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Section 10.2. Independence. Section 10.2 Objectives. Use a contingency table to find expected frequencies Use a chi-square distribution to test whether two variables are independent. Contingency Tables. r  c contingency table Shows the observed frequencies for two variables.

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Section 10.2

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Section 10 2

Larson/Farber 4th ed

Section 10.2

Independence


Section 10 2 objectives

Section 10.2 Objectives

  • Use a contingency table to find expected frequencies

  • Use a chi-square distribution to test whether two variables are independent

Larson/Farber 4th ed


Contingency tables

Contingency Tables

r c contingency table

  • Shows the observed frequencies for two variables.

  • The observed frequencies are arranged in r rows and c columns.

  • The intersection of a row and a column is called a cell.

Larson/Farber 4th ed


Contingency tables1

Contingency Tables

Example:

  • The contingency table shows the results of a random sample of 550 company CEOs classified by age and size of company.(Adapted from Grant Thornton LLP, The Segal Company)

Larson/Farber 4th ed


Finding the expected frequency

Finding the Expected Frequency

  • Assuming the two variables are independent, you can use the contingency table to find the expected frequency for each cell.

  • The expected frequency for a cell Er,cin a contingency table is

Larson/Farber 4th ed


Example finding expected frequencies

Example: Finding Expected Frequencies

Find the expected frequency for each cell in the contingency table. Assume that the variables, age and company size, are independent.

Larson/Farber 4th ed

marginal totals


Solution finding expected frequencies

Solution: Finding Expected Frequencies

Larson/Farber 4th ed


Solution finding expected frequencies1

Solution: Finding Expected Frequencies

Larson/Farber 4th ed


Solution finding expected frequencies2

Solution: Finding Expected Frequencies

Larson/Farber 4th ed


Chi square independence test

Chi-Square Independence Test

Chi-square independence test

  • Used to test the independence of two variables.

  • Can determine whether the occurrence of one variable affects the probability of the occurrence of the other variable.

Larson/Farber 4th ed


Chi square independence test1

Chi-Square Independence Test

For the chi-square independence test to be used, the following must be true.

  • The observed frequencies must be obtained by using a random sample.

  • Each expected frequency must be greater than or equal to 5.

Larson/Farber 4th ed


Chi square independence test2

Chi-Square Independence Test

  • If these conditions are satisfied, then the sampling distribution for the chi-square independence test is approximated by a chi-square distribution with (r – 1)(c – 1) degrees of freedom, where r and c are the number of rows and columns, respectively, of a contingency table.

  • The test statistic for the chi-square independence test is

    where O represents the observed frequencies and E represents the expected frequencies.

The test is always a right-tailed test.

Larson/Farber 4th ed


Chi square independence test3

Chi-Square Independence Test

In WordsIn Symbols

Identify the claim. State the null and alternative hypotheses.

Specify the level of significance.

Identify the degrees of freedom.

Determine the critical value.

State H0 and Ha.

Identify .

d.f. = (r – 1)(c – 1)

Use Table 6 in Appendix B.

Larson/Farber 4th ed


Chi square independence test4

Chi-Square Independence Test

In WordsIn Symbols

Determine the rejection region.

Calculate the test statistic.

Make a decision to reject or fail to reject the null hypothesis.

Interpret the decision in the context of the original claim.

If χ2 is in the rejection region, reject H0. Otherwise, fail to reject H0.

Larson/Farber 4th ed


Example performing a 2 independence test

Example: Performing a χ2 Independence Test

Using the age/company size contingency table, can you conclude that the CEOs ages are related to company size? Use α = 0.01. Expected frequencies are shown in parentheses.

Larson/Farber 4th ed


Solution performing a goodness of fit test

0.01

(2 – 1)(5 – 1) = 4

0.01

χ2

0

13.277

Solution: Performing a Goodness of Fit Test

  • H0:

  • Ha:

  • α =

  • d.f. =

  • Rejection Region

CEOs’ ages are independent of company size

CEOs’ ages are dependent on company size

  • Test Statistic:

  • Decision:

Larson/Farber 4th ed


Solution performing a goodness of fit test1

Solution: Performing a Goodness of Fit Test

Larson/Farber 4th ed


Solution performing a goodness of fit test2

0.01

(2 – 1)(5 – 1) = 4

0.01

χ2

0

13.277

Solution: Performing a Goodness of Fit Test

  • H0:

  • Ha:

  • α =

  • d.f. =

  • Rejection Region

CEOs’ ages are independent of company size

CEOs’ ages are dependent on company size

  • Test Statistic:

  • Decision:

χ2 = 77.9

Reject H0

There is enough evidence to conclude CEOs’ ages are dependent on company size.

77.9

Larson/Farber 4th ed


Section 10 2 summary

Section 10.2 Summary

  • Used a contingency table to find expected frequencies

  • Used a chi-square distribution to test whether two variables are independent

Larson/Farber 4th ed


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