Solution

Table 4.E.5 shows the observed values in the distribution with the respective relative frequencies in relation to the general total of the row. The calculation could also be done in relation to the general total of the column, arriving at the same result of the χ² statistic.

The data in Table 4.E.5 show the dependence between the variables. Assuming that there was no association between the variables, we would expect a ratio of 48% in relation to the total of the row of all three health insurance companies in the Dissatisfied column, 36% in the Neutral column, and 16% in the Satisfied column. The calculation of the expected values can be seen in Table 4.E.6. For example, the calculation of the first cell is 0.48 × 68 = 32.64.

To calculate the χ² statistic, we must apply expression (4.1) for the data in Tables 4.E.5 and 4.E.6. The calculation of each term $\frac{{\left(O_{\mathit{ij}}-E_{\mathit{ij}}\right)}^2}{E_{\mathit{ij}}}$ is shown in Table 4.E.7, jointly with the χ² measure resulting from the sum of the categories.

As we are going to study in Chapter 9, which discusses hypotheses tests, significance level α indicates the probability of rejecting a certain hypothesis when it is true. P-value, on the other hand, represents the probability associated to the sample observed value, indicating the lowest significance level that would lead to the rejection of the supposed hypothesis. In other words, P-value represents a decreasing reliability index of a result. The lower the value, the less we can believe in the assumed hypothesis.

In the case of the χ² statistic, whose test presupposes the nonassociation between the variables being studied, most statistical software, including SPSS and Stata, calculate the corresponding P-value. Thus, for a confidence level of 95%, if P-value < 0.05, the hypothesis is rejected and we can state that there is an association between the variables. On the other hand, if P-value > 0.05, we conclude that the variables are independent. All of these concepts will be studied in more detail in Chapter 9.

Excel calculates the P-value of the χ² statistic through the CHITEST or CHISQ.TEST (Excel 2010 and future versions) functions. In order to do that, we just need to select the set of cells corresponding to the observed or real values and the set of cells of the expected values.

• Solving the chi-square statistic on the SPSS software

Analogous to Example 4.1, calculating the chi-square statistic (χ²) on SPSS is also done on the tab Analyze → Descriptive Statistics → Crosstabs…. Once again, we are going to select the variable Agency in Row(s) and the variable Satisfaction in Column(s). Initially, to generate the observed values and the expected values in case of nonassociation between the variables (data in Tables 4.E.5 and 4.E.6), we must click on Cells… and select the options Observed and Expected in Counts, from the Crosstabs: Cell Display dialog box (Fig. 4.11). In the same box, to generate the adjusted standardized residuals, we must select the option Adjusted standardized in Residuals. The results can be seen in Fig. 4.12.

To calculate the χ² statistic, in Statistics…, we must select the option Chi-square (Fig. 4.13). Finally, we are going to click on Continue and OK. The result can be seen in Fig. 4.14.

Based on Fig. 4.14, we can see that the value of χ² is 15.861, similar to the one calculated in Table 4.E.7. We can also observe that the lowest significance level that would lead to the rejection of the nonassociation hypothesis between the variables (P-value) is 0.003. Since 0.003 < 0.05 (for a confidence level of 95%), the null hypothesis is rejected, which allows us to conclude that there is association between the variables.

• Solving the χ² statistic on the Stata software

In Section 4.2.1, we learned how to create contingency tables on Stata through the command tabulate, or simply tab. Besides the observed frequencies, this command also gives us the expected frequencies through the option expected, or simply exp, as well as the calculation of the χ² statistic using the option chi2, or simply ch. For the data in Example 4.1 available in the file HealthInsurance.dta, to obtain the observed and expected frequency distribution tables, jointly with the χ² statistic, we are going to use the following command:

tab agency satisfaction, exp ch

However, the command tab does not allow residuals to be generated in the output. As an alternative, the command tabchi was developed from a tabulation module created by Nicholas J. Cox, allowing the adjusted standardized residuals to be calculated too. In order for this command to be used, we must initially type:

findit tabchi

and install it in the link tab_chi from http://fmwww.bc.edu/RePEc/bocode/t. After doing this, we can type the following command:

tabchi agency satisfaction, a

The result is shown in Fig. 4.15 and is similar to those presented in Figs. 4.12 and 4.14 on the SPSS software. Note that, differently from the command tab, which requires the option exp so that the expected frequencies can be generated, the command tabchi already gives them to us automatically.

