Chi-Square Tests - The Chi-Square Tests We will cover three...

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The Chi-Square Tests We will cover three tests that are very similar in nature but differ in the conditions when they can be used. These are A) Goodness-of-tests B) Tests of homogeneity and C) Test of independence. Let’s start with the easiest one. A) Goodness-of-fit Test This is an extension of the one-population, one-sample, one-parameter problem where the random variable of interest is a categorical variable with 2 categories and the hypotheses were Ho: π = π o versus Ha: π ≠ π 0 . We now extend the above test to the case of a categorical random variable with k (k ≥ 2) categories. Suppose we have a random variable that has k = 3 categories. Then the hypotheses of interest will be Ho: π 1 = π 10 , π 2 = π 20 , π 3 = π 30 vs. Ha: At least one of π i ≠ π i0 Where π i are the proportion of population units in the i th category and π i0 are the values of π i specified by the null hypothesis. To test these hypotheses we select a random sample of size n and count the number of sample units observed in each category (denoted by O i ). Next, we calculate the expected number of observations (E i ) in each category assuming Ho to be true, using E i = n×π i0 . Finally we compare the observed frequencies with the expected frequencies using the test statistic ( 29 2 2 2 ( ) 1 ~ k i i df i i O E E χ χ = - = , where df = k – 1. Other steps of hypothesis testing are the same as before: 1) Assumptions a) Simple random samples from the population b) Categorical variable with k categories c) Large samples (O i ≥ 5 for all i) 2) Hypotheses: Ho: π 1 = π 10 , π 2 = π 20 , π 3 = π 30 vs. Ha: At least one of π i ≠ π i0 3) Test Statistic: ( 29 2 2 2 ( ) 1 ~ k i i df i i O E E χ χ = - = , with df = (k–1) STA 6126 Chap 8, Page 1 of 19
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4) The p-value = ( 29 2 2 ( ) df cal P χ χ 5) Decision Same rule as ever, Reject Ho if the p-value ≤ α. 6) Conclusion Same as before, explain the decision in simple English for the layman. Example: Suppose we suspect that a die (used in a Las Vegas Casino) is loaded. To see if this suspicion is warranted we roll the die 600 times and observe the frequencies given in Table 8.1. The hypotheses of interest are Ho: π 1 = π 2 = π 3 = π 4 = π 5 = π 6 = 1/6 vs. Ha: At least one of the π i ≠ 1/6. Let’s test these hypotheses. But first we need to check if all of the conditions are satisfied: 1) Assumptions Satisfied? a) Simple random samples from the population Yes b) Categorical variable with k categories Yes, k = 6 c) Large samples (O i ≥ 5 for all i) Yes, look at Table 8.1 2) Hypotheses: Ho: π 1 = π 2 = π 3 = π 4 = π 5 = π 6 = 1/6 vs. Ha: At least one of the π i ≠ 1/6. 3. Test Statistic: ( 29 2 6 2 2 (6 1) 1 ~ i i i i O E E χ χ - = - = 4. The p-value: For this we need to find the calculated value of the test statistic first. This is done in the following table (worksheet): Observed and Expected Values of 600 rolls of a die Categor y Observed (O i ) Expected (E i ) ( 29 i i O E - ( 29 2 i i O E - ( 29 2 i i i O E E - 1 115 100 15 225 2.25 2 97 100 – 3 9 0.09 3 91 100 – 9 81 0.81 4 101 100 1 1 0.01 5 110 100 10 100 0.10 6 86 100 – 14 196 1.96 Total 600 600 0 2 cal χ = 5.22 Then, ( 29 2 (5) 5.22 p value P χ - = . Now we need to look at table of the
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