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Chapter 12

12-1. Chapter 12. Chi-Square and Analysis of Variance (ANOVA). 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). Objectives. 12-3. Test a distribution for goodness of fit using chi-square.

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Chapter 12

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

  2. 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)

  3. 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.

  4. 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.

  5. 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:

  6. 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.

  7. 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.

  8. F r e q . C h e r r y S t r a w - O r a n g e L i m e G r a p e b e r r y O b s e r v e d 3 2 2 8 1 6 1 4 1 0 E x p e c t e d 2 0 2 0 2 0 2 0 12-2 Test for Goodness of Fit -Example 12-8 2 0

  9. 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.

  10. 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.

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

  12. 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.

  13. 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.

  14. 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.

  15. 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.

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

  17. 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.

  18. 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.

  19. 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.

  20. 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.

  21. 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.

  22. 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.

  23. 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.

  24. G r o u p P r e f e r P r e f e r N o n e w o l d p r e f e r e n c e p p p r r r o o o c c c e e e d d d u u u r r r e e e p r o c e d u r e p r o c e d u r e N u r s e s 1 0 0 8 0 2 0 D o c t o r s 5 0 1 2 0 3 0 12-3 Test for Independence -Example 12-24 G r o u p P r e f e r P r e f e r N o n e w o l d N u r s e s 1 0 0 8 0 2 0 D o c t o r s 5 0 1 2 0 3 0

  25. 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.

  26. 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.

  27. 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).

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

  29. 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.

  30. 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.

  31. 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 = … = pkH1: At least one proportion is different from the others.

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

  33. 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).

  34. 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.

  35. 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.

  36. 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.

  37. 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.

  38. 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.

  39. 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?

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

  41. 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).

  42. 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).

  43. 12-4 Analysis of Variance-Example 12-43 • 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.

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

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