slides_12_inferfinite

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Unformatted text preview: ∙ Start with the (unrealistic) case where the variance is known. Suppose for a given value , we want to test H : ≤ H 1 : ∙ We already know is the least favorable case. 59 ∙ As before, we use a statistic based on standardizing the sample average to have zero mean under H : , and variance one: T n X ̄ − / X ̄ − / sd X ̄ ∙ When , T Normal 0,1 . ∙ Suppose we have H : ≤ and H 1 : . Then the rejection rule is of the form T c and the choice of c is set once we have chosen . 60 ∙ In fact, we choose the critical value for a size test by requiring P Z c , where Z Normal 0,1 . ∙ As just a few examples, c .10 1.28, c .05 1.65, c .025 1.96, and c .01 2.33. ∙ As we shrink the probability of Type I error, the power decreases for all . ∙ The next graph shows the power functions for the 10% and 1% tests when 0 (the hypothesized mean value). 61 .2 .4 .6 .8 1-1-.5 .5 1 mu Power when Size = 10% Power when Size = 1% Power Functions with Null Mean Zero, Unit Variance, n = 25 62 ∙ If we reverse the inequality, so H : ≥ H 1 : then the rejection rule becomes with, say, .05, T − 1.65. In other words, we reject H only if X ̄ is “sufficiently” less than . ∙ The power function is the mirror image of the previous power function. 63 Two- Sided Alternative ∙ In many cases, one uses a simple null and a two-sided alternative: H : H 1 : ≠ Why? 1. It prevents us from looking at the data and then deciding on the relevant alternative. 2. It is harder to reject a null against a two-sided alternative, forcing us to find even stronger evidence against H before we reject it. 64 ∙ We use the same test statistic, T X ̄ − / n n X ̄ − , but our rejection rule changes. ∙ We want to have power against and . ∙ We use a symmetric rule: Reject H in favor of H 1 if | T | c for a suitable critical value c 0. ∙ Equivalently, reject H if T c or T − c . 65 ∙ As before, the choice of c is based based on the size of the test. If, for example, .05, we need to find the value c such that P | Z | c .05 where Z Normal 0,1 . Equivalently, P | Z | ≤ c .95 or P − c ≤ Z ≤ c .95 or c − − c .95 66 ∙ Now use the symmetry of the standard normal distribution: − c 1 − c . Plug in and solve to get c .975 or c − 1 .975 1.96 67 ∙ For a two-sided alternative, the critical value for a size 5% test is now the 97.5 percentile in a standard normal distribution. The rejection rule, at the 5% size, is | T | 1.96 ∙ Compared with a one-sided alternative, we require a larger statistic in...
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