1、&TwoDOE Class 90a1點圖 (Dot Diagram)&TwoDOE Class 90a2直方圖 (Histogram)&TwoDOE Class 90a3盒形圖 (Box Plot)&TwoDOE Class 90a4時間序列圖 (Time Series Plot)&TwoDOE Class 90a5期望值與變異數之公式n母體平均數(m ) = 隨機變數之期望值 E(X)n母體變異數(s 2) = 隨機變數之變異數 V(X)2 )( , ,)()(mxEXVdiscretexp(x)continuousdxxxfXExAll&TwoDOE Class 90a6期望值與變異數之公
2、式)()()()()()()()()( , 0),( ,),( ),(2)()()()()()()()(21212121212121212211212121211212121xExExxExExExxExVxVxxVxxCovxxxxExxCovxxCovxVxVxxVxVaaxVxbExaEbxaxE且則獨立若其中mm&TwoDOE Class 90a7Sample and Sampling&TwoDOE Class 90a8點估計(Point Estimation)n以抽樣得來之樣本資料, 依循某一公式計算出單一數值, 來估計母體參數, 稱為點估計.n好的點估計公式之條件:n不偏性n最小變
3、異n常用之點估計:n母體平均數(m)n母體變異數(s 2)nXXi1122nXXSnii&TwoDOE Class 90a9Central Limit Theorem&TwoDOE Class 90a10假設檢定(Hypothesis Testing)n“A person is innocent until proven guilty beyond a reasonable doubt.” 在沒有充分證據證明其犯罪之前, 任何人皆是清白的.n假設檢定H0: m = 50 cm/sH1: m 50 cm/snNull Hypothesis (H0) Vs. Alternative Hypothe
4、sis (H1)nOne-sided and two-sided HypothesesnA statistical hypothesis is a statement about the parameters of one or more populations.&TwoDOE Class 90a11About TestingnCritical RegionnAcceptance RegionnCritical Values&TwoDOE Class 90a12Errors in Hypothesis Testingn檢定結果可能為nType I Error(a): Reject H0 whi
5、le H0 is true.nType II Error(b): Fail to reject H0 while H0 is false.&TwoDOE Class 90a13 The Defendant isThe Jury finds thepersonInnocentGuiltyInnocentType II ErrorGuiltyType I Error )(:)(:10GuiltyHInnocentH有罪無辜&TwoDOE Class 90a14Making ConclusionsnWe always know the risk of rejecting H0, i.e., a, t
6、he significant level or the risk.nWe therefore do not know the probability of committing a type II error (b).nTwo ways of making conclusion:1. Reject H02. Fail to reject H0, (Do not say accept H0) or there is not enough evidence to reject H0.&TwoDOE Class 90a15Significant Level (a)na = P(type I erro
7、r) = P(reject H0 while H0 is true)n = 10, s = 2.5s/n = 0.79&TwoDOE Class 90a16&TwoDOE Class 90a17&TwoDOE Class 90a18&TwoDOE Class 90a19&TwoDOE Class 90a20The Power of a Statistical TestnPower = 1 - bnPower = the sensitivity of a statistical test&TwoDOE Class 90a211. From the problem context, identif
8、y the parameter of interest.2. State the null hypothesis, H0.3. Specify an appropriate alternative hypothesis, H1.4. Choose a significance level a.5. State an appropriate test statistic.6. State the rejection region for the statistic.7. Compute any necessary sample quantities, substitute these into
9、the equation for the test statistic, and compute that value.8. Decide whether or not H0 should be rejected and report that in the problem context.General Procedure for Hypothesis Testing&TwoDOE Class 90a22Inference on the Mean of a Population-Variance KnownnH0: m = m0H1: m m0 , where m0 is a specifi
10、ed constant. nSample mean is the unbiased point estimator for population mean.1 , 0 then),( trueis H if Therefore,., then , varianceand mean with ondistributi a fromdrawn samples are , If00002221NnXZnNXXXXnsmmmsmsm&TwoDOE Class 90a23Example 8-2Aircrew escape systems are powered by a solid propellant
11、. The burning rate of this propellant is an important product characteristic. Specifications require that the mean burning rate must be 50 cm/s. We know that the standard deviation of burning rate is 2 cm/s. The experimenter decides to specify a type I error probability or significance level of = 0.
