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

Chapter Outline 10.1 Goodness of Fit 10.2 Independence 10.3 Comparing Two Variances 10.4 Analysis of Variance

Section 10.1 Goodness of Fit

Section 10.1 Objectives Use the chi-square distribution to test whether a frequency distribution fits a claimed distribution

Properties of The Chi-Square Distribution All chi-square values χ2 are greater than or equal to zero. The chi-square distribution is a family of curves, each determined by the degrees of freedom. To find the critical values, use the χ2-distribution with degrees of freedom equal to one less than the sample size. d.f. = n – 1 Degrees of freedom The area under each curve of the chi-square distribution equals one.

Properties of The Chi-Square Distribution Chi-square distributions are positively skewed. chi-square distributions

Finding Critical Values for the χ2-Test Specify the level of significance . Determine the degrees of freedom d.f. = n – 1. The critical values for the χ2-distribution are found in Table 6 of Appendix B. To find the critical value(s) for a right-tailed test, use the value that corresponds to d.f. and . left-tailed test, use the value that corresponds to d.f. and 1 – . two-tailed test, use the values that corresponds to d.f. and ½ and d.f. and 1 – ½. 7 7

Finding Critical Values for the χ2-Test Right-tailed Left-tailed χ2 1 – α χ2 1 – α Two-tailed χ2 1 – α 8 8

Example: Finding Critical Values for χ2 Find the critical χ2-value for a left-tailed test when n = 11 and  = 0.01. Solution: Degrees of freedom: n – 1 = 11 – 1 = 10 d.f. The area to the right of the critical value is 1 –  = 1 – 0.01 = 0.99. χ2 From Table 6, the critical value is . 9 9

Example: Finding Critical Values for χ2 Find the critical χ2-value for a two-tailed test when n = 13 and  = 0.01. Solution: Degrees of freedom: n – 1 = 13 – 1 = 12 d.f. The areas to the right of the critical values are χ2 From Table 6, the critical values are and 10 10

Multinomial Experiments 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.

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 Classical 4% Oldies 2% Country 36% Pop 18% Gospel 11% Rock 29% Each outcome is classified into categories. The probability for each possible outcome is fixed.

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.

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.

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.

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 ith category is Ei = npi where n is the number of trials (the sample size) and pi is the assumed probability of the ith category.

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) Classical 8 Country 210 Gospel 72 Oldies 10 Pop 75 Rock 125

Solution: Finding Observed and Expected Frequencies Observed frequency: The number of radio music listeners naming a particular type of music Survey results (n = 500) Classical 8 Country 210 Gospel 72 Oldies 10 Pop 75 Rock 125 observed frequency

Solution: Finding Observed and Expected Frequencies Expected Frequency: Ei = npi Type of music % of listeners Observed frequency Expected frequency Classical 4% 8 Country 36% 210 Gospel 11% 72 Oldies 2% 10 Pop 18% 75 Rock 29% 125 500(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

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.

Chi-Square Goodness-of-Fit Test Chapter 10 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

Chi-Square Goodness-of-Fit Test Chapter 10 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

Chi-Square Goodness-of-Fit Test Chapter 10 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

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. Distribution of music preferences Classical 4% Country 36% Gospel 11% Oldies 2% Pop 18% Rock 29% Survey results (n = 500) Classical 8 Country 210 Gospel 72 Oldies 10 Pop 75 Rock 125

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 0.01 6 – 1 = 5 Test Statistic: Decision: Conclusion: 0.01 χ2 15.086

Solution: Performing a Goodness of Fit Test Chapter 10 Solution: Performing a Goodness of Fit Test Type of music Observed frequency Expected frequency Classical 8 20 Country 210 180 Gospel 72 55 Oldies 10 Pop 75 90 Rock 125 145 Larson/Farber 4th ed

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 0.01 6 – 1 = 5 Test Statistic: Decision: χ2 = 22.713 Reject H0 0.01 χ2 15.086 There is enough evidence to conclude that the distribution of music preferences differs from the claimed distribution. 22.713

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)

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 Color Frequency Brown 80 Yellow 95 Red 88 Blue 83 Orange 76 Green 78 n = 500

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 0.10 6 – 1 = 5 Test Statistic: Decision: Conclusion: 0.10 χ2 9.236

Solution: Performing a Goodness of Fit Test Chapter 10 Solution: Performing a Goodness of Fit Test Color Observed frequency Expected frequency Brown 80 83.3 Yellow 95 Red 88 Blue 83 Orange 76 Green 78 Larson/Farber 4th ed

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 0.01 6 – 1 = 5 Test Statistic: Decision: χ2 = 3.016 Fail to Reject H0 0.10 χ2 9.236 There is not enough evidence to dispute the claim that the distribution is uniform. 3.016

Section 10.1 Summary Used the chi-square distribution to test whether a frequency distribution fits a claimed distribution