Solution

Using the procedure described in the previous section, the value of the chi-square statistic is χ² = 18.214. Therefore:

$\mathit{Phi}=\sqrt{\frac{{\chi }^2}{n}}=\sqrt{\frac{18.214}{20}}=0.954$

Since both variables have four categories, in this case the condition 0 ≤ Phi ≤ 1 is not valid, making it difficult to interpret how strong the association is.

1. b) Contingency coefficient

The contingency coefficient (C), also known as Pearson’s contingency coefficient, is another measure of association for nominal variables based on the χ² statistic, being represented by the following expression:

$C=\sqrt{\frac{{\chi }^2}{n+{\chi }^2}}$

where n is the sample size.

The contingency coefficient (C) has as its lowest limit the value 0, indicating that there is no relationship between the variables; however, the highest limit of C varies depending on the number of categories, so:

$0\le C\le \sqrt{\frac{q-1}{q}}$

where:

$q=min\left(I,J\right)$

where I is the number of rows and J is the number of columns in a contingency table.

When $C=\sqrt{\frac{q-1}{q}}$, there is a perfect association between the variables; however, this limit never assumes the value 1. Hence, two contingency coefficients can only be compared if both are defined from tables with the same number of rows and columns.

Solution

We calculate C as follows:

$C=\sqrt{\frac{{\chi }^2}{n+{\chi }^2}}=\sqrt{\frac{18.214}{20+18.214}}=0.690$

Since the contingency table is 4 × 4 (q = min(4, 4) = 4), the values that C can assume are in the interval:

$0\le C\le \sqrt{\frac{3}{4}}\to 0\le C\le 0.866$

We can conclude that there is association between the variables.

1. c) Cramer’s V coefficient

Another measure of association for nominal variables based on the χ² statistic is Cramer's V coefficient, calculated by:

$V=\sqrt{\frac{{\chi }^2}{n.\left(q-1\right)}}$

where q = min(I, J), as presented in expression (4.5).

For 2 x 2 contingency tables, expression (4.6) is going to be $V=\sqrt{\frac{{\chi }^2}{n}}$, which corresponds to the Phi coefficient.

Cramer's V coefficient is an alternative to the Phi coefficient and to the contingency coefficient (C), and its value is always limited to the interval [0, 1], regardless of the number of categories in the rows and columns:

$0\le V\le 1$

Value 0 indicates that the variables do not have any kind of association and value 1 shows that they are perfectly associated. Therefore, Cramer's V coefficient allows us to compare contingency tables that have different dimensions.

Solution

$V=\sqrt{\frac{{\chi }^2}{n\left(q-1\right)}}=\sqrt{\frac{18.214}{20\times 3}}=0.551$

Since 0 ≤ V ≤ 1, there is association between the variables; however, it is not considered very strong.

• Solution of Examples 4.3, 4.4, and 4.5 (calculation of the Phi, contingency, and Cramer’s V coefficients) by using SPSS

In Section 4.2.1, we discussed how to create labels that correspond to the variable categories from the menu Data → Define Variable Properties…. The same procedure must be applied to the data in Table 4.E.8 (we cannot forget to define the variables as nominal). The file Market_Segmentation.sav gives us these data already tabulated on SPSS.

Similar to the calculation of the χ² statistic, calculating the Phi, contingency, and Cramer’s V coefficients on SPSS can also be done on the menu Analyze → Descriptive Statistics → Crosstabs…. We are going to select the variable Clothes in Row(s) and the variable Region in Column(s).

In Statistics…, we are going to select the options Contingency coefficient and Phi and Cramer’s V (Fig. 4.16). Note that these coefficients are calculated for nominal variables. The results of the statistics can be seen in Fig. 4.17.

For all three coefficients, the P-value of 0.033 (0.033 < 0.05) indicates that there is association between the variables being studied.

• Solution of Examples 4.3 and 4.5 (calculation of the Phi and Cramer’s V coefficients) by using Stata

Stata calculates the Phi and Cramer’s V coefficients through the command phi. Hence, they are going to be calculated for the data in Example 4.3 available in the file Market_Segmentation.dta.

In order for the phi command to be used, initially, we must type:

findit phi

and install it in the link snp3.pkg from http://www.stata.com/stb/stb3/. After doing this, we can type the following command:

phi clothes region

The results can be seen in Fig. 4.18. Note that the Phi coefficient on Stata is called Cohen’s w. Cramer's V coefficient, on the other hand, is called Cramer's phi-prime.

