Hypothesis Testing Problem. TMS-062: Lecture 5 Hypotheses Testing. Alternative Hypotheses. Test Statistic

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1 Hypothesis Testing Problem TMS-062: Lecture 5 Hypotheses Testing Same basic situation as befe: Data: random i. i. d. sample X 1,..., X n from a population and we wish to draw inference about unknown population parameter θ. But the question is different: Suppose we know that in the past θ = θ 0 but conditions have changed. Has θ changed too? Two cases, called hypotheses are to consider: Null hypothesis: H 0 : θ = θ 0 Alternative hypothesis: H A no change there is a change Alternative Hypotheses Test Statistic We use three fms of alternative hypothesis H A : { H A : θ > θ 0 One sided: H A : θ < θ 0 Two sided: H A : θ θ 0 We cannot observe θ directly, but we wish to decide from sample X = (X 1,..., X n ) which of H 0 and H A is true. So we compute a the value of some statistics, say T (X), called the test statistic and we decide H 0 if T (X) S 0 and H A if T (X) S A. Such a procedure is called Test of the hypothesis H 0 against H A.

2 Critical Region Types of Decision Errs Sample Space Domain of T Since our decision is based on a random value T (X) then there may be two associated types of errs: to reject H 0 when it is true (type I err) and to accept H 0 when H A holds (type II err). S 0 H 0 DECISION X T S A T (X) H A H 0 H A Traditionally, S A is called the critical region (CR). How to define it sensibly? REALITY H 0 OK Type I err H A Type II err OK Test Paradigm Usually it is considered me serious to reject H 0 when, in fact, it is true, so the probability of type I err is preset to a particular small value α (typically, 5% 1%). The parameter α is called the significance level of the test. Now a test with that level α is sought that minimises probability of type II err: { P { T (X) SA H 0 } = α P { T (X) S 0 H A } min 1 Specify H 0 and H A. 2 Choose a level α and a test statistic T. 3 Derive the critical region. 4 Compute the test statistic T. 5 Decide H A if T in CR, otherwise accept H 0. 6 Interpret your decision always citing the significance level α at which you conducted the test. 7 If we decide H A, then θ θ 0 and we usually estimate θ and construct a confidence interval f it.

3 Assume the distribution of T given θ is monotone with respect to θ, so that H A = {θ < θ 0 } implied that lesser values of T are me likely than those under H 0. This needs not always be the case, but it will be so in this course. Indeed many (but not all) of the test statistics we meet have the following fm: if H 0 = {θ = θ 0 } and θ(x) is an unbiased estimat of θ (i. e. the mean of θ(x) is θ) then T (X) = θ(x) θ 0 st.err( θ(x)). (1) if H A true Distribution of T α if H 0 true S A S 0 T 0 only this tail suppts H A If H A is one-sided, we chose S A to be as far on that side as possible to maximise its probability under H A. Power of the Test possible H A if H 0 true α/2 α/2 S A S 0 T 1 T 2 both tails suppt H A possible H A S A If H A is two sided, then H A distribution of T can be on either side. So S A has two pieces on both tails of the distribution of T under H 0. The test is constructed so as to protect H 0 against possibility to erroneously decide H A. Since we preset probability of this event to α (the err level of the test), probability of erroneously reject H 0 is α. But we do not have such a protection f H A : we do not know the prob. of acceptance of H 0 if H A is true f one-sided two-sided tests as H A specifies no value f θ. The function β(θ) = P { decide H A θ }, which is the probability to accept H A if the value of the parameter is θ, is called the power of the test.

