1、Copyright 2015 McGraw-Hill Education.All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.Estimation and Confidence IntervalsChapter 929-*Learning ObjectivesnLO9-1 Compute and interpret a point estimate of a population mean.nLO9-2 Compute and
2、 interpret a confidence interval for a population mean.nLO9-3 Compute and interpret a confidence interval for a population proportion.nLO9-4 Calculate the required sample size to estimate a population proportion or population mean.nLO9-5 Adjust a confidence interval for finite populations.9-估計的基本概念n
3、點估計(point estimation)：根據樣本資料求得一估計值，以推估未知的母體參數。n區間估計(interval estimation)：根據樣本資料所求出的點估計值(point estimate)，藉由點估計量抽樣分配的性質求出兩個數值而構成一區間，稱為區間估計值(interval estimate)，並利用此一區間推估未知母體參數的範圍。49-*點估計 Point Estimatesn點估計：由樣本中計算出一個代表（估計）母體數值的樣本數值nA point estimate is a single value(point)derived from a sample and used
4、 to estimate a population value.pssX22LO9-1 Compute and interpret a point estimate of a population mean.9-1-1點估計點估計 點點估估計計的的步步驟驟抽取具代表性的樣本選擇一個較佳的樣本統計量做為估計式計算樣本統計量的值以樣本統計量的值推論母體參數值並做決策9-統計估計分兩類：(p.280)n點估計點估計(point estimate)：是從一組樣本中根據樣本統計式所計算而得的數值，我們用此數值來推論母體參數值。e.g.某教授為了了解大學畢業生第一份工作的平均所得，就以簡單抽樣的方法，從全
5、台灣25萬大學畢業生這個母體當中，抽出50個為一組的隨機樣本。然後計算此樣本的平均薪資為2萬1千元。此教授若做以下結論：台灣的大學畢業生第一份工作的平均薪資為2萬1千元！9-點估計你覺得這個教授的結論可靠嗎？或者說，這個樣本統計值(50位大學畢業生第一份工作薪資的平均值)是否接近、代表母體參數(全台灣大學畢業生第一份工作薪水的平均值)？用點估計的結果去推估母體，不見得準確！只用一組樣本的樣本統計量去推估母體參數，非常不可靠！因為，再抽取另外一組樣本(50個樣本點)，得到的點估計值會跟前一組樣本的點估計不同，請問你要用哪一個點估計？以單單一組樣本得到的點估計值去估計母體參數值，缺點如下：(1)無
6、法得知是否真是母體參數的真值(2)如果不等，其誤差的大小無法得知89-*信賴區間估計Confidence Interval Estimatesn信賴區間估計(C.I.)：在某個特定機率下，由樣本資料推估母體參數落在那個區間內，亦為一種區間估計。這個特定機率又稱為信賴水準(level of confidence)nA confidence interval estimate is a range of values constructed from sample data so that the population parameter is likely to occur within tha
7、t range at a specified probability.The specified probability is called the level of confidence.LO9-2 Compute and interpret a confidence interval for a population mean.C.I.=點估計 誤差C.I.=point estimate margin of error9-信賴區間估計:步驟(1)透過樣本取得點估計值（平均數）e.g.大學畢業生第一年的平均薪資為2萬1千元(2)以此估計值為中心，根據特定的信賴水準，導出一個上下限的信賴區間e
8、.g.信賴區間為1萬8千元與2萬5千元之間 (3)常用的信賴水準為95或是99若是95的信賴水準，表示：我們有我們有9595的信心，母體參數會落在這個信賴區間之內的信心，母體參數會落在這個信賴區間之內e.g.大學畢業生第一年的薪資介於1萬8千元與2萬5千元之間，此區間包含母體平均數(全台大學畢業生平均薪資)的可信度(機率)為95。(有有95%95%的信心，母體參數會位於這個區間之內的信心，母體參數會位於這個區間之內)9-如何決定信賴區間的範圍？Factors Affecting Confidence Interval Estimates從樣本平均數推估母體平均數所在的信賴區間，有兩種情況：(1
9、)常態分配下常態分配下母體的標準差母體的標準差已知已知：(2)母體的標準差未知:用樣本標準差s取代母體標準差119-*已知，求母體平均的信賴區間Confidence Intervals for a Mean,Knownsample in the nsobservatio ofnumber thedeviation standard population thelevel confidence particular afor value-z mean sample nzx區間大小由信賴水準與平均數的標準差來決定。The width of the interval is determined by
10、 the level of confidence and the size of the standard error of the mean.而標準差受到下列二值的影響：The standard error is affected by two values:標準差(Standard deviation)樣本數 n(Number of observations in the sample)LO9-29-的抽樣分配為常態X註：註：z值為標準常態邊際值。值為標準常態邊際值。圖圖 P(Z/2Zz/2)=1(8-6)9-的抽樣分配為常態 n母體參數的信賴區間點估計量 抽樣誤差X2/2/2/2/2/2
