Chi-Square and Analysis of Variance (ANOVA) 12-1 Chapter 12 Chi-Square and Analysis of Variance (ANOVA) 1 1

Outline 12-2 12-1 Introduction 12-2 Test for Goodness of Fit 12-3 Tests Using Contingency Tables 12-4 Analysis of Variance (ANOVA) 2 2 2

Objectives 12-3 Test a distribution for goodness of fit using chi-square. Test two variables for independence using chi-square. Test proportions for homogeneity using chi-square. Use ANOVA technique to determine a difference among three or more means. 3 3

12-2 Test for Goodness of Fit 12-4 When one is testing to see whether a frequency distribution fits a specific pattern, the chi-square goodness-of-fit test is used. 4 4

12-2 Test for Goodness of Fit - Example 12-5 Suppose a market analyst wished to see whether consumers have any preference among five flavors of a new fruit soda. A sample of 100 people provided the following data: 5 5

12-2 Test for Goodness of Fit - Example 12-6 If there were no preference, one would expect that each flavor would be selected with equal frequency. In this case, the equal frequency is 100/5 = 20. That is, approximately 20 people would select each flavor. 6 6

12-2 Test for Goodness of Fit - Example 12-7 The frequencies obtained from the sample are called observed frequencies. The frequencies obtained from calculations are called expected frequencies. Table for the test is shown next. 7 7

12-2 Test for Goodness of Fit - Example 12-8 8 8

12-2 Test for Goodness of Fit - Example 12-9 The observed frequencies will almost always differ from the expected frequencies due to sampling error. Question: Are these differences significant, or are they due to chance? The chi-square goodness-of-fit test will enable one to answer this question. 9 9

12-2 Test for Goodness of Fit - Example 12-10 The appropriate hypotheses for this example are: H0: Consumers show no preference for flavors of the fruit soda. H1: Consumers show a preference. The d. f. for this test is equal to the number of categories minus 1. 10 10

12-2 Test for Goodness of Fit - Formula 12-11   O  E 2    2 E d . f .  number of categories  1 O  observed frequency E  expected frequency 11 11

12-2 Test for Goodness of Fit - Example 12-12 Is there enough evidence to reject the claim that there is no preference in the selection of fruit soda flavors? Let  = 0.05. Step 1: State the hypotheses and identify the claim. 12 12

12-2 Test for Goodness of Fit - Example 12-13 H0: Consumers show no preference for flavors (claim). H1: Consumers show a preference. Step 2: Find the critical value. The d. f. are 5 – 1 = 4 and  = 0.05. Hence, the critical value = 9.488. 13 13

12-2 Test for Goodness of Fit - Example 12-14 Step 3: Compute the test value.  = (32 – 20)2/20 + (28 is – 20)2/20 + … + (10 – 20)2/20 = 18.0. Step 4:Make the decision. The decision is to reject the null hypothesis, since 18.0 > 9.488. 14 14

12-2 Test for Goodness of Fit - Example 12-15 Step 5: Summarize the results. There is enough evidence to reject the claim that consumers show no preference for the flavors. 15 15

12-2 Test for Goodness of Fit - Example 12-16  9.488 16 16

12-2 Test for Goodness of Fit - Example 12-17 The advisor of an ecology club at a large college believes that the group consists of 10% freshmen, 20% sophomores, 40% juniors, and 30% seniors. The membership for the club this year consisted of 14 freshmen, 19 sophomores, 51 juniors, and 16 seniors. At  = 0.10, test the advisor’s conjecture. 17 17

12-2 Test for Goodness of Fit - Example 12-18 Step 1: State the hypotheses and identify the claim. H0: The club consists of 10% freshmen, 20% sophomores, 40% juniors, and 30% seniors (claim) H1: The distribution is not the same as stated in the null hypothesis. 18 18

12-2 Test for Goodness of Fit - Example 12-19 Step 2: Find the critical value. The d. f. are 4 – 1 = 3 and  = 0.10. Hence, the critical value = 6.251. Step 3: Compute the test value.   = (14 – 10)2/10 + (19 – 20)2/20 + … + (16 – 30)2/30 = 11.208. 19 19

12-2 Test for Goodness of Fit - Example 12-20 Step 4:Make the decision. The decision is to reject the null hypothesis, since 11.208 > 6.251. Step 5: Summarize the results. There is enough evidence to reject the advisor’s claim. 20 20

12-3 Tests Using Contingency Tables 12-21 When data can be tabulated in table form in terms of frequencies, several types of hypotheses can be tested using the chi-square test. Two such tests are the independence of variables test and the homogeneity of proportions test. 21 21

