© The McGraw-Hill Companies, Inc., 2000 12-1 Chapter 12 Chi-Square.

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Chapter 12Chapter 12

Chi-SquareChi-Square

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12-212-2 OutlineOutline

12-1 Introduction

12-2 Test for Goodness of Fit

12-3 Tests Using Contingency

Tables

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12-312-3 ObjectivesObjectives

Test a distribution for goodness of fit using chi-square.

Test two variables for independence using chi-square.

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When one is testing to see whether a frequency distribution fits a specific pattern, the chi-square chi-square goodness-of-fit testgoodness-of-fit test is used.

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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 -12-2 Test for Goodness of Fit - Example

Cherry Straw-berry

Orange Lime Grape

32 28 16 14 10

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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.

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The frequencies obtained from the sample are called observed observed frequenciesfrequencies.

The frequencies obtained from calculations are called expected expected frequenciesfrequencies.

Table for the test is shown next.

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Example

Freq. Cherry Straw-berry

Orange Lime Grape

Observed 32 28 16 14 10

Expected 20 20 20 20 20

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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.

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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.

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Formula

2

2

1

O E

E

d f number of categories

O observed frequency

E frequency

. .

expected

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Is there enough evidence to reject the claim that there is no preference in the selection of fruit soda flavours? Let = 0.05.

Step 1:Step 1: State the hypotheses and identify the claim.

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H0: Consumers show no preference for flavours (claim).

H1: Consumers show a preference. Step 2:Step 2: Find the critical value. The d. f.

are 5 – 1 = 4 and = 0.05. Hence, the critical value = 9.488.

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Step 3:Step 3: Compute the test value. = (32 – 20)2/20 + (28 is – 20)2/20 + … + (10 – 20)2/20 = 18.0.

Step 4:Step 4:Make the decision. The decision is to reject the null hypothesis, since 18.0 > 9.488.

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Step 5:Step 5: Summarize the results. There is enough evidence to reject the claim that consumers show no preference for the flavours.

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Example

9.488

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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.

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Step 1: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.

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Step 2:Step 2: Find the critical value. The d. f. are 4 – 1 = 3 and = 0.10. Hence, the critical value = 6.251.

Step 3:Step 3: Compute the test value. = (14 – 10)2/10 + (19 – 20)2/20 + … + (16 – 30)2/30 = 11.208.

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Step 4:Step 4:Make the decision. The decision is to reject the null hypothesis, since 11.208 > 6.251.

Step 5:Step 5: Summarize the results. There is enough evidence to reject the advisor’s claim.

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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 independence of variablesvariables test and the homogeneity of homogeneity of proportionsproportions test.

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The test of independence of variablestest of independence of variables is used to determine whether two variables are independent when a single sample is selected.

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Suppose a new postoperative procedure is administered to a number of patients in a large hospital.

Question: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.

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Example

Group Prefernew

procedure

Preferold

procedure

Nopreference

Nurses 100 80 20

Doctors 50 120 30

Group Prefernew

procedure

Preferold

procedure

Nopreference

Nurses 100 80 20

Doctors 50 120 30

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The null and the alternative hypotheses are as follows:

H0: The opinion about the procedure is independentindependent of the profession.

H1: The opinion about the procedure is dependentdependent on the profession.

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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.

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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).

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From the MINITAB output, the P-value = 0.000. Hence, the null hypothesis will be rejected (P<0.0005).

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.

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