 30 observations, the central limit theorem guarantees the sampling distribution of the mean follows a normal distribution. 若母體標準差已知，樣本平均的抽樣分配的z值如下： When the population standard deviation is known, a z-statistic for the sampling distribution of the sample mean is calculated as: 8-*

LO8-5 採用樣本平均數的抽樣分配：範例 p.271 Using the Sampling Distribution of the Sample Mean – Example 可樂公司的品管部門紀錄大瓶裝可樂的注入量，每一瓶的量或多或少有些出入，但差距不能太大，他們既不願量過少，也不願量過多，紀錄顯示裝瓶量為常態分配，平均值為31.2盎司，母體標準差為0.4盎司。 The Quality Assurance Department for Cola, Inc. maintains records regarding the amount of cola in its jumbo bottle. The actual amount of cola in each bottle is critical, but varies a small amount from one bottle to the next. Cola, Inc. does not wish to underfill the bottles. On the other hand, it cannot overfill each bottle. Records indicate that the amount of cola follows the normal probability distribution. The mean amount per bottle is 31.2 ounces and the population standard deviation is 0.4 ounces. 今天上午8點品管技師由生產線隨機選了16瓶，平均值為31.38盎司，請問這是否不尋常？此生產線是否裝填過量的可樂？換句話說，樣本誤差為0.18盎司是否不尋常？ At 8 a.m. today the quality technician randomly selected 16 bottles from the filling line. The mean amount of cola contained in the bottles is 31.38 ounces. Is this an unlikely result? Is it likely the process is putting too much soda in the bottles? To put it another way, is the sampling error of 0.18 ounces unusual? 8-*

LO8-5 採用樣本平均數的抽樣分配：範例 p.271 Using the Sampling Distribution of the Sample Mean – Example Step 1: 先算樣本平均31.38的z值為多少？Find the z-value corresponding to the sample mean of 31.38. 8-*

LO8-5 採用樣本平均數的抽樣分配：範例 p.271 Using the Sampling Distribution of the Sample Mean – Example Step 2: 再看z≧1.8的機率為多少？ Find the probability of observing a Z equal to or greater than 1.80. 8-*

LO8-5 採用樣本平均數的抽樣分配：範例 p.271 Using the Sampling Distribution of the Sample Mean – Example 樣本平均31.38的確不尋常，因為根據抽樣分配，這個樣本平均發生的機率低於4％，所以，這條生產線裝得太滿了！ What do we conclude? It is unlikely, less than a 4 percent chance, we could select a sample of 16 observations from a normal population with a mean of 31.2 ounces and a population standard deviation of 0.4 ounces and find the sample mean equal to or greater than 31.38 ounces. We conclude the process is putting too much cola in the bottles. 8-*

例題1： 假設拉拉山某農場生產的水密桃的重量呈現常態分配，，平均重量為300g，標準差30g，家樂福想批發採買1000箱，但為了避免批購損失，下單前先採樣檢測。問： 1. 隨機抽1個，重量超過330g的機率為何？ 2. 隨機抽12個，平均重量超過330g的機率為何？ 3. 將12個裝成一盒，一盒的重量不足3.5kg的機率為何？ 4. 隨機抽取50個裝箱，一箱的重量在15±0.5kg的機率為何？

例題1：解答 X~N(300,302) 1. P(X > 330) = P(z > (330-300)/30) = P(z > 1) = 0.5-0.3413= 0.1587 2. 抽樣分配~ N(300, 900/12) , n = 12, mean = 300, s2 = 302 /12 s = 8.66, z = (330-300)/8.66 = 3.464 P(mean > 330) = P(z > 3.46) = 0.0003 3. 12粒水密桃裝一盒，每粒的重量都是常態分配，故採用加法定理 E(ΣX) = 12μ = 12*300=3600 V(ΣX) = 12σ2 = 12*900 = 10800  s = 103.923 重量不足X=3.5kg=3500g  z = (3500-3600)/103.923 = -0.9623 P(X < 3500) = P(z < -0.96) = 0.1685 4. n = 50 >30 中央極限定理 E(ΣX) = 50μ = 50*300=15000 V(ΣX) = 50σ2 = 50*900 = 45000  s = 212.132 重量介於(15-0.5kg, 15+0.5kg)=(14500, 15500)g  z = ±500/212.132= -2.357 P(14500< X < 15500) = P(-2.36 < z < 2.36) =0.9818

例題2： 甲、乙、丙、丁、戊5人合夥開餐廳，由乙、丙2人負責經營，最低消費額訂為200元，客滿時每日營收3萬元，丁有時也到餐廳用餐，每次他去時都幾乎客滿，有時還得等候入座，一年下來，公司的損益報表顯示：平均每日營收18000元，標準差3000元，此與丁的認知差距極大，因此他懷疑乙、丙2人中飽私囊，於是聯合甲、戊，3人抽樣調查，採分層隨機抽樣，共抽36個營業日，樣本平均營收為26000元，請問乙、丙2人是否有中飽私囊？

例題2：解答 公司報表：μ = 18000，σ = 3000 n = 36, 抽樣分配的mean = μ = 18000 標準差 = 500 標準差 = 500 P( ≧26000) = P(z≧ ) = P(z ≧16) = 0 若公司母體平均營收為18000，且標準差為3000，則3人隨機抽樣，能抽出平均營收為26000的機率近乎於0，故負責經營的乙、丙2人有中飽私囊。

例題3： 台北市房價高，30坪的公寓價格約等於薪水階級15～20年薪資收入的總和，想買房子得拼命賺錢、存錢外，還得積極詢價，假設目前正待售的房價呈現右偏分配，平均數為900萬元，標準差170萬元，某人想買房子，他從房市待售屋中抽取500間，問： 1. 樣本平均與母體平均數差距在20萬元以內的機率為何？ 2. 房屋平均價格低於850萬元的機率為何？

例題3：解答 令 為500間房子的平均價格，n=500 ＞30 中央極限定理  為常態分配 = μ = 900萬, = 7.6萬 = μ = 900萬, = 7.6萬 (880-900)/7.6 = -2.63, (850-900)/7.6 = - 6.58 1. P(880 < < 920) = P(-2.63 < z < 2.63) = 0.9914 2. P( ≦850) = P(z≦-6.58) = 0

樣本平均數之抽樣分配型態 圖7.1 　抽樣分配的型態

樣本平均數之抽樣分配的性質 1/2 設隨機樣本X1,X2,…,Xn，係自無限母體抽樣（如抽後放回），則此抽樣分配的期望值、變異數與標準差分別為 式中μ與σ分別表示母體的平均數與標準差，而N為母體大小，n則為樣本大小。 (7-1)

樣本平均數之抽樣分配的性質 2/2 設隨機樣本X1,X2,…,Xn是自有限母體抽樣（如抽後不放回），則此抽樣分配的的期望值、變異數與標準差分別為 式中μ與σ分別表示母體的平均數與標準差，而N 為母體大小，n 則為樣本大小。 若 n/N < 0.05時，有限母體的抽樣可視為無限母體的情況來處理。 (7-2)

的抽樣分配