ch+9+抽样分配.ppt

1、Sampling Distributions,Chapter 9,AU,Ch 9- 2,Outline,9.1 Sampling Distribution of the Mean 9.2 Sampling Distribution of the Proportion 9.3 Sampling Distribution of the Difference between Two Means (9.4 From Here to Inference),AU,Ch 9- 3,Introduction,In real life calculating parameters of populations

2、is prohibitive because populations are very large. Rather than investigating the whole population, we take a sample, calculate a statistic related to the parameter of interest, and make an inference. The sampling distribution of the statistic is the tool that tells us how close is the statistic to t

3、he parameter.,AU,Ch 9- 4,9.1 Sampling Distribution of the (sample) Mean,例題 丟一個骰子無限次 隨機變數 X 表示骰子出現的點數 X 的機率分配 (表與圖), 平均數, 變異數為：,E(X) = m = 1(1/6) + 2(1/6) + 3(1/6)+ + 6(1/6) = 3.5 s2 = V(X) = (1-3.5)2(1/6) + (2-3.5)2(1/6) + + (6-3.5)2(1/6) = 2.92,AU,Ch 9- 5,Throwing a die twice sample mean,假設只有上帝知道X

4、的機率分配, 平均數m是3.5, 變異數s2是2.92, 而你不知道。 你希望丟骰子兩次(樣本數n=2), 藉由樣本平均點數 估計平均數m 。,對於以樣本平均數 來估計 m, 我們想重複抽樣很多次, 甚至是抽出所有可能情形, 看有怎麼樣的結果發生.,AU,Ch 9- 6,Throwing a die twice sample mean,AU,Ch 9- 7,The distribution of when n = 2,的平均數 = 1.0(1/36)+ 1.5(2/36)+.= 3.5 的變異數 = (1.0-3.5)2(1/36)+ (1.5-3.5)2(2/36). = 1.46,AU,C

5、h 9- 8,Sampling Distribution of the Mean,教科書第308頁表9.2 教科書第308頁圖9.2 如表9.2與圖9.2所示，樣本平均數也有自己的機率分配，在此將樣本平均數的機率分配改名為(樣本)平均數的抽樣分配。 教科書第309頁第11-14行,AU,Ch 9- 9,Sampling Distribution of the Mean,AU,Ch 9- 10,Sampling Distribution of the Mean,Notice that is smaller than s2x. The larger the sample size the smal

6、ler . Therefore, tends to fall closer to m, as the sample size increases.,抽樣分配的標準差改名為標準誤(standard error),課本第309頁最後一段至311頁第一段,AU,Ch 9- 11,Sampling Distribution of the Mean,Demonstration: The variance of the sample mean is smaller than the variance of the population.,1,2,3,Mean = 1.5,Mean = 2.5,Mean =

7、 2.,Population,1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2,2,2,2,2,2,2,2,2,2,2,Compare the variability of the population to the variability of the sample mean.,Let us take samples of two observations,AU,Ch 9- 12,Also, Expected value of th

8、e population = (1 + 2 + 3)/3 = 2,Expected value of the sample mean = (1.5 + 2 + 2.5)/3 = 2,Sampling Distribution of the Mean,AU,Ch 9- 13,Sampling Distribution of the Mean,In terms of mean and variance, We have known the relationship between population and distribution of sample mean. How about the s

9、hape? See figure 9.2. What have you found?,AU,Ch 9- 14,If a random sample is drawn from any population, the sampling distribution of the sample mean is approximately normal for a sufficiently large sample size. The larger the sample size, the more closely the sampling distribution of will resemble a

10、 normal distribution.,The Central Limit Theorem(中央極限定理),AU,Ch 9- 15,Finite Populations (有限母體),For infinitely large populations:,For finite populations(母體個數是有限個):,當N比n大於20倍以上時, 我們視該母體為無限母體,AU,Ch 9- 16,Sampling Distribution of the Sample Mean,The standard deviation of is called the standard error of t