11、/21/1XPzznP XzXznnXzznnXzn (8-9)9-常態母體資料的標準差已知Q:如何從樣本平均值推估未知未知的母體平均值？回想上章學到的已知分配為常態(分配未知但樣本夠大)，母體平均值、標準差已知已知，當要推估樣本平均值樣本平均值抽樣分配所涵蓋的機率時，很容易將其標準化因為我們知道：N(ux,x2/N)如果 P(-A A)=95%，求A為多少？=P(-1.96 Z 1.96)=95%=P(-1.96 (-=P(-1.96 (-x x)/()/(x x/n/n)1.96)=95%)1.96)=95%=P(x x-1.96(x/n)x x 1.96(x/n)）95然而，在這章中，有
12、個已知的 ，但 x x 未知未知：=P(-1.96(=P(-1.96(x x/n/n)x x 1.96(1.96(x x/n/n)）9595此為95%信賴區間，包含母體平均值的機率為95XXXXXXX9-信賴水準P(-1.96(P(-1.96(x x/n/n)x x 1.96(1.96(x x/n/n)）9595信賴水準95：此信賴區間，包含母體平均值的機率為95n若是信賴水準是若是信賴水準是9999？P(-b b)=99%=P(-2.58 Z 2.58)=99%P(P(-2.58(-2.58(x x/n/n)x x 2.58(2.58(x x/n/n)）9999n若是信賴水準是若是信賴水準是
13、9090？P(P(-1.64(-1.64(x x/n/n)x x 1.64(1.64(x x/n/n)）9090XXXXXXX169-*影響信賴區間估計的因素：Factors Affecting Confidence Interval Estimates信賴區間的大小受到下列因素的影響：The width of a confidence interval is determined by:樣本大小樣本大小(The sample size),n 母體的變異度母體的變異度/離散度離散度(The variability in the population,usually estimated by s
14、)信賴水準信賴水準 (The desired level of confidence)LO9-29-信賴水準的意義(p.282)n95%信賴水準下，母體參數的信賴區間為 1.96(1.96(x x/n/n)信賴區間的大小會隨著樣本數樣本數不同而有所不同，若將全部可能的樣本樣本平均值平均值計算出所有的區間，若以95為信賴水準，則表示：在所有的區間中有95的區間會包含母體參數，有5的區間不包含母體參數。e.g.如果選出100組樣本（各內含50位大學畢業生薪資)，然後計算各樣本的平均值。從各樣本平均值再推出各個相對應的信賴區間，共有100個信賴區間。95的這些信賴區間會涵蓋到母體母體平均值注意：以上
15、都是以隨機抽樣的方式抽出所有的樣本。95信賴區間表示母體參數有95的機率（我們相信）落在此區間內X9-的抽樣分配為常態n信賴區間 為一可能包含之真實值的隨機區間。n機率 ，可解釋為，多次重複抽樣下之相對次數，亦即大約有(1)100%個區間包含。n只要由所觀察的樣本求得後，隨機區間 即可視為的(1)100%信賴區間。X/2/2/,/Xzn Xzn/2/2(/)1P XznXzn/2(xz/2/,/)n xzn9-的抽樣分配為常態X鏈結圖鏈結圖 8.6圖圖 母體平均數母體平均數的的95%信賴區間，信賴區間，100個區間個區間中有中有94個包含個包含209-*區間估計：闡釋Interval Esti
16、mates-Interpretation95信賴區間：表示若將所有95信賴區間全都估算出來，有95的區間內包含母體參數，或在特定的n下，有95的樣本平均值落在假定母體平均值1.96個標準差的範圍內。For a 95%confidence interval about 95%of similarly constructed intervals will contain the parameter being estimated.Also 95%of the sample means for a specified sample size will lie within 1.96 standard
17、 deviations of the hypothesized population.LO9-2未知！219-*已知，求母體平均的信賴區間（例子）p.285Confidence Interval for a Mean,Known-Example美國經理協會調查零售業中級經理的平均年薪，他們隨機抽了49個樣本，樣本平均值為$45,420，母體標準差為$2,050The American Management Association surveys middle managers in the retail industry and wants to estimate their mean ann
18、ual income.A random sample of 49 managers reveals a sample mean of$45,420.The standard deviation of this population is$2,050.母體平均值的最佳點估計為？What is the best point estimate of the population mean?母體平均數值的合理範圍為？What is a reasonable range of values for the population mean?上述計算結果的意涵為？What do these results
19、mean?LO9-2229-*已知，求母體平均的信賴區間（例子）p.285Confidence Interval for a Mean,Known-Example美國經理協會調查零售業中級經理的平均年薪，他們隨機抽了49個樣本，樣本平均值為$45,420，母體標準差為$2,050The American Management Association surveys middle managers in the retail industry and wants to estimate their mean annual income.A random sample of 49 managers
20、 reveals a sample mean of$45,420.The standard deviation of this population is$2,050.母體平均值的最佳點估計為？What is the best point estimate of the population mean?因為我們不知道（也無從得知）母體平均值，於是，最佳估計值就是隨機抽樣的樣本統計值Our best estimate of the unknown population mean is the corresponding sample statistic.因此，樣本平均值$45,420為未知母體平
