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Cross Tabulation and Chi-Square Testing
Cross-Tabulation While a frequency distribution describes one variable at a time, a cross-tabulation describes two or more variables simultaneously. Cross-tabulation results in tables that reflect the joint distribution of two or more variables with a limited number of categories or distinct values, e.g., Table 16.3. Copyright © 2012 Pearson Education, Inc. Chapter 16 - 2
Chapter 16 - 3 Copyright © 2012 Pearson Education, Inc. Usage GENDER FemaleMaleRaw Total Light Users14519 Medium Users5510 Heavy Users51116 Column Total2421
Two Variables Cross-Tabulation Since two variables have been cross classified, percentages could be computed either column wise, based on column totals (Table 16.4), or row wise, based on row totals (Table 16.5). The general rule is to compute the percentages in the direction of the independent variable, across the dependent variable. The correct way of calculating percentages is as shown in Table 16.4. Copyright © 2012 Pearson Education, Inc. Chapter 16 - 4
Usage GENDER FemaleMale Light Users58.4%23.8% Medium Users20.8%23.8% Heavy Users20.8%52.4% Column Total100.0% Chapter 16 - 5 Copyright © 2012 Pearson Education, Inc.
Usage GENDER FemaleMaleRaw Total Light Users73.7%26.3%100.0% Medium Users50.0% 100.0% Heavy Users31.2%68.8%100.0% Chapter 16 - 6 Copyright © 2012 Pearson Education, Inc.
Statistics Associated with Cross- Tabulation Chi-Square To determine whether a systematic association exists, the probability of obtaining a value of chi-square as large or larger than the one calculated from the cross-tabulation is estimated. An important characteristic of the chi-square statistic is the number of degrees of freedom (df) associated with it. That is, df = (r - 1) x (c -1). The null hypothesis (H 0 ) of no association between the two variables will be rejected only when the calculated value of the test statistic is greater than the critical value of the chi- square distribution with the appropriate degrees of freedom, as shown in Figure 16.10. Copyright © 2012 Pearson Education, Inc. Chapter 16 - 7
2 Level of Significance, α Chapter 16 - 8 Copyright © 2012 Pearson Education, Inc.
The chi-square statistic ( 2 ) is used to test the statistical significance of the observed association in a cross- tabulation. The expected frequency for each cell can be calculated by using a simple formula: where = total number in the row = total number in the column = total sample size Chapter 16 - 9 Copyright © 2012 Pearson Education, Inc.
For the data in Table 16.3, for the six cells from left to right and top to bottom = (24 x 19)/45 = 10.1 = (21 x 19)/45 =8.9 = (24 x 10)/45 = 5.3 = (21 x 10)/45 =4.7 = (24 x 16)/45 = 8.5 = (21 x 16)/45 =7.5 Chapter 16 - 10 Copyright © 2012 Pearson Education, Inc.
= (14 -10.1) 2 + (5 – 8.9) 2 10.1 8.9 + (5 – 5.3) 2 + (5 – 4.7) 2 5.3 4.7 + (5 – 8.5) 2 + (11 – 7.5) 2 8.5 7.5 = 1.51 + 1.71 + 0.02 + 0.02 + 1.44 + 1.63 = 6.33 Chapter 16 - 11 Copyright © 2012 Pearson Education, Inc.
Statistics Associated with Cross-Tabulation Chi-Square (Cont.) The chi-square distribution is a skewed distribution whose shape depends solely on the number of degrees of freedom. As the number of degrees of freedom increases, the chi-square distribution becomes more symmetrical. Table 3 in the Statistical Appendix contains upper-tail areas of the chi-square distribution for different degrees of freedom. For 2 degrees of freedom, the probability of exceeding a chi-square value of 5.991 is 0.05. For the cross-tabulation given in Table 16.3, there are (3-1) x (2-1) = 2 degrees of freedom. The calculated chi-square statistic had a value of 6.333. Since this is greater than the critical value of 5.991, the null hypothesis of no association is rejected indicating that the association is statistically significant at the 0.05 level. Copyright © 2012 Pearson Education, Inc. Chapter 16 - 12
Statistics Associated with Cross-Tabulation Phi Coefficient The phi coefficient ( ) is used as a measure of the strength of association in the special case of a table with two rows and two columns (a 2 x 2 table). The phi coefficient is proportional to the square root of the chi-square statistic: It takes the value of 0 when there is no association, which would be indicated by a chi-square value of 0 as well. When the variables are perfectly associated, phi assumes the value of 1 and all the observations fall just on the main or minor diagonal. Copyright © 2012 Pearson Education, Inc. Chapter 16 - 13
Statistics Associated with Cross-Tabulation Contingency Coefficient While the phi coefficient is specific to a 2 x 2 table, the contingency coefficient (C) can be used to assess the strength of association in a table of any size. The contingency coefficient varies between 0 and 1. The maximum value of the contingency coefficient depends on the size of the table (number of rows and number of columns). For this reason, it should be used only to compare tables of the same size. Copyright © 2012 Pearson Education, Inc. Chapter 16 - 14
Statistics Associated with Cross- Tabulation Cramer’s V Cramer's V is a modified version of the phi correlation coefficient, , and is used in tables larger than 2 x 2. or. Chapter 16 - 15 Copyright © 2012 Pearson Education, Inc.
Cross-Tabulation in Practice While conducting cross-tabulation analysis in practice, it is useful to proceed along the following steps: Test the null hypothesis that there is no association between the variables using the chi-square statistic. If you fail to reject the null hypothesis, then there is no relationship. If H 0 is rejected, then determine the strength of the association using an appropriate statistic (phi-coefficient, contingency coefficient, or Cramer's V), as discussed earlier. If H 0 is rejected, interpret the pattern of the relationship by computing the percentages in the direction of the independent variable, across the dependent variable. Draw marketing conclusions. Copyright © 2012 Pearson Education, Inc. Chapter 16 - 16
Construct the Cross-Tabulation Data Calculate the Chi-Square Statistic, Test the Null Hypothesis of No Association Reject H 0 ? Interpret the Pattern of Relationship by Calculating Percentages in the Direction of the Independent Variable No Association Determine the Strength of Association Using an Appropriate Statistic NOYES Chapter 16 -17 Copyright © 2012 Pearson Education, Inc.
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