ch21NormaltestofHypothesesonmu

ch21NormaltestofHypothesesonmu - Chapter 21: Chapter 21...

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Unformatted text preview: Chapter 21: Chapter 21 Normal Test of Hypotheses on μ A. General Concepts When we carry out a research project or experiment, we may have reservations about the validity of certain existing theories, or facts, or beliefs. We then collect data in the hope of establishing a new theory, presented in the form of a hypothesis (statement). There are thus two hypotheses for us to decide between: The old, existing and authoritative one is called the null hypothesis, denoted by H 0 The new, daring or challenging one is called the The new, daring or challenging one is called the alternative hypothesis, denoted as H A = H1 They are not of equal standing. H0 The old one, , is in a privileged position, as it has been established and used for some time, perhaps for centuries. HA The new one, , is fragile. It is essentially a trouble maker. People would not mind giving it up if evidence is not strong enough to support it. This is in accord with some western legal points of view. In a court case, the prosecutor (or the plaintiff) In a court case, the prosecutor (or the plaintiff) proposes an hypothesis: HA :Mr A. has committed a crime, X The judge in the first place already assumes an hypothesis H0 : Mr A. is innocent of X. HA H0 This is proposed to nullify (to validate) . The trial (i.e. the test) then goes on. H0 The verdict might turn out to be or . HA There are two types of errors: There are two types of Correspondingly, there are two types of risks in the test: α = T ype I risk = P ( Commit Type I error) β = T ype I I risk = P( Commit Type I I error) I deally, the trial procedure (i.e. the test) can be Ideally, designed so that both α and βare small. If however, the judge cannot keep both If however, the judge cannot keep both αandβsmall, then he chooses to give up β. He, in the first place, must keepαas small as as possible, so as to protect the defendants’ rights. I n other words, the court does not want to wrong an innocent person. an Thus, the focus is on the control of α. Usually, the choice of the α-value depends on -value the nature of X. the (i) If X = illegal parking, then perhaps, α= 0.05 (i) If X = illegal parking, then perhaps, (ii) If X = speeding, then perhaps,α= 0.01 (iii) If X = murdering, then α= 0.001, or less. (iii) H Such a test of hypotheses (which accepts A only Such when evidence is significantly strong or significantly significantly very strong so that the chance of H significantly wronging is small or very small) is called a significance test. significance H0 The procedure gives priority to . The 0 L et us carry over the above ideas to the tests of Let statistical hypotheses, which concern numbers. statistical B. Test of statistical hypotheses B. Test of statistical hypotheses Suppose, by a large scale survey, it was established that, last year, the mean income of the people in a certain industry was $9000 per month. Does this figure still stand, or has it increased this year? There are two hypotheses in the question for us to test: H0 :μ= 9,000 (More generally, H 0 : µ = µ0 ) H A : µ > µ0 ) H A : μ> 9,000 (More generally, where μ=this year’s income. where To test H 0 versus H A , we take a sample, x1 , x2 ,........., xn (i.e. we interview n persons to obtain data x1 , 2 ,........., xn ) andxcompute the sample mean: ∑x x= n i Suppose H 0 is correct, i.e. suppose Suppose Then by N.S.T., σ2 X ~ N (µ , ) 0 µ = µ0 . n As a first-stage discussion, let us assume that σ is As known. Then P( x > µ + 1.645 σ ) = 0.05 Then 0 n x − µ0 P( > 1.645) = 0.05 σn P( Z > 1.645) = 0.05 x − µ0 where the standardized quantity is called the Z= test statistic. σn If we set α= 0.05, then the decision rule that we“ Accept H A 0.05, whenever z >1.645” will lead to the effect that the H probability of accepting H A when 0 is true, is 0.05. probability i.e. P( Commit Type I error) = 0.05. H This is exactly what we want concerning the protection of0 This Example 1: Example 1 It was established that last year , the monthly income earned by people in a certain industry has a normal distribution, with mean (in $1000) 9.0 and s.d. 1.5. This year a small scale survey of 10 people of the industry yielded a mean of 9.9. Assuming that the s.d. is still 1.5 this year, is it justified to say that the mean income this year has increased , at the 5% risk level? Solution: Solution Let X=monthly income of people in a certain industry Let μdenote this year’s mean income. denote 1. Hypotheses: We want to test 1. We H 0 : µ = 9 versus H : µ > 9.0 versus A 2. Data: n=10, σ=1.5, 2. Data: =1.5, 3. Test statistic: “Under Test “Under x = 9.9 H0 ”, µ0 = 9 Z T .S x − µ0 9.9 − 9 = = = 1.8974 σ n 1.5 10 4. Tables: 4. From Table B, the cut­off value for z is 1.645 when α= 0.05 0.05 5. Conclusion: 5. Conclusion Since the test statistic valueZ = 1.8974 is greater than the cut-off value from the tables,Z C = 1.645 we accept H A at the α=5% level of significance level. T .S Note 1: Note 1 The cut­off value, 1.645, is specifically called the critical value. The test, which uses a standardized normal variable Z, is called a normal test, or z­test. HA The above conclusion of accepting at the 0.05 risk level may be rephrased as: “The mean income this year has significantly increased at the 5% significance level (S.L). ” “Significantly increased” means “not marginally increased”. Example 2: Example 2 Everything is as in Example 1 except that x = 9. 