21、均的點估計The sample mean of$45,420 is the point estimate of the unknown population mean.LO9-2239-*已知，求母體平均的信賴區間（例子）p.285Confidence Interval for a Mean,Known-Example美國經理協會調查零售業中級經理的平均年薪，他們隨機抽了49個樣本，樣本平均值為$45,420，母體標準差為$2,050The American Management Association surveys middle managers in the retail industr
22、y and wants to estimate their mean annual income.A random sample of 49 managers reveals a sample mean of$45,420.The standard deviation of this population is$2,050.母體平均數值的合理範圍為？What is a reasonable range of values for the population mean?Suppose the association decides to use the 95 percent level of
23、confidence.若信賴區間訂為95，我們用下式來求區間估計範圍：LO9-2249-*給定信賴水準後，如何查表找到z值 p.283How to Obtain a z-value for a Given Confidence Level查表得知：要求95的信賴水準，對應的z值為1.96（因為0.475*2=0.95)The 95 percent confidence refers to the middle 95 percent of the observations.Therefore,the remaining 5 percent are equally divided between
24、the two tails.Following is a portion of Appendix B.3.LO9-2259-*已知，求母體平均的信賴區間（例子）p.285Confidence Interval for a Mean,Known-Example美國經理協會調查零售業中級經理的平均年薪，他們隨機抽了49個樣本，樣本平均值為$45,420，母體標準差為$2,050The American Management Association surveys middle managers in the retail industry and wants to estimate their m
25、ean annual income.A random sample of 49 managers reveals a sample mean of$45,420.The standard deviation of this population is$2,050.95%的信賴區間估計為：The 95 percent confidence interval estimate is:LO9-2269-*已知，求母體平均的信賴區間（例子）p.285Confidence Interval for a Mean,Known-Example Confidence Interval for a Mean I
26、nterpretation美國經理協會調查零售業中級經理的平均年薪，他們隨機抽了49個樣本，樣本平均值為$45,420，母體標準差為$2,050The American Management Association surveys middle managers in the retail industry and wants to estimate their mean annual income.A random sample of 49 managers reveals a sample mean of$45,420.The standard deviation of this popu
27、lation is$2,050.如何解釋信賴上下限：如何解釋信賴上下限：$45,846 與與$45,994？What is the interpretation of the confidence limits$45,846 and$45,994?若我們隨機抽若我們隨機抽4949個經理作樣本，且抽出許多個經理作樣本，且抽出許多組樣本（每組都是組樣本（每組都是4949個樣本），每組樣本都個樣本），每組樣本都算出樣本平均值，然後計算出各組的算出樣本平均值，然後計算出各組的9595信信賴區間，我們可以預期大約賴區間，我們可以預期大約9595的區間內包的區間內包含母體平均值含母體平均值（雖然我們並不知
28、道（雖然我們並不知道？）？），且也可預期，且也可預期5 5的區間不包含母體平均值的區間不包含母體平均值 。If we select many samples of 49 managers,and for each sample we compute the mean and then construct a 95 percent confidence interval,we could expect about 95 percent of these confidence intervals to contain the population mean.Conversely,about 5 p
29、ercent of the intervals would not contain the population mean annual income,.LO9-29-P.286-288 電腦模擬範例Town Bank多年經營汽車出租業，知道租4年合約的平均行駛距離為5萬英里，標準差為5千英里，以上為母體參數，若他想實驗看看抽樣來估算母體平均，於是，他抽30個樣本觀察值，用信賴區間來估計母體平均，據此實驗，我們想看看是否95個區間會包含母體平均值，若抽60組，應該有57組會包含5萬英里。（為方便計算，單位：千英里）(Sol):用統計軟體抽60組隨機樣本(n=30)，/n=5/30=0.9139
30、5%信賴區間列於287頁的表中，表中顯示：共有4組的95信賴區間不包含5萬英里，故而僅56組（佔93.33）包含母體平均值。由此更可看出抽樣誤差的存在，或更進一步說：用特定一組隨機樣本來代表母體，可能還是會產生偏誤的問題，且即使用區間估計也是如此。9-範例假設花蓮吉安鄉的碾米廠根據過去的經驗，分裝大包裝的白米時，每包重量標準差是9公斤。今隨機抽樣100包白米秤重，平均每包105公斤。請問：在95的信賴水準下，白米平均重量(ux)的信賴區間是多少？P(z*/n x +z*/n)=95%P(105 1.96*9/100 x 105+1.96*9/10)=95%P(103.236 x 106.764
