lecture12 - ISYE 2028 A and B Lecture 12 Confidence...

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Unformatted text preview: ISYE 2028 A and B Lecture 12 Confidence Intervals and Hypothesis Testing ’cont. Dr. Kobi Abayomi March 25, 2009 We have looked at hypothesis testing generally, but we have used only the specific example of a test for the population mean. For instance, if X ∼ μ,σ 2 is a random variable [model], and we collect some data x = ∑ n i x i . Then, the hypotheses H : μ = μ vs. H a : μ 6 = μ we use in a two sided test of the population mean. You will recall that we use the sampling distribution x ∼ N ( μ,σ 2 /n ) to construct the test statistic: Z = x- μ p σ 2 /n which has the standard normal distribution N (0 , 1). This setup is often sufficient: the Z statistic is the deviation of the data from the null hypothesis, over its standard deviation. In words: Z ≡ obs- exp S.D ( obs ) is the statistic we want to use if we want to test the proportion of people who vote for Pedro, the mean income of Njoroge’s in Kisumu, if the sample mean is representative of our population mean. 1 Situations often arise where the sample mean cannot sufficiently describe, or test for, im- portant hypothetical differences in populations. We must appeal to other distributions, to other quantifications of difference, to test other hypothesis. A useful alternative is... 1 The Chi Squared Distribution and associated Hy- potheses Tests Recall this example from Lecture 10: Say we are interested in the fairness of a die. Here is the observed distribution after 120 tosses: Die Face 1 2 3 4 5 6 Obs. Count 30 17 15 23 24 21 The appropriate test statistic here is the Chi-square . 1.1 The Chi-Square test for Goodness of Fit Formally, here, we are going to test H : The die is fair vs. H a : The die is not fair In general the hypotheses tests are H : π i = n i n , for all i vs. H a : π i 6 = n i n , for at least 1 i Remember here, our observed test statistic is χ 2 o = (25- 20) 2 20 + ··· + (16- 20) 2 20 = 18 . 00. The number of degrees of freedom n- 1, here 6- 1 = 5. Notice that the total number of 2 observations is fixed, that is how we calculate the expected frequency. Once the total is set we lose a degree of freedom. A χ 2 . 95 , 5 = 11 . 70 Here χ 2 o > χ 2 . 95 , 5 = 11 . 70 so we reject the null hypothesis. We conclude the die is unfair. 1 . 1.2 The Chi-Square test for independence, the Two-Way layout The Chi-Square test is useful for the contingency table, or two-way setup. Remember the contingency table from Lecture 10: a variable on rows, a variable on columns, each cell has the observed counts for each bivariate value of the variable. We used this example: Fashion Level Classlevel Low Middle High Total Graduate 6 4 1 11 PhD. 5 1 2 8 Pre-K 30 25 75 130 Total 41 30 78 149 The formal hypotheses tests are H : π ij = n ij n .....
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This note was uploaded on 11/08/2009 for the course ISYE 2028 taught by Professor Shim during the Spring '07 term at Georgia Tech.

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lecture12 - ISYE 2028 A and B Lecture 12 Confidence...

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