Statistical Inference

Transcription

1 Statistical Inference Classical and Bayesian Methods Class 7 AMS-UCSC Tue 31, 2012 Winter Session 1 (Class 7) AMS-132/206 Tue 31, / 13

2 Topics Topics We will talk about... 1 Hypothesis testing Winter Session 1 (Class 7) AMS-132/206 Tue 31, / 13

3 Significance level/size of the test Definition Level of significance/size of a test Which test is the best? We would like to have a test δ with low values of its power function for θ Ω 0 and high values of its power function for θ Ω 1. These two goals can not be reached simultaneously. Normally you select a number 0 < α 0 < 1 such that: π(θ α) α 0, for all θ Ω 0 The value of α 0 is the level of the test and we say that the test has a level of significance α 0 Size of a test The size of a test is defined as: α(δ) = sup θ Ω0 π(θ δ). A test is a level α 0 test if and only if its size is at most α 0. If the null hypothesis is simple (H 0 : θ = θ 0 ), the size of δ will be α(δ) = π(θ 0 δ) Winter Session 1 (Class 7) AMS-132/206 Tue 31, / 13

4 Significance level/size of the test Example 1 Testing hypothesis about the Mean of a Normal Distribution with Known Variance In the previous class (page 11) our test δ was to reject H 0 : µ = µ 0 if X n µ 0 c. The variable Y = X n µ 0 has a Normal distribution with zero mean and variance σ 2 /n when µ = µ 0. For each α 0 we can find the value of c such that Pr( X n µ 0 c) = α 0. Due to the symmetry of the Normal distribution, the value of c must be the 1 α 0 /2 quantile of the distribution of Y. This quantile is c = Φ 1 (1 α 0 /2)σ/n 1/2 We can rewrite this test in terms of the statistic Z = n 1/2 X n µ 0 σ and reject H 0 if Z Φ 1 (1 α 0 /2) Winter Session 1 (Class 7) AMS-132/206 Tue 31, / 13

5 The p-value Definition Calculating p-values Instead of choosing a value of α 0 in advance, we can report the value of the Z statistic and the smallest level α 0 such that we would reject the null hypothesis at level α 0 with the observed data. This value is the p-value. p-value Is the smallest level α 0 such that we would reject the null hypothesis at level α 0 with the observed data An experimenter who rejects a null hypothesis if and only if the p-value is at most α 0 is using a test with level of significance α 0. An experimenter who wants a level α 0 test, will reject the null hypothesis if and only if the p-value is at most α 0. For this reason the p-value is sometimes called the observed level of significance. Because the p-value is calculated as a probability in the upper tail of the distribution of a test statistics T, it is sometimes called the tail area. Winter Session 1 (Class 7) AMS-132/206 Tue 31, / 13

6 The p-value Example Calculating p-value: Mean of a Normal distribution with known variance Suppose you fix α 0 = 0.05, you would reject H 0 if Z > Φ 1 (1 0.05/2) = If the observed value of Z = 2.78 we would reject H 0. On another hand, the hypothesis H 0 would be rejected for every level of significance α 0 such that 2.78 Φ 1 (1 α 0 /2). This inequality translate into α This value p = is called the p-value. You would also reject H 0 at level 0.01 since 0.01 > Winter Session 1 (Class 7) AMS-132/206 Tue 31, / 13

7 Example 2 Testing hypothesis about the Uniform Distribution Let X 1, X 2,..., X n be a sample from a Uniform distribution in the interval [0, θ] with θ unknown. We want to test the hypothesis: H 0 : 3 θ 4 H 1 : θ < 3 or θ > 4 The MLE of θ is Y n = max{x 1, X 2,..., X n }. Let δ be the test: Do not reject H 0 if 2.9 < Y n < 4, and rejects H 0 if Y n does not lie in this interval. Critical region of the test: All values of X 1, X 2,..., X n for which either Y n 2.9 or Y n 4. (R : (, 2.9] [4, )) Power function of the test: This is specified by the relation: π(θ δ) = Pr(Y n 2.9 θ) + Pr(Y n 4 θ) Size of the test: α(δ) = sup 3 θ 4 π(θ δ) Winter Session 1 (Class 7) AMS-132/206 Tue 31, / 13

8 Example 2 Testing hypothesis about the Uniform Distribution (Cont.) Power Function Calculation We need to calculate π(θ δ) for all values of θ. If θ 2.9: Pr(Y n 2.9 θ) = 1 and Pr(Y n 4 θ) = 0, therefore π(θ δ) = 1 If 2.9 < θ 4: Pr(Y n 2.9 θ) = (2.9/θ) n and Pr(Y n 4 θ) = 0, therefore π(θ δ) = (2.9/θ) n (*) If θ > 4: Pr(Y n 2.9 θ) = (2.9/θ) n and Pr(Y n 4 θ) = 1 (4/θ) n, therefore π(θ δ) = (2.9/θ) n + 1 (4/θ) n (*): Remember that if X 1, X 2,..., X n is a set of iid random variables, Pr(Y n y θ) = Pr(X 1 y θ).pr(x 2 y θ)... Pr(X n y θ) = F (y θ) n Winter Session 1 (Class 7) AMS-132/206 Tue 31, / 13

9 Example 2 Testing hypothesis about the Uniform Distribution (Cont.) The size of δ is α(δ) = sup 3 θ 4 π(θ δ). Since the function is decreasing in the interval 3 θ 4, α(δ) = π(δ θ) = (2.9/3) n. If n = 68 then the size of δ = (2.9/3) 68 = δ is a level α 0 test for every level of significance α Winter Session 1 (Class 7) AMS-132/206 Tue 31, / 13

10 Example 2 Testing hypothesis about the Uniform Distribution (Cont.) Sketch of the Power Function Winter Session 1 (Class 7) AMS-132/206 Tue 31, / 13

11 Example 3 Testing hypothesis about a Bernoulli parameter Let X 1, X 2,..., X n be a sample from a Bernoulli distribution with parameter p unknown. We want to test the hypothesis: H 0 : p p 0 H 1 : p > p 0 Let Y = n i=1 X i. Y has a binomial distribution with parameters n and p. If p is large, Y will be large. Suppose we choose δ the test that rejects H 0 if Y c for some constant c. For a test of size α 0, c should be the smallest number such that Pr(Y c p = p 0 ) α 0. Winter Session 1 (Class 7) AMS-132/206 Tue 31, / 13

12 Example 3 Testing hypothesis about a Bernoulli parameter (Cont.) If n = 10, p 0 = 0.3 and α 0 = 0.1 we must choose c > 5. The reason for this is that 10 Pr(Y = y p = 0.3) = and 10 y=5 y=6 Pr(Y = y p = 0.3) = Note: In this problem the p value is the Pr(Y y p = p 0 ). If Y = 6 is observed, then Pr(Y 6 p = 0.3) = is the p value. Winter Session 1 (Class 7) AMS-132/206 Tue 31, / 13

13 Thanks for your attention... Winter Session 1 (Class 7) AMS-132/206 Tue 31, / 13