Solution

To calculate Spearman’s coefficient, first, we are going to assign rankings to each category of each variable depending on their respective values, as shown in Table 4.E.10.

Applying expression (4.8), we have:

$r_{\mathit{sp}}=1-\frac{6\sum _{k=1}^n{d}_k^2}{n\left(n^2-1\right)}=1-\frac{6\times 40}{10\times 99}=0.7576$

Value 0.758 indicates a strong positive association between the variables.

• Calculating Spearman’s coefficient using SPSS software

File Grades.sav shows the data from Example 4.6 (rankings in Table 4.E.9) tabulated in an ordinal scale (defined in the environment Variable View).

Similar to the calculation of the χ² statistic and the Phi, contingency, and Cramer’s V coefficients, Spearman’s coefficient can also be generated by SPSS from the menu Analyze → Descriptive Statistics → Crosstabs…. We are going to select the variable Simulation in Row(s) and the variable Finance in Column(s).

In Statistics…, we are going to select the option Correlations (Fig. 4.20). We are going to click on Continue and then, finally, on OK. The result of Spearman’s coefficient is shown in Fig. 4.21.

The P-value 0.011 < 0.05 (under the hypothesis of nonassociation between the variables) indicates that there is a correlation between the grades in Simulation and Finance, with 95% confidence.

Spearman’s coefficient can also be calculated in the menu Analyze → Correlate → Bivariate…. We must select the variables that interest us, in addition to Spearman’s coefficient, as shown in Fig. 4.22. We are going to click on OK, resulting in Fig. 4.23.

• Calculating Spearman’s coefficient by using Stata software

In Stata, Spearman's coefficient is calculated using the command spearman. Therefore, for the data in Example 4.6, available in the file Grades.dta, we must type the following command:

spearman simulation finance

The results can be seen in Fig. 4.24.

Solution

Applying expression (4.9), we have:

$cov\left(X,Y\right)=\frac{\left(7.6-7.08\right)\left(1,961-1,856.22\right)+\cdots +\left(5.4-7.08\right)\left(775-1,856.22\right)}{95}=\frac{72,326.93}{95}=761.336$

The covariance can be calculated in Excel by using the COVARIANCE.S (sample) function.

In the following section, we are also going to discuss how the covariance can be calculated on SPSS, jointly with Pearson’s correlation coefficient. SPSS considers the same expression presented in this section.

Solution

Calculating Pearson’s correlation coefficient through expression (4.10) is as follows:

$\rho =\frac{cov\left(X,Y\right)}{S_X\cdot S_Y}=\frac{761.336}{970.774\times 1.009}=0.777$

This calculation could also be done by using expression (4.11), which does not depend on the sample size. The result indicates a strong positive correlation between the variables Family Income and Years of Education.

Excel also calculates Pearson’s correlation coefficient through the PEARSON function.

• Solution of Examples 4.8 and 4.9 (calculation of the covariance and Pearson’s correlation coefficient) on SPSS

Once again, open the file Income_Education.sav. To calculate the covariance and Pearson’s correlation coefficient on SPSS, we are going to click on Analyze → Correlate → Bivariate…. The Bivariate Correlations window will open. We are going to select the variables Family Income and Years of Education, in addition to Pearson’s correlation coefficient, as shown in Fig. 4.36. In Options…, we must select the option Cross-product deviations and covariances, according to Fig. 4.37. We are going to click on Continue and then on OK. The results of the statistics are presented in Fig. 4.38.

Analogous to Spearman's coefficient, Pearson’s correlation coefficient can also be generated on SPSS from the menu Analyze → Descriptive Statistics → Crosstabs… (option Correlations in the Statistics button…).

• Solution of Examples 4.8 and 4.9(calculation of the covariance and Pearson’s correlation coefficient) on Stata

To calculate Pearson’s correlation coefficient on Stata, we must use the command correlate, or simply corr, followed by the list of variables we are interested in. The result is the correlation matrix between the respective variables.

Once again, open the file Income_Education.dta. Thus, for the data in this file, we can type the following command:

corr income education

The result can be seen in Fig. 4.39.

To calculate the covariance, we must use the option covariance, or only cov, at the end of the command correlate (or simply corr). Thus, to generate Fig. 4.40, we must type the following command:

corr income education, cov