1、Chapter 7   假設檢驗 Hypothesis testing 統計假設檢驗是統計推斷的另一個核心內容 . 數理統計中稱有關總體分布的論斷為統計假設 . 統計假設檢驗就是根據來自總體的樣本來判斷統計假設是否成立 . 例如 : 我們根據以往經驗判斷某地區人均年收入為 8000 元 , 這個判斷就是統計假設 ; 那么這個假設是否成立 ? 抽取一定量的樣本 , 通過樣本數據來檢驗 , 這就是統計假設檢驗 . 假設檢驗在理論研究和實際應用上都占有重要地位 . 本章主要介紹假設檢驗的基本概念和正態總體假設檢驗方法 . Statistical hypothesis testing is

2、another core issue in statistical inference. We called the thesis about population distribution the statistical assumptions in Mathematical Statistics. A statistical hypothesis is an assertion about the distribution of one or more random variables. A test of statistical hypothesis is a rule which, w

3、hen the experimental sample values have been obtained, leads to a decision to accept or to reject the hypothesis under consideration. Hypothesis testing owns an important status both in theoretical research and practical applications. In this chapter we will introduce the basic concepts of hypothesi

4、s testing and the hypothesis testing methods in normal population station. § 7.1 假設檢驗的基本概念 §7.1 The basic concepts of hypothesis testing關于總體分布的論斷有兩種形式 . 一種是關于總體分布類型的論斷 , 稱為非參數假設 ; 另一種是分布類型已知 , 關于該分布中未知參數的論斷 , 稱為參數假設 . 對一個樣本進行考察 , 從而決定它能否合理地被認為與假設相符 , 這一過程叫做假設檢驗 . 本章只涉及參數假設的假設檢驗 . The these

5、s about the population distribution have two forms. One is the thesis about the population distribution types, known as non-parametric hypothesis; another is that the type of population distribution is known but with unknown parameters, we called the thesis about the unknown parameters parametric hy

6、pothesis. We will only introduce parametric hypothesis testing in this chapter. 一、問題的提出1 An example 我們通過一個例子來引出假設檢驗的一些重要概念 . In this section, we will introduce some important concepts through an example of hypothesis test. 例 7.1 從某地區 2002 年的新生兒中隨機抽取 20 個 , 測得其平均體重為 3160g, 樣本標準差為 300g, 根據過去的統計資料 , 新生

7、兒平均體重為 3140g. 設新生兒體重服從正態分布 . 問 2002 年的新生兒體重與過去有無顯著差異 ?Example 7.1 We selected 20 newborns randomly from a region in 2002, the average weight of them is 3160g, the sample standard deviation of the weight is 300g, and based on past statistics, the average weight of newborn is 3140g. If the weight of o

8、beys normal distribution. Is there any significant difference about the weight between the newborn in 2002 and the old ones?用隨機變量 Let denote the weight of newborn, then based on the hypothesis, we have., 問題是我們想通過樣本來判The problem is that whether the mean of population是否等于 3140g, 統計上表述為 is equal to 314

9、0g or not, which can be expressed as ,This稱此假設為零假設或原假設 . 判斷結果或者是接受該假設 , 或者拒絕該假設 , 如出現后者 , 就得到該假設的對立tis hypothesis is called zero hypotheses or the original hypothesis. If is not correct, so is correct. This稱此假設為零假設或原假設 . 判斷結果或者是接受該假設 , 或者拒絕該假設 , 如出現后者 , 就得到該假設的對立tis hypothesis could be called the al

10、ternative hypothesis. The above hypothesis testing problem is often expressed as .2 二、顯significant test 從統計資料看,過去的新生兒平均體重為 3140g,2002 年的樣本新生兒平均體重為 3160g, 有 20g 的差異 , 這 20g 差異可能產生于不同的情況 , 一是 2002 年的新生兒體重與過去無差異 , 20g 的差異是由抽樣的隨機性造成的 ; 另一種情況是抽樣的隨機性不可能造成 20g 這樣大的差異 ,2002 年的新生兒體重與過去確實有差異 . 所以問題的關鍵是 20g 的差

11、異能否用抽樣的隨機性來解釋 . According to statistics, in the past, the average weight of newborns is 3140g, while in 2002, the average weight of newborn samples is 3160g, a difference in 20g, this difference may arise in two situations. One is that there is no essential difference in them, the difference in 20g

12、is only caused by the randomness of the sample; another is caused by the essential difference in them. So the point is whether the difference can be explained by the randomness of sample or not. 樣本均值是總體均值的良好估計 , 若 The sample mean is a good estimation of the population mean , if , 應該比較小 , 我們應該設立一個合理的

13、界限should be relatively small, we should establish a reasonable limit , 當 When 時 , 我們接受原假設,we accept the zero hypothesis ;Otherwise ,we will accept alternative hypothesis. We know that,Setting =0.01, if the observation of satisfies ,that is to say the small probability event occur. Generally speaking

14、, small probability events will not occur in one experiment, so we believe is unreasonable, we call critical region. Otherwise, we have not enough evidence to reject ,so we accept .Which is called significant test. In this example, andSo we cannot reject ,thus we can believe no significant differenc

15、e between the newborn in 2002 and the old ones.Noticing is equivalent to ,i.e.,Afterward, we can make judgment only by the sample value of .3三、兩類錯誤333 Two types of errors 如前所說 , 顯著性檢驗是根據小概率事件的實際不可能原理進行判斷的 , 然而 , 由于小概率事件即使其概率很小 , 還是可能發生的 , 因此 , 利用上述方法進行假設檢驗 , 仍有可能作出錯誤的判斷 , 有兩種情況 : As mentioned above,

