Download presentation

Presentation is loading. Please wait.

Published byTabitha Payne Modified over 4 years ago

1
Chapter 10 Chi-Square Tests and the F- Distribution 1 Larson/Farber 4th ed

2
Chapter Outline 10.1 Goodness of Fit 10.2 Independence 10.3 Comparing Two Variances 10.4 Analysis of Variance 2 Larson/Farber 4th ed

3
Section 10.1 Goodness of Fit 3 Larson/Farber 4th ed

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

5
Multinomial Experiments 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. 5 Larson/Farber 4th ed

6
Multinomial Experiments 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. 6 Larson/Farber 4th ed

7
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. 7 Larson/Farber 4th ed

8
Chi-Square Goodness-of-Fit Test 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. 8 Larson/Farber 4th ed

9
Chi-Square Goodness-of-Fit Test 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. 9 Larson/Farber 4th ed

10
Chi-Square Goodness-of-Fit Test 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. 10 Larson/Farber 4th ed

11
Example: Finding Observed and Expected Frequencies 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 11 Larson/Farber 4th ed

12
Solution: Finding Observed and Expected Frequencies 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 12 Larson/Farber 4th ed

13
Solution: Finding Observed and Expected Frequencies 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 13 Larson/Farber 4th ed

14
Chi-Square Goodness-of-Fit Test 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. 14 Larson/Farber 4th ed

15
Chi-Square Goodness-of-Fit Test 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. 15 Larson/Farber 4th ed

16
Chi - Square Goodness - of - Fit Test 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 16 Larson/Farber 4th ed

17
Chi - Square Goodness - of - Fit Test 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 17 Larson/Farber 4th ed

18
Example: Performing a Goodness of Fit Test 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% 18 Larson/Farber 4th ed

19
Solution: Performing a Goodness of Fit Test 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 19 Larson/Farber 4th ed

20
Solution: Performing a Goodness of Fit Test Type of music Observed frequency Expected frequency Classical820 Country210180 Gospel7255 Oldies10 Pop7590 Rock125145 20 Larson/Farber 4th ed

21
Solution: Performing a Goodness of Fit Test 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 21 Larson/Farber 4th ed

22
Example: Performing a Goodness of Fit Test 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) 22 Larson/Farber 4th ed

23
Example: Performing a Goodness of Fit Test 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 23 Larson/Farber 4th ed n = 500

24
Solution: Performing a Goodness of Fit Test 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 24 Larson/Farber 4th ed

25
Solution: Performing a Goodness of Fit Test Color Observed frequency Expected frequency Brown8083.3 Yellow9583.3 Red8883.3 Blue8383.3 Orange7683.3 Green7883.3 25 Larson/Farber 4th ed

26
Solution: Performing a Goodness of Fit Test 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 26 Larson/Farber 4th ed

27
Section 10.1 Summary Used the chi-square distribution to test whether a frequency distribution fits a claimed distribution 27 Larson/Farber 4th ed

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google