Chapter7假设检验Hypothesistesting.

Chapter 7 Hypothesis testingStatistical hypothesis testing is another core issue in statistical inference . Wecalled 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, when 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 hypothesis testing and the hypothesis testing methods in normal population station .§7.1 The basic concepts of hypothesis testingThe theses about the population distribution have two forms. One is the thesis aboutthe population distribution types, known as non-parametric hypothesis ; another is thatthe type of population distribution is known but with unknown parameters, we called the thesis about the unknown parameters parametric hypothesis. We will only introduce parametric hypothesis testing in this chapter .1 An exampleIn this section, we will introduce some important concepts th rough an example ofhypothesis test.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 weigh t is 300g, and based on past statistics, the average weight of newborn is 3140g. If the weight of X obeys normal distribution. Is there any significant difference about the weight between the newborn in 2002 and the old ones?Let X denote the weight of newborn, then based on the hypothesis , wehave 2~(,)X N μσ.The problem is that whether the mean of population is equal to 3140g or not, which can be expressed as0:3140H μ=,This hypothesis is called zero hypotheses or the original hypothesis. If 0:3140H μ=is not correct, so 1:3140H μ≠ is correct. This hypothesis could be called the alternativehypothesis . The above hypothesis testing problem is often expressed as01:3140:3140H H μμ=↔≠.2 significant testAccording 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. O ne is that there is no essential difference in them, the difference in 20g 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.The sample mean is a good estimation of the population mean , if 3140μ=,|3140|X -should be relatively small, we should establish a reasonable limit C , When|3140|X C -<,we accept the zero hypothesis ;Otherwise ,we will accept alternativehy pothesis .We know that~(1)X T t n =-,Setting α=0.01,0.005(||(1))0.01P T t n α≥-==, if the observation of ||T satisfies||t 0.005(1)t n ≥-,that is to say the small probability event 0.005{||(1)}T t n ≥- occur.Generally speaking, small probability events will not occur in one experiment, so we believe 0H is unreasonable, we call 0.005{||(1)}t t n ≥- critical region . Otherwise, we have not enough evidence to reject 0H ,so we accept 0H .Which is called significant test . In this example,0.00520,(19) 2.861,n t ==and 0.298 2.861,x t ==<So we cannot reject 0H ,thus wecan believe no significant difference between the newborn in 2002 and the old ones.Noticing 0.005||(1)T t n ≥-is equivalent to 0.005||(1)X t n S μ-≥-0.005(1)C t n S =-,Afterward, we can make judgment only by the sample value of ||T .3 Two types of errorsAs mentioned above, 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. The original hypothesis is actually correct, but the test result wrongly reject it, which commit "abandoning true" errors, often referred to as type I error.2. The original hypothesis is not right, but the test result wrongly accept it, which commit "maintaining false " errors, often referred to as type II error.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 significance 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, We can not reduce the probability of committing two types of errors simultaneously, 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 single normal population hypothesis testingLet 1(,,)n X X denote a random sample from a normal distribution 2~(,)X N μσ,α issignificance level.1 Test of mean1.1 Variance knownThe problem as follows0010::H H μμμμ=↔≠When 0H is true, then~(0,1)X U N =,thus/2(||)P U u αα≥=,So the critical region is /2{||}u u α≥.Example 7.2 Let us assume 2(4.55,0.108)X N ,we have nine random samples,and themean is 4.484,suppose the variance have no change, testing the following problem with the significance level 0.05,α=01: 4.55: 4.55H H μμ=↔≠.0.025u =1.96, 0.108, 4.484,9x n σ===，we get4.484 4.55|||| 1.83 1.960.108/3u -==<,Then 0H can be accepted.1.2 Variance unknownThe problem as follows0010::H H μμμμ=↔≠,When 0H is true, then~(1)X T t n =-,Thus/2(||(1))P T t n αα≥-=,So the critical region is /2{||(1)}t t n α≥-.Example 7.3 Let us assume X have normal distribution,we have 25 random samples,and themean is 