4 Z -test f Mean σ is Known Critical Region Settings: X 1,..., X n is an i. i. d. sample from N (µ, σ 2 ) distribution with a known σ. To test: (i) H 0 = {µ = µ 0 } vs. H A = {µ µ 0 } (ii) H 0 = {µ = µ 0 } vs. H A = {µ < µ 0 } (iii) H 0 = {µ = µ 0 } vs. H A = {µ > µ 0 } Test statistic Z = X µ 0 n(x σ/ n = µ0 ). σ Given the critical level α, an obvious choice of CR is (i) (, Z α/2 ) (Z α/2, + ) (ii) (, Z α ) (iii) (Z α, + ) (see Figure above) where Z α is such that P{N (0, 1) > Z α } = α. Then if H 0 holds true, Z N (0, 1). Example Solution From past recds, a mail der company have found that the value of an der is nmally distributed with mean SEK and st.dev. 15 SEK. Recent economic reverses have caused them to reassess this view. A random sample of 25 ders is chosen and the average is SEK. 1 Does this mean that the mean value of an der is now different from 56 SEK? 2 Suppose you have been asked if the mean is now less than 56 SEK what is your response? 1 Take α = 5%, f instance. From tables, Z 0.05 = and Z = F the first question, the zero hypothesis: H 0 = {µ = 56}, alternative hypothesis H A = {µ 56} (two-sided). Critical region SA 2 = (, 1.96) (1.96, + ). The value of the test statistic 25( )/15 = 1.8 SA 2 so there is no sufficient evidence at the 5% level to suggest that the value has changed. 2 In contrast, the claim that the mean is now less than 56 SEK is suppted with 5% err level as the value 1.8 of the test statistic does fall in (, 1.645) = SA 1 the CR f one-sided alternative hypothesis H A = {µ < 56}.

5 t-test f Mean σ is Unknown Critical Region The settings, H 0, H A are the same, but σ is unknown. Therefe replacing it with its estimate sample variance, leads to statistic n(x µ0 ) t(x) =, S where S 2 = 1 n 1 ( n i=1 X 2 i nx 2) If H 0 is true, t(x) has t(n 1)-distribution with n 1 degrees of freedom. CR s are as above with Z α replaced by t α, where t α is such that P{t > t α } = α f t having t(n 1) distribution. As σ is rarely known, t-test is most often used. However, f large samples (n 30) by the CLT, t(n 1) is close to the standard nmal distribution N (0, 1) so that Z -test is also possible even if the data are not nmally distributed. Test f Proption s Value Test Statistic Settings: total population proption possessing a certain feature is p, a random sample proption of size n is ˆp. We assume here that n is large (n 50). To test: (i) H 0 = {p = p 0 } vs. H A = {p p 0 } (ii) H 0 = {p = p 0 } vs. H A = {p < p 0 } (iii) H 0 = {p = p 0 } vs. H A = {p > p 0 } Let X i = 1 if i-th individual in the sample possesses the feature and X i = 0 otherwise. Then p = S n /n, where S n = X X n. But S n Bin(n, p), so that E S n = np and var S n = np(1 p). Therefe by the CLT, the following test statistic ˆp p 0 Z =. p0 (1 p 0 ) is asymptotically N (0, 1) provided H 0 is true. So the CR is as f Z -test. n

6 p-value Up to now we first decided upon the critical level α, then constructed the cresponding CR, finally computed T = T (X) and checked whether its value falls into CR. Alternatively, instead of specifying α and CR at the beginning, we could derive the maximal value of err α at which H 0 can still be accepted given T. This value is called p-value of T = P{obtaining a sample as extreme as T } P{T T θ = θ 0 } if H A = {θ > θ 0 } = P{T T θ = θ 0 } if H A = {θ < θ 0 } 2P{T T θ = θ 0 } if H A = {θ θ 0 } Note the coefficient 2 f symmetrical statistics in two-sided case. Distribution of T under H 0 H A = {θ > θ 0 } p-value T T Example Notice: T CR p-value < α. Therefe, an alternative procedure is to compute the p-value of the observed test statistic and if p-value < α decide H A if p-value > α decide H 0. Statistical packages generally quote p-values when tests are perfmed as they do not have pre-assigned significance levels and let the users decide upon their own α. The danger here is to decide after getting the value whether it is big small. Mail ders (revisited): the average is 56 SEK and the alternative hypothesis is µ < 56. The observed value of Z -statistic is As f Z N (0, 1) we have P(Z 1.8) = 0.036, then we can accept H 0 only at the level of 3.6%. p-value is thus here. As it is less than our pre-set value of 5%, we reject H 0.