12-3 Tests Using Contingency Tables 12-22 The test of independence of variables is used to determine whether two variables are independent when a single sample is selected. The test of homogeneity of proportions is used to determine whether the proportions for a variable are equal when several samples are selected from different populations. 22 22

12-3 Test for Independence - Example 12-23 Suppose a new postoperative procedure is administered to a number of patients in a large hospital. Question: Do the doctors feel differently about this procedure from the nurses, or do they feel basically the same way? Data is on the next slide. 23 23

12-3 Test for Independence - Example 12-24 24 24

12-3 Test for Independence - Example 12-25 The null and the alternative hypotheses are as follows: H0: The opinion about the procedure is independent of the profession. H1: The opinion about the procedure is dependent on the profession. 25 25

12-3 Test for Independence - Example 12-26 If the null hypothesis is not rejected, the test means that both professions feel basically the same way about the procedure, and the differences are due to chance. If the null hypothesis is rejected, the test means that one group feels differently about the procedure from the other. 26 26

12-3 Test for Independence - Example 12-27 Note: The rejection of the null hypothesis does not mean that one group favors the procedure and the other does not. The test value is the  2 value (same as the goodness-of-fit test value). The expected values are computed from: (row sum)(column sum)/(grand total). 27 27

12-3 Test for Independence - Example 12-28 28 28

12-3 Test for Independence - Example 12-29 From the MINITAB output, the P-value = 0. Hence, the null hypothesis will be rejected. If the critical value approach is used, the degrees of freedom for the chi-square critical value will be (number of columns –1)(number of rows – 1). d.f. = (3 –1)(2 – 1) = 2. 29 29

12-3 Test for Homogeneity of Proportions 12-30 Here, samples are selected from several different populations and one is interested in determining whether the proportions of elements that have a common characteristic are the same for each population. 30 30

12-3 Test for Homogeneity of Proportions 12-31 The sample sizes are specified in advance, making either the row totals or column totals in the contingency table known before the samples are selected. The hypotheses will be: H0: p1 = p2 = … = pk H1: At least one proportion is different from the others. 31 31

12-3 Test for Homogeneity of Proportions 12-32 The computations for this test are the same as that for the test of independence. 32 32

12-4 Analysis of Variance (ANOVA) 12-33 When an F test is used to test a hypothesis concerning the means of three or more populations, the technique is called analysis of variance (ANOVA). 6 6

12-4 Assumptions for the F Test for Comparing Three or More Means 12-34 The populations from which the samples were obtained must be normally or approximately normally distributed. The samples must be independent of each other. The variances of the populations must be equal. 7 7

12-4 Analysis of Variance 12-35 Although means are being compared in this F test, variances are used in the test instead of the means. Two different estimates of the population variance are made. 8 8

12-4 Analysis of Variance 12-36 Between-group variance - this involves computing the variance by using the means of the groups or between the groups. Within-group variance - this involves computing the variance by using all the data and is not affected by differences in the means. 9 9

12-4 Analysis of Variance 12-37 The following hypotheses should be used when testing for the difference between three or more means. H0:   =  = … = k H1: At least one mean is different from the others. 10 10

12-4 Analysis of Variance 12-38 d.f.N. = k – 1, where k is the number of groups. d.f.D. = N – k, where N is the sum of the sample sizes of the groups. Note: The formulas for this test are tedious to work through, so examples will be done in MINITAB. See text for formulas. 11 11

12-4 Analysis of Variance -Example 12-39 A marketing specialist wishes to see whether there is a difference in the average time a customer has to wait in a checkout line in three large self-service department stores. The times (in minutes) are shown on the next slide. Is there a significant difference in the mean waiting times of customers for each store using  = 0.05? 12 12

12-4 Analysis of Variance -Example 12-40 13 13

12-4 Analysis of Variance -Example 12-41 Step 1: State the hypotheses and identify the claim. H0:   =  H1: At least one mean is different from the others (claim). 14 14

12-4 Analysis of Variance -Example 12-42 Step 2: Find the critical value. Since k = 3, N = 18, and  = 0.05, d.f.N. = k – 1 = 3 – 1= 2, d.f.D. = N – k = 18 – 3 = 15. The critical value is 3.68. Step 3: Compute the test value. From the MINITAB output, F = 2.70. (See your text for computations). 15 15

12-4 Analysis of Variance -Example 12-43 12-4 Analysis of Variance -Example Step 4: Make a decision. Since 2.70 < 3.68, the decision is not to reject the null hypothesis. Step 5: Summarize the results. There is not enough evidence to support the claim that there is a difference among the means. The ANOVA summary table is given on the next slide. 16 16

12-4 Analysis of Variance -Example 12-44 17 17