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GOODNESS OF FIT Larson/Farber 4th ed 1 Section 10.1

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Section 10.1 Objectives Larson/Farber 4th ed 2 Use the chi-square distribution to test whether a frequency distribution fits a claimed distribution

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Multinomial Experiments Larson/Farber 4th ed 3 Multinomial experiment A probability experiment consisting of a fixed number of trials in which there are more than two possible outcomes for each independent trial. A binomial experiment had only two possible outcomes. The probability for each outcome is fixed and each outcome is classified into categories.

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Multinomial Experiments Larson/Farber 4th ed 4 Example: A radio station claims that the distribution of music preferences for listeners in the broadcast region is as shown below. Distribution of music Preferences Classical4%Oldies2% Country36%Pop18% Gospel11%Rock29% Each outcome is classified into categories. The probability for each possible outcome is fixed.

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Chi-Square Goodness-of-Fit Test Larson/Farber 4th ed 5 Chi-Square Goodness-of-Fit Test Used to test whether a frequency distribution fits an expected distribution. The null hypothesis states that the frequency distribution fits the specified distribution. The alternative hypothesis states that the frequency distribution does not fit the specified distribution.

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Chi-Square Goodness-of-Fit Test Larson/Farber 4th ed 6 Example: To test the radio station’s claim, the executive can perform a chi-square goodness-of-fit test using the following hypotheses. H 0 : The distribution of music preferences in the broadcast region is 4% classical, 36% country, 11% gospel, 2% oldies, 18% pop, and 29% rock. (claim) H a : The distribution of music preferences differs from the claimed or expected distribution.

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Chi-Square Goodness-of-Fit Test Larson/Farber 4th ed 7 To calculate the test statistic for the chi-square goodness-of-fit test, the observed frequencies and the expected frequencies are used. The observed frequency O of a category is the frequency for the category observed in the sample data.

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Chi-Square Goodness-of-Fit Test Larson/Farber 4th ed 8 The expected frequency E of a category is the calculated frequency for the category. Expected frequencies are obtained assuming the specified (or hypothesized) distribution. The expected frequency for the i th category is E i = np i where n is the number of trials (the sample size) and p i is the assumed probability of the i th category.

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Example: Finding Observed and Expected Frequencies Larson/Farber 4th ed 9 A marketing executive randomly selects 500 radio music listeners from the broadcast region and asks each whether he or she prefers classical, country, gospel, oldies, pop, or rock music. The results are shown at the right. Find the observed frequencies and the expected frequencies for each type of music. Survey results (n = 500) Classical8 Country210 Gospel72 Oldies10 Pop75 Rock125

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Solution: Finding Observed and Expected Frequencies Larson/Farber 4th ed 10 Observed frequency: The number of radio music listeners naming a particular type of music Survey results (n = 500) Classical8 Country210 Gospel72 Oldies10 Pop75 Rock125 observed frequency

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Solution: Finding Observed and Expected Frequencies Larson/Farber 4th ed 11 Expected Frequency: E i = np i Type of music % of listeners Observed frequency Expected frequency Classical 4%8 Country36%210 Gospel11%72 Oldies 2%10 Pop18%75 Rock29%125 n = 500 500(0.04) = 20 500(0.36) = 180 500(0.11) = 55 500(0.02) = 10 500(0.18) = 90 500(0.29) = 145

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Chi-Square Goodness-of-Fit Test Larson/Farber 4th ed 12 For the chi-square goodness-of-fit test to be used, the following must be true. 1. The observed frequencies must be obtained by using a random sample. 2. Each expected frequency must be greater than or equal to 5.

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Chi-Square Goodness-of-Fit Test Larson/Farber 4th ed 13 If these conditions are satisfied, then the sampling distribution for the goodness-of-fit test is approximated by a chi-square distribution with k – 1 degrees of freedom, where k is the number of categories. The test statistic for the chi-square goodness-of-fit test is where O represents the observed frequency of each category and E represents the expected frequency of each category. The test is always a right-tailed test.