4 Is it still justified to claim that the mean income this year has increased, at the 5% significance level? Solution for Example 2: Solution for Example 2 Let X=monthly income of people in a certain industry Let μdenote this year’s mean income. denote 1. Hypotheses: We want to test 1. We H 0 : µ = 9 versus H A : µ > 9.0 versus 2. Data: n=10, σ=1.5, 2. Data: =1.5, 3. Test statistic: “Under Test “Under x = 9.4 H0 ”, µ0 = 9 Z T .S x − µ0 9.4 − 9 = = = 0.8433 σ n 1.5 10 4. Tables: 4. From Table B, the cut­off value for z is 1.645 when α= 0.05 5. Conclusion: 5. Conclusion Since the test statistic valueZ = 0.8433 is less than the ZC = cut-off value from the tables,1.645 we accept H 0 at the α=5% level of significance level. T .S Note 2: Note 2 H A : µ > µ0 H 0 : µ = µ0 For testing versus under the assumption of a normal population with known σ, the test the statistic is statistic Z T .S The decision rules are: x − µ0 = σn The values 1.645, 1.96, 2.326 and 2.576 are called the The critical values (C.V) at the 0.05, 0.025, 0.01 and critical 0.0005 S.L.’s respectively. 0.0005 The test is called a right-tailed z-test (right-tailed normal The z-test test.) test.) We may appreciate the structure of the test statistic x−µ in a more direct and intuitive way. Z= T .S 0 Note 3: Note σ n x − µ0 The numerator, , informs us by how much the observed value exceeds the proposed target 0 x µ It is the signal. Is this signal marginally bigger than zero, or significantly bigger than zero? To know the answer, we have to compare it with the x volatility of ,i.e. the noise of which is , by x N.S.T., σ se( X ) = n Thus Thus ZT .S = x − µ 0 Signal = = S N ratio σ Noise n If this ratio is not appreciably bigger than 1, we do not regard the signal as big enough to be distinguishable from the noise: we cannot hear clearly. The above two examples say that the S/N ratio must be HA at least 1.645 to justify the claim If z exceeds 1.96, 2.326 or 2.576, then the signal is HA more and more distinctly hearable, i.e. is more and more significantly established. Note 4: Note 4 At this first level, we always consider a simple null H :µ = µ hypothesis, , which asserts that μ takes a single value,µ0 0 0 HA The alternative hypothesis is often a composite one, such as H : µ > µ µ =9 H0 Suppose is . HA Then can mean μ= 9.1,9.2,…..,100,……just anything bigger than 9.0. HA Therefore is a collection of many, many simple hypotheses. A 0 HA µ > µ0 The hypotheses is that is right­sided. The hypotheses is that is It requires a right­tailed test, whose critical region (i.e. acceptance region for ) is on the right. HA HA It is possible that in other problems, , is left­sided, H i.e. A : µ < µ0 Then we need a left­tailed test : H Accept if and only if z < ­1.645 (atα= 0.05) A I t is also possible that H A is two-sided, ii.e. H A : µ ≠ µ0 two-sided .e. Then we need a two-tailed test: two-tailed A ccept H A if and only if z < -1.96 or z > 1.96, for α= 0.05. 0.05. Example 3: Example 3 From a population ~N(μ, 21²), a sample of size 18 is taken. The sample mean is 44. Test the null hypotheses that against the alternative hypotheses that 58 H0 : µ = H A : µ ≠ 58 , at the 0.01 significance level. Solution: Test Statistic is 0.01 Z T .S = x − µ 0 44 − 58 = = −2.828 σ n 21 18 C.V = ±2.576 Since , we REJECT at 0.01 significance level . Note 5: Note 5 In Example 1, we computed the test statistic to be z = 1.8974 and compared it with the 0.05 critical value: C.V0.05 = 1.645 Then we accepted at 0.05 risk level, meaning that if H0 H0 is true, the chance of wronging it is 0.05. Actually, the exact chance of wronging is less than 0.05. It is P = P( Z > 1.8974) = 0.5 – 0.45415 = 0.02889 This is called the P­value of the test. HA Nowadays, many people prefer to use the P­value as a Nowadays, many people prefer to use the HA criteria for accepting (when P > 5%) or H0 (when P ≤ 5%). 5%). The P­value is often given as additional information (for judgment) on a computer report. ...
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This note was uploaded on 02/22/2010 for the course FBE STAT0302 taught by Professor Unknown during the Spring '10 term at HKU.

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