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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

12-1

Chapter 12

Chi-Square and Analysis of Variance (ANOVA)

outline
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
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.
12 2 test for goodness of fit
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.
12 2 test for goodness of fit example
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:
12 2 test for goodness of fit example6
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.
12 2 test for goodness of fit example7
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.
12 2 test for goodness of fit example9
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.
12 2 test for goodness of fit example10
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.
12 2 test for goodness of fit formula
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 2 test for goodness of fit example12
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 2 test for goodness of fit example13
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.
12 2 test for goodness of fit example14
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.
12 2 test for goodness of fit example15
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.
12 2 test for goodness of fit example17
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.
12 2 test for goodness of fit example18
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.
12 2 test for goodness of fit example19
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.
12 2 test for goodness of fit example20
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.
12 3 tests using contingency tables
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.
12 3 tests using contingency tables22
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.
12 3 test for independence example
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.
12 3 test for independence example25
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.
12 3 test for independence example26
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.
12 3 test for independence example27
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).
12 3 test for independence example29
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.
12 3 test for homogeneity of proportions
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.
12 3 test for homogeneity of proportions31
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.
12 3 test for homogeneity of proportions32
12-3 Test for Homogeneity of Proportions

12-32

  • The computations for this test are the same as that for the test of independence.
12 4 analysis of variance anova
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).
12 4 assumptions for the f test for comparing three or more means
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.
12 4 analysis of variance
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.
12 4 analysis of variance36
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.
12 4 analysis of variance37
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.
12 4 analysis of variance38
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.
12 4 analysis of variance example
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 4 analysis of variance example41
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).
12 4 analysis of variance example42
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).
12 4 analysis of variance example43
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.