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Chi-Square Goodness-of-Fit Test for Music Preferences

Perform a chi-square goodness-of-fit test to determine if the distribution of music preferences differs from the claimed distribution in a radio station's broadcast region.

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Chi-Square Goodness-of-Fit Test for Music Preferences

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

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

  4. Section 10.1 Objectives • Use the chi-square distribution to test whether a frequency distribution fits a claimed distribution 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. 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. Each outcome is classified into categories. The probability for each possible outcome is fixed. Larson/Farber 4th ed

  7. Chi-Square Goodness-of-Fit Test Chi-Square Goodness-of-FitTest • 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. 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. H0: The distribution of music preferences in the broadcast region is 4% classical, 36% country, 11% gospel, 2% oldies, 18% pop, and 29% rock. (claim) Ha: The distribution of music preferences differs from the claimed or expected distribution. 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 Oof a category is the frequency for the category observed in the sample data. Larson/Farber 4th ed

  10. Chi-Square Goodness-of-Fit Test • The expected frequency Eof a category is the calculated frequency for the category. • Expected frequencies are obtained assuming the specified (or hypothesized) distribution. The expected frequency for the ith category isEi = npi where n is the number of trials (the sample size) and piis the assumed probability of the ith category. 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. 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 observed frequency Larson/Farber 4th ed

  13. Solution: Finding Observed and Expected Frequencies Expected Frequency:Ei = npi 500(0.04) = 20 500(0.36) = 180 500(0.11) = 55 500(0.02) = 10 500(0.18) = 90 500(0.29) = 145 n = 500 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. • The observed frequencies must be obtained by using a random sample. • Each expected frequency must be greater than or equal to 5. 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. Larson/Farber 4th ed

  16. Chi-Square Goodness-of-Fit Test In Words In Symbols • Identify the claim. State the null and alternative hypotheses. • Specify the level of significance. • Identify the degrees of freedom. • Determine the critical value. State H0 and Ha. Identify . d.f. = k – 1 Use Table 6 in Appendix B. Larson/Farber 4th ed

  17. Chi-Square Goodness-of-Fit Test In Words In Symbols • Determine the rejection region. • Calculate the test statistic. • Make a decision to reject or fail to reject the null hypothesis. • Interpret the decision in the context of the original claim. If χ2 is in the rejection region, reject H0. Otherwise, fail to reject H0. 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. Larson/Farber 4th ed

  19. 0.01 6 – 1 = 5 0.01 χ2 0 15.086 Solution: Performing a Goodness of Fit Test • H0: • Ha: • α = • d.f. = • Rejection Region 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 • Test Statistic: • Decision: • Conclusion: Larson/Farber 4th ed

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

  21. 0.01 6 – 1 = 5 0.01 χ2 0 15.086 Solution: Performing a Goodness of Fit Test • H0: • Ha: • α = • d.f. = • Rejection Region 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 • Test Statistic: • Decision: χ2 = 22.713 Reject H0 There is enough evidence to conclude that the distribution of music preferences differs from the claimed distribution. 22.713 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) Larson/Farber 4th ed

  23. Example: Performing a Goodness of Fit Test • 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 Larson/Farber 4th ed

  24. 0.10 6 – 1 = 5 0.10 χ2 0 9.236 Solution: Performing a Goodness of Fit Test • H0: • Ha: • α = • d.f. = • Rejection Region 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 • Test Statistic: • Decision: • Conclusion: Larson/Farber 4th ed

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

  26. 0.01 6 – 1 = 5 0.10 χ2 0 9.236 Solution: Performing a Goodness of Fit Test • H0: • Ha: • α = • d.f. = • Rejection Region 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 • Test Statistic: • Decision: χ2 = 3.016 Fail to Reject H0 There is not enough evidence to dispute the claim that the distribution is uniform. 3.016 Larson/Farber 4th ed

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

  28. Section 10.2 Independence Larson/Farber 4th ed

  29. Section 10.2 Objectives • Use a contingency table to find expected frequencies • Use a chi-square distribution to test whether two variables are independent Larson/Farber 4th ed

  30. Contingency Tables r c contingency table • Shows the observed frequencies for two variables. • The observed frequencies are arranged in r rows and c columns. • The intersection of a row and a column is called a cell. Larson/Farber 4th ed

