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## PowerPoint Slideshow about 'Chapter 12' - niveditha

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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.
- 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

12-4

- 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

12-5

- 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 -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 - 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 -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 - 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

12-11

2

O

E

2

E

d

.

f

.

number

of

categories

1

O

observed

frequency

E

expected

frequency

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

12-21

- 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 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

12-23

- 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 - 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 -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 - 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 -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

12-30

- 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 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 Proportions

12-32

- The computations for this test are the same as that for the test of independence.

12-4 Analysis of Variance (ANOVA)

12-33

- 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

12-34

- 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

12-35

- 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 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 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 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

12-39

- 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-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-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-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.

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