11、he mean,AU,Ch 9- 17,補充幾個圖, 下學期會經常用到,AU,Ch 9- 18,樣本平均數的平均數和標準差,母體,m, s2,. . .,. . .,這些樣本平均數 的平均數會剛好等於母體平均數 m 。 即,這些樣本平均數 的變異數等於母體變異數 s2 除以 n 。 標準差等於母體標準差 s 除以根號 n 。即,SRS: 簡單隨機樣本,AU,Ch 9- 19,樣本平均數的平均數和標準差,抽樣分配的變異數( ) 小於我們抽樣母體的變異數 (s),指這些資料的變異性,指這些平均結果的變異性,AU,Ch 9- 20,樣本平均數抽樣分配的形狀,關於形狀 情形一：母體是常態分配, 也是常

12、態分配,常態分布,常態分配,AU,Ch 9- 21,樣本平均數抽樣分配的形狀,關於形狀 情形二：母體不是常態分配, 當n夠大, 近似常態分配,AU,Ch 9- 22,Example 9.1 The amount of soda pop in each bottle is normally distributed with a mean of 32.2 ounces and a standard deviation of 0.3 ounces. Find the probability that a bottle bought by a customer will contain more th

13、an 32 ounces. Solution The random variable X is the amount of soda in a bottle.,m = 32.2,0.7486,x = 32,Sampling Distribution of the Sample Mean,AU,Ch 9- 23,Find the probability that a carton of four bottles will have a mean of more than 32 ounces of soda per bottle. Solution Define the random variab

14、le as the mean amount of soda per bottle.,0.9082,Sampling Distribution of the Sample Mean,AU,Ch 9- 24,Example 9.2 Deans claim: The average weekly income of B.B.A graduates one year after graduation is $800. Suppose the distribution of weekly income has a standard deviation of $100. What is the proba

15、bility that 25 randomly selected graduates have an average weekly income of less than $750? Solution,Sampling Distribution of the Sample Mean,Suppose the weekly income, X, is not extremely nonnormal distributed.,AU,Ch 9- 25,Example 9.2 continued If a random sample of 25 graduates actually had an ave

16、rage weekly income of $750, what would you conclude about the validity of the claim that the average weekly income is 800? Solution With m = 800 the probability of observing a sample mean as low as 750 is very small (0.0062). The claim that the mean weekly income is $800 is probably unjustified (不正當

17、的). It will be more reasonable to assume that m is smaller than $800, because then a sample mean of $750 becomes more probable.,Sampling Distribution of the Sample Mean,AU,Ch 9- 26,Using Sampling Distributions for Inference,推導第317頁倒數第6個行的公式,AU,Ch 9- 27,To make inference about population parameters w

18、e use sampling distributions (as in Example 9.2). The symmetry of the normal distribution along with the sample distribution of the mean lead to:,- Z.025,Z.025,Using Sampling Distributions for Inference,AU,Ch 9- 28,Using Sampling Distributions for Inference,-1.96,-1.96,0,.025,.025,.025,.025,Standard

19、 normal distribution Z,Normal distribution of,Z,m,m=800,AU,Ch 9- 29,Conclusion There is 95% chance that the sample mean falls within the interval 760.8, 839.2 if the population mean is 800. Since the sample mean was 750, the population mean is probably not 800.,Using Sampling Distributions for Infer

20、ence,AU,Ch 9- 30,9.2 Sampling Distribution of a Proportion,老師對真理大學工管系學生做一項調查。 母體是真理大學工管系所有學生。調查訪問10位二年級同學，問題是你的手機是不是Moto? 受訪樣本中有 位回答是。,本題最有意義的參數為何? 統計量應為何? 對於真大所有工管系同學，你認為Moto手機熱門嗎?,1. 真大工管系同學持有Moto手機的比例,2. 10位同學持有Moto手機的比例,AU,Ch 9- 31,比例的符號,經常一項問題的結果只有兩種情形。 習慣上，會將感興趣的那個結果稱為成功，另一個稱為失敗。 母體中屬於成功那類佔全部的