31、)=95%在95的信賴水準下，每包白米平均重量在103.236公斤及106.764公斤之間。XX9-p.289 Q.7XN(?,2.32),n=60.=8.6問：a.點估計？8.6b.在=99%，P(_ x 5n5N-1N-1609-*A Confidence Interval for a Population Proportion,LO9-39-母體比例的區間估計/2/2/2/2/2/2/2/2(1)/()11(1)/(1)/(1)/1(1)/,(1)/PPZPPnPzZzPPPzzPPnP PzPPnPPzPPnPzPPn PzPPn 629-*A Confidence Interval
32、for a Population Proportion,-Example(p.299)BBA工會想與TU工會合併，但必須至少3/4成員同意才能合併，在BBA工會內抽2000名會員調查，有1600人同意合併，母體比例為何？The union representing the Bottle Blowers of America(BBA)is considering a proposal to merge with the Teamsters Union.According to BBA union bylaws,at least three-fourths of the union members
33、hip must approve any merger.A random sample of 2,000 current BBA members reveals 1,600 plan to vote for the merger proposal.What is the estimate of the population proportion?95信賴區間？根據此抽樣調查，你能否下結論說：已有足夠會員同意合併？為什麼？Develop a 95 percent confidence interval for the population proportion.Basing your decis
34、ion on this sample information,can you conclude that the necessary proportion of BBA members favor the merger?Why?LO9-3First,compute the sample proportion:pxn1,6002000 0.80Compute the 95%Confidence Interval pzp(1p)n0.801.96.80(1.80)2,000.80.018 (0.782,0.818)Conclude:The merger proposal will likely p
35、ass because the interval estimate includes values greaterthan 75 percent of the union membership.9-樣本比例的應用：選舉民調 p.299-300某人要競選國會議員，必須取得過半數選票才能當選，他抽樣500位選民調查，結果有275人說會投他，請問他是否能當選？（分析結果）：樣本比例 p=275/500=0.55點估計值 0.5，能當選？95%區間估計：p zp(1-p)/n =0.55 1.96.55(1-.55)/500=0.55 0.044=(0.506,0.594)因為此信賴區間的下限大於0.
36、5故他非常可能會當選！9-EX 9-16 p.3009-EX 9-16 p.300a.0.75,found by 300/400.點估計b.Between 0.694 and 0.806,found by(99%,z=2.5762.58)c.We are reasonably sure the population proportion is between 69 and 81 percent.We expect 99%of similar constructed samples to contain the true population proportion.669-*估計該抽多少樣本？適
37、當的樣本大小：Selecting an Appropriate Sample Size下列3種因素可決定樣本大小，但都與母體大小無關：There are 3 factors that determine the size of a sample,none of which has any direct relationship to the size of the population:n 需要多大的信賴水準？信賴水準越大，樣本數就越大The level of confidence desired n 能容忍多大的誤差？誤差越小，樣本數就越大The margin of error the re
38、searcher will toleraten 母體變異有多大？母體離散度越大，樣本數就越大The variation in the population being studiedLO9-4 Calculate the required sample size to estimate a population proportion or population mean.679-*選擇適當的樣本大小：若母體標準差未知呢？Selecting an Appropriate Sample Size:What if the Population Standard Deviation is not Kn
39、own?如何估算？有3種可能的估算法：1.Conduct a pilot study抽一組小樣本，計算樣本標準差，來替代2.Use a comparable study若有類似研究，用該研究所估算的標準差，作為此處的3.Use a range-based approach用全距除以6，來替代LO9-49-平均數的估計誤差(margin of error)：EnE 乃是信賴區間公式中用來估算誤差項的式子：Ezn222E nz2znE699-*估計該抽多少樣本？Sample Size for Estimating the Population Mean2 EznLO9-4709-*估計該抽多少樣本
40、？例1(p.302)Sample Size for Estimating Population Mean Example 1(p.302)公行系學生想知道大城市的市議員平均月薪是公行系學生想知道大城市的市議員平均月薪是多少，她想用多少，她想用95的信賴區間來估計，且誤差的信賴區間來估計，且誤差要小於要小於100，她用勞工部的資料估計標準差為，她用勞工部的資料估計標準差為1000，請問她的樣本應該多大？，請問她的樣本應該多大？A student in public administration wants to determine the mean amount members of city