66.5,and the standard deviation of sample is 15, testing the following problem with the significance level 0.05,α=01:70:70H H μμ=↔≠.25n =，0.025(24)t =2.06,66.5x =,we get66.570|||| 1.167 2.0615/5t -==<,Then 0H can be accepted.There are still two other kinds of problem of testing mean as follows0010::H H μμμμ=↔>; 0010::H H μμμμ=↔<.Using similar methods above to discuss,we can get the result in table 7.12 Test of varianceThe problem as follows22220010::H H σσσσ=↔≠,When 0H is true, then2222(1)~(1)n S n χχσ-=-,thus22221/2/2((1)(1))P n n ααχχχχα-≤-≥-=或,So the critical region is 22221/2/2{(1)}{(1)}n n ααχχχχ-≤-≥-.Example 7.4 Let us assume X have normal distribution,we have 5 random samples,1.32, 1.55, 1.36, 1.40, 1.44,and the standard deviation is 0.048σ=, testing the following problem with the significance level 0.05,α=222201:0.048:0.048H H σσ=↔≠.5n =,220.0250.975(4)11.14,(4)0.484χχ==,Fromthe givensample,we have20.00778S =,hence2240.0077813.5111.140.048χ⨯==>,So we reject 0H .There are still two other kinds of problem of testing variance as follows22220010::H H σσσσ=↔>; 22220010::H H σσσσ=↔<.Using similar methods above to discuss,we can get the result in table 7.2§7.3 double normal population hypothesis testingLet 112(,,...,)n X X X denote a random sample of size 1n from a distribution that is211(,),N μσ 212(,,...,)n Y Y Y denote a random sample of size 2n from a distributionthat is 222(,),N μσwhere 211,μσ ,222,μσ are unknown parameters, the two random samples are independent1.2212σσ=, testing of 12μμ-The problem as follows012112::H a H a μμμμ-=↔-≠,When 0H is true, then12~(2)X Y T t n n =+-,Thus/212(||(2))P T t n n αα≥+-=,So the critical region is /212{||(2)}T t n n α≥+-.Example 7.4 Let us assume X and Y both have normal distribution with equalvariances,we have the following random samples,X 24.3, 20.8, 23.7, 21.3, 17.4; Y 18.2, 16.9, 20.2, 16.7.testing the following problem with the significance level 0.05,α=012112::H H μμμμ=↔≠.0a =.125,4n n ==,0.025(7) 2.365t =,221221.5,7.505,18, 2.593X S Y S ====,hence2.324S ω==,| 2.245 2.365|T =<=,So we accept 0H .Using similar methods above to discuss the following problems012112::H a H a μμμμ-≤↔->; 012112::H a H a μμμμ-≥↔-<.we can get the result in table 7.3.2 testing of 2212/σσThe problem as follows2222012112::H H σσσσ=↔≠,When 0H is true, then221212(1,1)F S S F n n =--,hence1212212((1,1)(1,1))P F F n n F F n n ααα-≤--≥--=或,So the critical region is 112212{(1,1)(1,1)}F F n n F F n n αα-≤--≥--或. Using similar methods above to discuss the following problems,2222012112::H H σσσσ≤↔>,we can get the result in table 7.4.1(1,F n -1(1,F n -Example 7.5 There are two team A and B participate a paper contest, A team have 9 people, andB 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 team ’s concerning the significance level 0.05?Testing the following problem2222012112::,H H σσσσ≤↔>0.025(8,7) 3.73F =,so the critical region is { 3.73}F ≥,from the given sample we have2212240.75,40.98S S ==,hence21225.875 3.73S F S ==>,So we can believe the variance of B team is significantly greater than A team ’s.Exercises1.Let us assume that the life of a tire in mile, say X ,is normally distributed with mean θ and standard deviation 5000.Past experience indicates that 3000θ=.The manufacturer claims that the tires made by a new process have mean 3000θ>,we observe 10 independent values of X ,the sample mean is 5100,under the significance level 0.05,can we believe the manufacturer ’s 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 believe the temperature of this substance is 1277?(α=0.05).3.Let us assume that the life of a electronic product is normally distributed, and standard deviation 1.6.Now the technique is improved, we draw 9 products in the new products, the mean life is 52.8,sample standard deviation is 1.19,can we believe the variance of the life is changed? (α=0.05).4. Seeds of a particular variety of plant were randomly assigned to either a nutritionally rich environment (the treatment) or standard conditions (the control). After a predetermined period, allProviding that the two sample are from normal population, concluding that is there any difference between the mean weights due to environmental conditions? (α=0.05) 5.Let population (,9)XN μ,μ is unknown parameter, 125(,,)X X is a samplefrom population, considering the testing problem:0010::H H μμμμ=↔≠,Critical region is 1250{(,,)|||}C x x x c μ=-≥.Please find the constant c ,thesignificance level is 0.05.。