2 Optimal tests First we note that minimization of Pr(x E Co) for all 0 E e1 is equivalent to maximizing pr(x∈Cn) for all 6∈61 Definition 1 The probability of reject Ho when false at some point 01 E O1, i.e. Pr(x E C1: 0=01) is called the power of the test at 0=01 Note that Pr(x∈C1;b=61)=1-Pr(x∈Co;6=61)=1-B(61) In the above example we can define the power of the test at some 01 Ee sayb=54, to be pr[(|Xn-60)/1.265≥1.96:6=54 Under the alternative hypothesis that 0= 54, then it is true that 1651 N(O, 1). We would like to know that the probability of the statistics constructed under the null hypothesis that xnx60 would fall in the rejection re is, the power of the test at 0=54 to be P 1,265≥1.96;6 +P/.205-196~(54-60) 54 1.265 ≥1.96~(54 60) =0.993 1 1.265 Hence, the power of the test defined by Ci above is indeed very high for 6=54 From this we know that to calculate the power of a test we need to know the distribution of the test statistics T(x) under the alternative hypothesis. In this case it is the distribution of Xn=54 In the example above, the test statistic T(x) have a standard normal distribution under both the null and the alternative hypothesis. However, it is quite often the case when it happen that a test statistics have a different distribution under the null and the alternative hypotheses For example, the unit root test. See Chapter 212 Optimal Tests First we note that minimization of Pr(x ∈ C0) for all θ ∈ Θ1 is equivalent to maximizing Pr(x ∈ C1) for all θ ∈ Θ1. Definition 1: The probability of reject H0 when false at some point θ1 ∈ Θ1, i.e. Pr(x ∈ C1; θ = Θ1) is called the power of the test at θ = θ1. Note that Pr(x ∈ C1; θ = θ1) = 1 − Pr(x ∈ C0; θ = θ1) = 1 − β(θ1). In the above example we can define the power of the test at some θ1 ∈ Θ1, say θ = 54, to be Pr[(|X¯ n − 60|)/1.265 ≥ 1.96; θ = 54]. Under the alternative hypothesis that θ = 54, then it is true that X¯n−54 1.265 ∼ N(0, 1). We would like to know that the probability of the statistics constructed under the null hypothesis that X¯n−60 1.265 would fall in the rejection region; that is, the power of the test at θ = 54 to be Pr  |X¯ n − 60| 1.265 ≥ 1.96; θ = 54 = Pr  |X¯ n − 54| 1.265 ≤ −1.96 − (54 − 60) 1.265  +Pr  |X¯ n − 54| 1.265 ≥ 1.96 − (54 − 60) 1.265  = 0.993. Hence, the power of the test defined by C1 above is indeed very high for θ = 54. From this we know that to calculate the power of a test we need to know the distribution of the test statistics τ (x) under the alternative hypothesis. In this case it is the distribution of X¯n−54 1.265 . 1 1 In the example above, the test statistic τ (x) have a standard normal distribution under both the null and the alternative hypothesis. However, it is quite often the case when it happen that a test statistics have a different distribution under the null and the alternative hypotheses. For example, the unit root test. See Chapter 21. 7