sw318 social work statistics slide 1 chi-square test of independence practice problem – 1 this...

SW318 Social Work Statistics Slide

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Chi-Square Test of Independence Practice Problem – 1

This question asks you to use a chi-square test of independence to answer the question. First of all, the level of measurement for the independent and the dependent variable can be any level that defines groups (dichotomous, nominal, ordinal, or grouped interval). "Marital status" [marstat] is nominal and "sex" [sex] is dichotomous, so the level of measurement requirements are satisfied.


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Chi-Square Test of Independence in SPSS (1)

You can conduct a chi-square test of independence in crosstabulation of SPSS by selecting:

Analyze > Descriptive Statistics > Crosstabs…


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First, select and move the variables for the question to “Row(s):” and “Column(s):” list boxes.

The variable mentioned first in the problem, sex, is used as the independent variable and is moved to the “Column(s):” list box.

The variable mentioned second in the problem, marital status, is used as the dependent variable and is moved to the “Row(s)” list box.

Second, click on “Statistics…” button to request the test statistic.


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Second, click on “Continue” button to close the Statistics dialog box.

First, click on “Chi-square” to request the chi-square test of independence.


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Now click on “Cells…” button to specify the contents in the cells of the crosstabs table.


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Make sure both “Observed” and “Expected” in the “Counts” section in “Crosstabs: Cell Display” dialog box are checked and click on “Continue” and “OK” buttons.


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The chi-square test of independence requires that the expected frequency for all cells be 5.0 or higher.

The crosstabulated table contains the count and expected counts for each cell in the table.

We can see that none of the expected counts are less than 5, and the smallest expected count is 5.9.


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In the table Chi-Square Tests result, SPSS also tells us that “0 cells have expected count less than 5 and the minimum expected count is 5.93”.

The sample size requirement for the chi-square test of independence is satisfied.


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The probability of the chi-square test statistic (chi-square=114.748) was p<0.001, less than the alpha level of significance of 0.05.

The null hypothesis that differences in "marital status" are independent of differences in "sex" (the actual frequencies are equal to the expected frequencies in the cross-tabulated table) is rejected.

In other words, the research hypothesis that differences in “sex" are related to differences in "marital status" is supported by this analysis.

The answer to the question is True.


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Chi-Square Test of Independence Practice Problem – 2

This question asks you to use a chi-square test of independence to answer the question. First of all, the level of measurement for the independent and the dependent variable can be any level that defines groups (dichotomous, nominal, ordinal, or grouped interval). “Employment status" [empstat] is nominal and "sex" [sex] is dichotomous, so the level of measurement requirements are satisfied.


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Conduct a chi-square test of independence in crosstabulation of SPSS by selecting:

Analyze > Descriptive Statistics > Crosstabs…


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Select and move [empstat] to Row(s): box and make sure [sex] is still in the Column(s): box .Then, click OK.


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The chi-square test of independence requires that the expected frequency for all cells be 5.0 or higher.

Since the Chi-Square Tests result shows that “2 cells have expected count less than 5” , the sample size requirement for the chi-square test of independence is not satisfied.

The chi-square statistic may lead to erroneous conclusions, so it should not be interpreted.

The answer to the question is incorrect application of a statistic.


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Steps in solving chi-square test of independence problems - 1

The following is a guide to the decision process for answering homework problems about chi-square test of independence problems:

Is the dependent and independent variable nominal, ordinal, dichotomous, or grouped interval?

Incorrect application of

a statistic

Yes

No


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Steps in solving chi-square test of independence problems - 2

Yes

Expected cell counts less than 5?

No

Incorrect application of

a statistic

Compute the Chi-Square test of independence

Is the p-value for the chi-square test of independence <= alpha?

Yes

False No

True

sw318 social work statistics slide 1 chi-square test of independence practice problem – 1 this...

Documents

crosstabulation of spss

table chisquare tests

sex sex

expected counts

expected frequency

cells button

independent variable

minimum expected count