21、比例稱為母體成功比例，或母體比例，用符號 p 表示。 例如參數是 “真大工管系持有Moto手機的比例”，符號是 p 。 樣本中屬於成功那類佔全部的比例稱為樣本成功比例，或樣本比例，用符號 表示。 例如統計量是 “受訪同學持有Moto手機的比例”，符號是,因為不知道參數 p 的值，因此用統計量估計它。,AU,Ch 9- 32,The estimator of p =,The parameter of interest for nominal data is the proportion of times a particular outcome (success) occurs. To esti

22、mate the population proportion p we use the sample proportion.,9.2 Sampling Distribution of a Proportion,AU,Ch 9- 33,Since X is binomial, probabilities about can be calculated from the binomial distribution. (因為 X 是二項分配，所以 也是二項分配) Yet, for inference about we prefer to use normal approximation to the

23、 binomial. (但是在推論時，我們還是喜歡用常態分配；很幸運地，二項分配在某些情況下可以近似到常態分配),9.2 Sampling Distribution of a Proportion,AU,Ch 9- 34,Normal approximation to the Binomial,Normal approximation to the binomial works best when the number of experiments (sample size) is large, and the probability of success, p, is close to 0.

24、5. For the approximation to provide good results two conditions should be met: np 5; n(1 - p) 5,AU,Ch 9- 35,常態作為二項的近似分配,想將二項分配近似到常態分配，最好要滿足： 實驗次數(即樣本數)很大 成功機率(p) 接近 0.5 整合上述兩條件而成為：np=5 ; n(1-p)=5 二項分配是離散型分配，常態分配是連續型分配；故用常態分配求近似機率時，須將 X 的範圍加以轉換 (目的是可以求面積),AU,Ch 9- 36,Normal approximation to the Binom

25、ial,Example Approximate the binomial probability P(x=10) when n = 20 and p = 0.5 The parameters of the normal distribution used to approximate the binomial are: m = np; s2 = np(1 - p),AU,Ch 9- 37,Normal Approximation to Binomial,Binomial distribution with n=20 and p=.5 似乎可以近似到常態分配,AU,Ch 9- 38,10,P(X

26、Binomial = 10) =,m = np = 20(0.5) = 10; s2 = np(1 - p) = 20(0.5)(1 - 0.5) = 5 s = 51/2 = 2.24,.176,Let us build a normal distribution to approximate the binomial P(X = 10).,Normal approximation to the Binomial,AU,Ch 9- 39,More examples of normal approximation to the binomial,4,14,P(X 14) ,P(Y 4.5),P

27、(Y 13.5),Normal approximation to the Binomial,P(X 4) ,P(X 14) ?,這裡的0.5被稱為連續校正因子,AU,Ch 9- 40,Approximate Sampling Distribution of a Sample Proportion,From the laws of expected value and variance, it can be shown that E( ) = p and V( ) =p(1-p)/n 證明 If both np = 5 and n(1-p) = 5, then 當計算 X的機率時, 必須作連續性

28、修正(將X範圍加或減0.5);同理, 在計算 的機率時, 也作連續性修正 ( 範圍加或減 1/(2n) ) 但在 情形中, 因為有無作修正的機率相差較小, 故可不作連續性修正,AU,Ch 9- 41,Approximate Sampling Distribution of a Sample Proportion,If both np = 5 and n(1-p) = 5, then If both np = 5 and n(1-p) = 5, then Z is approximately standard normally distributed.,The standard deviatio

29、n of is called the standard error of the proportion,AU,Ch 9- 42,樣本比例的形成,AU,Ch 9- 43,Example 9.2 A state representative received 52% of the votes in the last election. One year later the representative wanted to study his popularity. If his popularity has not changed, what is the probability that mor

30、e than half of a sample of 300 voters would vote for him?,AU,Ch 9- 44,Example 9.2 Solution The number of respondents who prefer the representative is binomial with n = 300 and p = 0.52. Thus, np = 300(0.52) = 156 andn(1-p) = 300(1-0.52) = 144 (both greater than 5),AU,Ch 9- 45,9.3 Sampling Distribution of the Difference Between Two Means,Independent samples are drawn from each of two normal populations Were interested in the sampling distribution of the