Chap 9: Testing Hypotheses & Assessing Goodness of Fit Section 9.1: INTRODUCTION In section 8.2, we fitted a Poisson dist’n to counts. This chapter will.

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Chap 9: Testing Hypotheses & Assessing Goodness of Fit Section 9.1: INTRODUCTION In section 8.2, we fitted a Poisson dist’n to counts. This chapter will provide us with tools in testing (Neyman-Pearson Paradigm) and assessing how good was such a fit.

9.2: The Neyman-Pearson Paradigm Definitions: Null Hypothesis vs Alternative (one-sided or two-sided) Hypothesis. Decision to reject in favor of is based on a statistic function of the sample values Acceptance region vs Rejection region. Type I error vs Type II error. Significance level & Power of the test.

9.3: Optimal Tests: The Neyman-Pearson Lemma Ideal: in a class of tests at a level of significance, we would like to select the most powerful one. Lemma: (simple null vs simple alternative) Assume an LRT at level that rejects when Then any other test at level will be less or equally powerful than that LRT. Rationale: The LR (Likelihood Ratio) measures the relative plausibilities of the null and the alternative. The LRT (Likelihood Ratio Test) is optimal and rejects for small values of the LR.

9.4: The Duality of Confidence Intervals & Hypothesis Testing Thm A: Then the set is a confidence region for, where denotes the acceptance region of the test. Theorem B: Suppose that is a confidence region for ; that is, Then an acceptance region for a test at level of is

9.5:Generalized Likelihood Ratio Tests Test statistic: Theorem: Under smoothness conditions on the PMF or PDF, the null dist’n of tends to a chi-square dist’n with degrees of freedom equal to as the sample size tends to infinity.

9.6: Likelihood Ratio Tests for Multinomial Distribution Problem: A generalized LRT of the Goodness of Fit of a model for Multinomial cell probabilities will be derived. Here, the large sample dist’n of is a chi-square with m-k-1 degrees of freedom.

9.7: The Poisson Dispersion Test Here, the LRT statistic resumes to: The Poisson Dispersion Test is: which tends to a chi-square with degrees of freedom as the sample size tends to infinity.

9.9: Probability Plots Let the ordered sample values be denoted by the order statistics Page 155 #17 implies that If the underlying dist’n is uniform, then the plot of the ordered observations against their expected values should look linear. For examples, please visit MINITAB.

9.10: Tests for Normality A goodness-of-fit test can be based on the coefficients of SKEWNESS or KURTOSIS but their sampling distributions are difficult to evaluate in closed form. We will base our goodness-of-fit test on the linearity of the Probability Plot, as measured by the correlation coefficient, r, of the x and y components. Such a test rejects for small values of r.

9.11: Conclusion Estimation (Chap 8) & Hypotheses Testing (Chap 9) were introduced when fitting probability distributions and testing models based on LRT (if not a chi-square dist’n as a large-sample approximation). estimating parameter from data testing hypotheses about parameter value As graphical method, we discussed the Probability Plot technique.

STEPS for Testing Hypotheses: 1.Formulate hypotheses 2.State test statistic & form RR=rejection region 3.With a specified level, determine the RR 4.Calculate the test statistic from the data 5.Draw a conclusion: either REJECT the null hypothesis at level or FAIL to REJECT the null hypothesis at level INTREPRET the conclusion in the context of the problem CALCULATE the p-value to strengthen the conclusion.

EXAMPLE: 5-week weight loss program ! Subscriptions for a new diet program state that the participants are expected to lose over 22 pounds in five weeks. From the data of 56 participants, the sample mean and the sample standard deviation are found to be 23.5 pounds and 10.2 pounds, respectively. Could the statement in the brochure be substantiated on the basis of these findings? Test at level 5%. Calculate the p-value and interpret the result.

SOLUTION: 0. Let denote the population mean weight loss from the five weeks of participation in the program. 1. Formulation 2a. Test statistic 2b. Since it’s a 1-sided test 3. c will be found (next slide) from the definition of 4.

Solutions (cont’d) In step 3, the specified level of confidence 5% determines the critical value c such that :

Solutions (cont’d) 5. Conclusion: C1: since the observed value z = 1.10 is NOT in the Rejection Region, then we fail to reject the null hypothesis in favor of the alternative hypothesis. C2 : The data do not provide evidence to reject the null.

TESTS about pop’n MEAN: Case 1:

TESTS about pop’n MEAN: Case 2: Large Sample (n > 30) Random samples come from any pop’n dist’n with unknown mean & variance. Table is the same for Case 1 & Case 2. Test statistics are different for Case 1 & Case 2.

TESTS about pop’n MEAN: Case 3:

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