41、councils in large cities earn per month.She would like to estimate the mean with a 95%confidence interval and a margin of error of less than$100.The student found a report by the Department of Labor that estimated the standard deviation to be$1,000.What is the required sample size?Given in the probl
42、em:E,the maximum allowable error,is$100,The value of z for a 95 percent level of confidence is 1.96,The estimate of the standard deviation is$1,000.故應該抽故應該抽385385個樣本個樣本nzE2(1.96)($1,000)$1002(19.6)2384.16385LO9-49-估計該抽多少樣本？例1(p.302)Sample Size for Estimating Population Mean Example 1n若她要提高信賴水準到99，要抽
43、多少樣本才夠呢？n=664 增加了279個觀察值(72%)n若她要降低誤差到50呢？n=1537 增加了1152個觀察值(299%)n若母體誤差提高到1200呢？n=554 增加了169個觀察值(44%)222.576 1000663.58100znE221.96 10001536.6450znE221.96 1200553.19100znE729-*估計該抽多少樣本？例2Sample Size for Estimating Population Mean Example 2消費者團體要估算單一家庭在七月份的平均電費，誤差不超過5，信賴水準99，根據類似研究，標準差估計為20，應該抽多少樣本？
44、A consumer group would like to estimate the mean monthly electricity charge for a single family house in July within$5 using a 99 percent level of confidence.Based on similar studies,the standard deviation is estimated to be$20.00.How large of a sample is required?n(2.58)(20)52107LO9-4nzE29-估計母體比例母體
45、比例應抽多少樣本？的估計誤差(margin of error)：EnE 乃是信賴區間公式中用來估算誤差項的式子：(1)Ezn22(1)E nz2(1)znE749-*估計母體比例應抽多少樣本？Sample Size for Estimating a Population Proportionwhere:n is the size of the samplez is the standard normal value corresponding to the desired level of confidenceE is the maximum allowable errorLO9-4n(1)z
46、E29-估計母體比例應抽多少樣本？n若不知道若不知道，可用樣本比例取代，可用樣本比例取代n若無法估計樣本比例，可用若無法估計樣本比例，可用0.5取代取代769-*估計母體比例應抽多少樣本？例1 Sample Size for Estimating Population Proportion Example 1美國養狗俱樂部想估算兒童養寵物狗的比例有多少，若他美國養狗俱樂部想估算兒童養寵物狗的比例有多少，若他們不想誤差超過母體比例的們不想誤差超過母體比例的3，他們應該訪問多少兒童才，他們應該訪問多少兒童才夠？假設夠？假設95的信賴水準，且他們估計有的信賴水準，且他們估計有30兒童養狗。兒童養狗。
47、The American Kennel Club wants to estimate the proportion of children that have a dog as a pet.If the club wants the estimate to be within 3%of the population proportion,how many children would they need to contact?Assume a 95%level of confidence and that the club estimated that 30%of the children h
48、ave a dog as a pet.89703.96.1)70)(.30(.2nLO9-4n(1)zE2779-*估計母體比例應抽多少樣本？例1(p.303)Sample Size for Estimating Population Proportion Example 2想研究有多少比例的城市由私人收集清運垃圾，調想研究有多少比例的城市由私人收集清運垃圾，調查人員希望誤差小於母體比例的查人員希望誤差小於母體比例的0.10，信賴水準為，信賴水準為90，但無法估算母體比例。請問應該抽多少樣本？，但無法估算母體比例。請問應該抽多少樣本？A study needs to estimate the
49、proportion of cities that have private refuse collectors.The investigator wants the margin of error to be within.10 of the population proportion,the desired level of confidence is 90 percent,and no estimate is available for the population proportion.What is the required sample size?LO9-4n(1)zE2n(.5)
50、(1.5)1.645.102 67.65n 68 cities9-p.304 ex 26Past surveys reveal that 30%of tourists going to Las Vegas to gamble spend more than$1,000.The Visitors Bureau of Las Vegas wants to update this percentage.a.The new study is to use the 90%confidence level.The estimate is to be within 1%of the population p