  31. Contingency Tables Example: • The contingency table shows the results of a random sample of 550 company CEOs classified by age and size of company.(Adapted from Grant Thornton LLP, The Segal Company) Larson/Farber 4th ed

  32. Finding the Expected Frequency • Assuming the two variables are independent, you can use the contingency table to find the expected frequency for each cell. • The expected frequency for a cell Er,cin a contingency table is Larson/Farber 4th ed

  33. Example: Finding Expected Frequencies Find the expected frequency for each cell in the contingency table. Assume that the variables, age and company size, are independent. marginal totals Larson/Farber 4th ed

  34. Solution: Finding Expected Frequencies Larson/Farber 4th ed

  35. Solution: Finding Expected Frequencies Larson/Farber 4th ed

  36. Solution: Finding Expected Frequencies Larson/Farber 4th ed

  37. Chi-Square Independence Test Chi-square independence test • Used to test the independence of two variables. • Can determine whether the occurrence of one variable affects the probability of the occurrence of the other variable. Larson/Farber 4th ed

  38. Chi-Square Independence Test For the chi-square independence test to be used, the following must be true. • The observed frequencies must be obtained by using a random sample. • Each expected frequency must be greater than or equal to 5. Larson/Farber 4th ed

  39. Chi-Square Independence Test • If these conditions are satisfied, then the sampling distribution for the chi-square independence test is approximated by a chi-square distribution with (r – 1)(c – 1) degrees of freedom, where r and c are the number of rows and columns, respectively, of a contingency table. • The test statistic for the chi-square independence test is where O represents the observed frequencies and E represents the expected frequencies. The test is always a right-tailed test. Larson/Farber 4th ed

  40. Chi-Square Independence Test In Words In Symbols • Identify the claim. State the null and alternative hypotheses. • Specify the level of significance. • Identify the degrees of freedom. • Determine the critical value. State H0 and Ha. Identify . d.f. = (r – 1)(c – 1) Use Table 6 in Appendix B. Larson/Farber 4th ed

  41. Chi-Square Independence Test In Words In Symbols • Determine the rejection region. • Calculate the test statistic. • Make a decision to reject or fail to reject the null hypothesis. • Interpret the decision in the context of the original claim. If χ2 is in the rejection region, reject H0. Otherwise, fail to reject H0. Larson/Farber 4th ed

  42. Example: Performing a χ2 Independence Test Using the age/company size contingency table, can you conclude that the CEOs ages are related to company size? Use α = 0.01. Expected frequencies are shown in parentheses. Larson/Farber 4th ed

  43. 0.01 (2 – 1)(5 – 1) = 4 0.01 χ2 0 13.277 Solution: Performing a Goodness of Fit Test • H0: • Ha: • α = • d.f. = • Rejection Region CEOs’ ages are independent of company size CEOs’ ages are dependent on company size • Test Statistic: • Decision: Larson/Farber 4th ed

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

  45. 0.01 (2 – 1)(5 – 1) = 4 0.01 χ2 0 13.277 Solution: Performing a Goodness of Fit Test • H0: • Ha: • α = • d.f. = • Rejection Region CEOs’ ages are independent of company size CEOs’ ages are dependent on company size • Test Statistic: • Decision: χ2 = 77.9 Reject H0 There is enough evidence to conclude CEOs’ ages are dependent on company size. 77.9 Larson/Farber 4th ed

  46. Section 10.2 Summary • Used a contingency table to find expected frequencies • Used a chi-square distribution to test whether two variables are independent Larson/Farber 4th ed

  47. Section 10.3 Comparing Two Variances Larson/Farber 4th ed

  48. Section 10.3 Objectives • Interpret the F-distribution and use an F-table to find critical values • Perform a two-sample F-test to compare two variances Larson/Farber 4th ed

  49. F-Distribution • Let represent the sample variances of two different populations. • If both populations are normal and the population variances are equal, then the sampling distribution of is called an F-distribution. Larson/Farber 4th ed

  50. Properties of the F-Distribution • The F-distribution is a family of curves each of which is determined by two types of degrees of freedom: • The degrees of freedom corresponding to the variance in the numerator, denoted d.f.N • The degrees of freedom corresponding to the variance in the denominator, denoted d.f.D • F-distributions are positively skewed. • The total area under each curve of an F-distribution is equal to 1. Larson/Farber 4th ed

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