12、05. He selects a random sample of n = 25 and obtains a sample average of the burning rate of x = 51.3 cm/s. What conclusions should be drawn?&TwoDOE Class 90a241.The parameter of interest is m, the meaning burning rate.2.H0: m = 50 cm/s3.H1: m 50 cm/s4.a = 0.055.The test statistics is:6.Reject H0 if
13、 Z0 1.96 or Z0 1.96, we reject H0: m = 50 at the 0.05 level of significance. We conclude that the mean burning rate differs from 50 cm/s, based on a sample of 25 measurements. In fact, there is string evidence that the mean burning rate exceeds 50 cm/s. nxZ/00sm25. 325/2503 .510Z&TwoDOE Class 90a25P
14、-Values in Hypothesis TestsnWhere Z0 is the test statistic, and (z) is the standard normal cumulative function.&TwoDOE Class 90a26The Sample Size (I)nGiven values of a and d, find the required sample size n to achieve a particular level of b.02222/2/2/2/2/ re whe / Then, Let 0 when / / Sincemmddssdb
15、dsdsdsdbbaabbaaaZZnnZZZnZnZnZ&TwoDOE Class 90a27The Operating Characteristic Curves- Normal test (z-test)nUse to performing sample size or type II error calculations.nThe parameter d is defined as:so that it can be used for all problems regardless of the values of m0 and s.n課本41頁之公式為兩平均數差之假設檢定所需之樣本數
16、公式。sdsmm|0d&TwoDOE Class 90a28&TwoDOE Class 90a29Construction of the C.I.nFrom Central Limit Theory, . , ,25 and , 22nNXnXIfsmsmnUse standardization and the properties of Z, asmsasmasmaaaaaa1/11 and 2/2/2/2/2/2/nzXnzXPznXzPzZzPnXZ&TwoDOE Class 90a30Inference on the Mean of a Population-Variance Unkn
17、ownnLet X1, X2, , Xn be a random sample for a normal distribution with unknown mean m and unknown variance s2. The quantityhas a t distribution with n - 1 degrees of freedom.&TwoDOE Class 90a31Inference on the Mean of a Population-Variance UnknownnH0: m = m0H1: m m0 , where m0 is a specified constan
18、t. nVariance unknown, therefore, use s instead of s in the test statistic.nIf n is large enough ( 30), we can use Z-test. However, n is usually small. In this case, T0 will not follow the standard normal distribution.&TwoDOE Class 90a32Inference for the Difference in Means-Two Normal Distributions a
19、nd Variance UnknownnWhy? &TwoDOE Class 90a33&TwoDOE Class 90a34 have, we still and S with S and placing 22212221Re is distributed approximately as t with degrees of freedom given by&TwoDOE Class 90a35C.I. on the Difference in Means&TwoDOE Class 90a36C.I. on the Difference in Means&TwoDOE Class 90a37
20、Paired t-TestnWhen the observations on the two populations of interest are collected in pairs.nLet (X11, X21), (X12, X22), , (X1n, X2n) be a set of n paired observations, in which X1j(m1, s12) and X2j(m2, s22) and Dj = X1j X2j, j = 1, 2, , n. Then, to test H0: m1= m2 is the same as performing a one-
21、sample t-test H0: mD = 0 since mD = E(X1-X2) = E(X1)-E(X2) = m1 - m2 &TwoDOE Class 90a38&TwoDOE Class 90a39Inference on the Variance of a Normal Population (I)nH0: s2 = s02H1: s2 s02 , where s02 is a specified constant. nSampling from a normal distribution with unknown mean m and unknown variance s2
22、, the quantityhas a Chi-square distribution with n-1 degrees of freedom. That is, 2221sSn212221nSns&TwoDOE Class 90a40Inference on the Variance of a Normal Population (II)nLet X1, X2, , Xn be a random sample for a normal distribution with unknown mean m and unknown variance s2. To test the hypothesi
23、sH0: s2 = s02H1: s2 s02 , where s02 is a specified constant.We use the statisticnIf H0 is true, then the statistic has a chi-square distribution with n-1 d.f.&TwoDOE Class 90a41 kkkxexkxfxkk2 addition,In freedom. of degrees ofnumber theis 0 2/21 :ondistributi square-chi of PDF22/12/2/sm&TwoDOE Class
24、 90a42The ReasoningnFor H0 to be true, the value of 02 can not be too large or too small.nWhat values of 02 should we reject H0? (based on a value) What values of 02 should we conclude that there is not enough evidence to reject H0?&TwoDOE Class 90a43&TwoDOE Class 90a44Example 8-11An automatic filli
25、ng machine is used to fill bottles with liquid detergent. A random sample of 20 bottles results in a sample variance of fill volume of s2 = 0.0153 (fluid ounces)2. If the variance of fill volume exceeds 0.01 (fluid ounces)2, an unacceptable proportion of bottles will be underfilled and overfilled. I
26、s there evidence in the sample data to suggest that the manufacturer has a problem with underfilled and overfilled bottles? Use a = 0.05, and assume that fill volume has a normal distribution.&TwoDOE Class 90a451. The parameter of interest is the population variance s2.2. H0: s2 = 0.013. H1: s2 0.01
27、4. a = 0.055. The test statistics is 6. Reject H0 if 7. Computations:8. Conclusions: Since , we conclude that there is no strong evidence that the variance of fill volume exceeds 0.01 (fluid ounces)2.14.3007.29219,05. 02007.2901. 00153. 0192014.30219,05. 020202201ssn &TwoDOE Class 90a46Hypothesis Te
28、sting on Variance - Normal PopulationH1Test StatisticReject H0 ifs2 s0221,2/12021,2/20or nnaas2 s02 21,20nas2 s02202201sSn 21,120na&TwoDOE Class 90a47&TwoDOE Class 90a48&TwoDOE Class 90a49&TwoDOE Class 90a50&TwoDOE Class 90a51精品课件精品课件！&TwoDOE Class 90a52精品课件精品课件！&TwoDOE Class 90a53Hypothesis Testing on the Ratio of Two Variances