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Chi - Square Goodness - of - Fit Test Larson/Farber 4th ed 14 1.Identify the claim. State the null and alternative hypotheses. 2.Specify the level of significance. 3.Identify the degrees of freedom. 4.Determine the critical value. State H 0 and H a. Identify . Use Table 6 in Appendix B. d.f. = k – 1 In WordsIn Symbols

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Chi - Square Goodness - of - Fit Test Larson/Farber 4th ed 15 If χ 2 is in the rejection region, reject H 0. Otherwise, fail to reject H 0. 5.Determine the rejection region. 6.Calculate the test statistic. 7.Make a decision to reject or fail to reject the null hypothesis. 8.Interpret the decision in the context of the original claim. In WordsIn Symbols

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Example: Performing a Goodness of Fit Test Larson/Farber 4th ed 16 Use the music preference data to perform a chi- square goodness-of-fit test to test whether the distributions are different. Use α = 0.01. Survey results (n = 500) Classical8 Country210 Gospel72 Oldies10 Pop75 Rock125 Distribution of music preferences Classical4% Country36% Gospel11% Oldies2% Pop18% Rock29%

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Solution: Performing a Goodness of Fit Test Larson/Farber 4th ed 17 H 0 : H a : α = d.f. = Rejection Region Test Statistic: Decision: Conclusion: 0.01 6 – 1 = 5 0.01 χ2χ2 0 15.086 music preference is 4% classical, 36% country, 11% gospel, 2% oldies, 18% pop, and 29% rock music preference differs from the claimed or expected distribution

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Solution: Performing a Goodness of Fit Test Larson/Farber 4th ed 18 Type of music Observed frequency Expected frequency Classical820 Country210180 Gospel7255 Oldies10 Pop7590 Rock125145

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Solution: Performing a Goodness of Fit Test Larson/Farber 4th ed 19 H 0 : H a : α = d.f. = Rejection Region Test Statistic: Decision: 0.01 6 – 1 = 5 0.01 χ2χ2 0 15.086 music preference is 4% classical, 36% country, 11% gospel, 2% oldies, 18% pop, and 29% rock music preference differs from the claimed or expected distribution χ 2 = 22.713 22.713 There is enough evidence to conclude that the distribution of music preferences differs from the claimed distribution. Reject H 0

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Example: Performing a Goodness of Fit Test Larson/Farber 4th ed 20 The manufacturer of M&M’s candies claims that the number of different-colored candies in bags of dark chocolate M&M’s is uniformly distributed. To test this claim, you randomly select a bag that contains 500 dark chocolate M&M’s. The results are shown in the table on the next slide. Using α = 0.10, perform a chi- square goodness-of-fit test to test the claimed or expected distribution. What can you conclude? (Adapted from Mars Incorporated)

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Example: Performing a Goodness of Fit Test Larson/Farber 4th ed 21 ColorFrequency Brown80 Yellow95 Red88 Blue83 Orange76 Green78 Solution: The claim is that the distribution is uniform, so the expected frequencies of the colors are equal. To find each expected frequency, divide the sample size by the number of colors. E = 500/6 ≈ 83.3 n = 500

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Solution: Performing a Goodness of Fit Test Larson/Farber 4th ed 22 H 0 : H a : α = d.f. = Rejection Region Test Statistic: Decision: Conclusion: 0.10 6 – 1 = 5 0.10 χ2χ2 0 9.236 Distribution of different-colored candies in bags of dark chocolate M&Ms is uniform Distribution of different-colored candies in bags of dark chocolate M&Ms is not uniform

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Solution: Performing a Goodness of Fit Test Larson/Farber 4th ed 23 Color Observed frequency Expected frequency Brown8083.3 Yellow9583.3 Red8883.3 Blue8383.3 Orange7683.3 Green7883.3

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Solution: Performing a Goodness of Fit Test Larson/Farber 4th ed 24 H 0 : H a : α = d.f. = Rejection Region Test Statistic: Decision: 0.01 6 – 1 = 5 0.10 χ2χ2 0 9.236 χ 2 = 3.016 3.016 There is not enough evidence to dispute the claim that the distribution is uniform. Distribution of different-colored candies in bags of dark chocolate M&Ms is uniform Distribution of different-colored candies in bags of dark chocolate M&Ms is not uniform Fail to Reject H 0

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Section 10.1 Summary Larson/Farber 4th ed 25 Used the chi-square distribution to test whether a frequency distribution fits a claimed distribution

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