16、 a significant test is based on the fact that small probability event will not occur in one experiment. However, small probability events still may occur, therefore using the above hypothesis testing method still may make wrong judgments, there are two situations: 1. 原假設 1. The original hypothesis 實

17、際上是正確的 , 但檢驗結果卻錯誤地拒絕了它 , 這是犯了 ” 棄真 ” 錯誤 , 通常稱為第一類錯誤 . is actually correct, but the test result wrongly reject it, which commit "abandoning true" errors, often referred to as type I error. 2. 原假設 2. The original hypothesis is not right, but the test result wrongly accept it, which commit &q

18、uot;maintaining false " errors, often referred to as type II error. 由于樣本是隨機的 , 所以 , 我們總是以一定的概率犯以上兩類錯誤 . 在統計學中 , 我們把犯第一類錯誤的概率 As the sample is random, so we are committing two types of errors on certain probability. In statistics, we call the probability of committing type I error the significan

19、ce level, abbreviated as the level. Naturally, people desire the probability of committing two types of errors as small as possible, but for a given sample size, 不能同時做到減少犯兩類錯誤的概率 , 所以往往先固定犯第一類錯誤的概率的上限 , 再選擇犯第二類錯誤概率較小的檢驗 . We can not reduce the probability of committing two types of errors simultaneo

20、usly, Commonly, we often fix the upper bound of the probability of committing type I error, and then select a test with smaller probability of committing type II error. § 7.2 單個正態總體的假設檢驗 §7.2 single normal population hypothesis testing Let denote a random sample from a normal distribution,

21、 is significance level.一、均值1 Test of mean 1. 方1.1 Variance known已知 , 均值kikkkknThe檢驗問題為The problem as follows ,When is true, then,thus,So the critical region is.例7.2已知某煉鐵廠鐵水含炭量服從正態分布Example 7.2 Let us assume ,we have nine random samples,and the mean is 4.484,suppose the variance have no change, testi

22、ng the following problem with the significance level .=1.96, ，we get,Then can be accepted.1.2 Variance unknownThe檢驗問題為The problem as follows ,When is true, then ,Thus,So the critical region is .例7.2已知某煉鐵廠鐵水含炭量服從正態分布Example 7.3 Let us assume have normal distribution,we have 25 random samples,and the

23、mean is 66.5,and the standard deviation of sample is 15, testing the following problem with the significance level .，2.06,we get,Then can be accepted.There are still two other kinds of problem of testing mean as follows;.Using similar methods above to discuss,we can get the result in table 7.1Table

24、7.1 assumptionstatisticcritical region known known known unknown unknown unknown2 Test of varianceThe檢驗問題為The problem as follows ,When is true, then,thus,So the critical region is .例7.2已知某煉鐵廠鐵水含炭量服從正態分布Example 7.4 Let us assume have normal distribution,we have 5 random samples, 1.32, 1.55, 1.36, 1.4

25、0, 1.44,and the standard deviation is , testing the following problem with the significance level .,From the given sample,we have ,hence,So we reject .There are still two other kinds of problem of testing variance as follows;.Using similar methods above to discuss,we can get the result in table 7.2T

26、able 7.2 assumptionstatisticcritical region unknown unknown unknown例 7.4 某公司生產的發動機部件的直徑服從正態分布 . 該公司稱它的標準差§ 7.3 兩個正態總體的假設檢驗 §7.3 double normal population hypothesis testing 設總體Let denote a random sample of size from a distribution that is denote a random sample of size from a distribution t

27、hat is where , are unknown parameters, the two random samples are independent1.一、, testing of The檢驗問題為The problem as follows ,When is true, then,Thus,So the critical region is .例7.2已知某煉鐵廠鐵水含炭量服從正態分布Example 7.4 Let us assume and both have normal distribution with equal variances,we have the following

28、 random samples, X 24.3, 20.8, 23.7, 21.3, 17.4; 乙礦 18.2,  16.9,  20.2,  16.7.YYYYYYY 18.2, 16.9, 20.2, 16.7. testing the following problem with the significance level .,hence,So we accept .Using similar methods above to discuss the following problems;.we can get the result in ta

29、ble 7.3.Table 7.3 assumptionstatisticcritical region2 一、testing of The檢驗問題為The problem as follows,When is true, then,hence,So the critical region is .Using similar methods above to discuss the following problems,we can get the result in table 7.4.Table 7.4 statisticcritical regionExample 7.5 There a

30、re two team A and B participate a paper contest, A team have 9 people, and B team have 8 people, the score as follows:A team 85, 59, 66, 81, 35, 57, 76, 63, 78,B team 65, 72, 69, 65, 58, 68, 52, 64,Can we believe the variance of B team is significantly greater than A teams concerning the significanc

31、e level 0.05?Testing the following problem,so the critical region is ,from the given sample we have ,hence,So we can believe the variance of B team is significantly greater than A teams.Exercises1.Let us assume that the life of a tire in mile, say ,is normally distributed with mean and standard devi

32、ation 5000.Past experience indicates that .The manufacturer claims that the tires made by a new process have mean ,we observe 10 independent values of ,the sample mean is 5100,under the significance level 0.05,can we believe the manufacturers claim?2. Let us assume the temperature of certain substance is normally distributed, we have 5 sample values:1250, 1265, 1245, 1260, 1275,Can we