From: Measuring agreement between healthcare survey instruments using mutual information
Procedure | Scenario 1 | Scenario 2 | Scenario 3 | |||
---|---|---|---|---|---|---|
Data (\( {x}_{11} \), \( {x}_{12} \), \( {x}_{21} \), \( {x}_{22} \)) | (10, 5, 5, 20) | (5, 10, 20, 5) | (5, 10, 5, 20) | |||
Step 1 Contingency table | 10 5 | 5 20 | 5 20 | 10 5 | 5 5 | 10 20 |
Step 2 | \( {I_{agreement}}_{+\left(\frac{20}{40}\right){ \log}_2\frac{\left(\frac{20}{40}\right)}{\left(\frac{25}{40}\right)\left(\frac{25}{40}\right)}}^{=\left(\frac{10}{40}\right){ \log}_2\frac{\left(\frac{10}{40}\right)}{\frac{(15)}{40}\left(\frac{15}{40}\right)}} \) | \( {I_{agreement}}_{+\left(\frac{5}{40}\right){ \log}_2\frac{\left(\frac{5}{40}\right)}{\left(\frac{15}{40}\right)\left(\frac{25}{40}\right)}}^{=\left(\frac{5}{40}\right){ \log}_2\frac{\left(\frac{5}{40}\right)}{\frac{(15)}{40}\left(\frac{25}{40}\right)}} \) | \( {I_{agreement}}_{+\left(\frac{20}{40}\right){ \log}_2\frac{\left(\frac{20}{40}\right)}{\left(\frac{30}{40}\right)\left(\frac{25}{40}\right)}}^{=\left(\frac{5}{40}\right){ \log}_2\frac{\left(\frac{5}{40}\right)}{\frac{(15)}{40}\left(\frac{10}{40}\right)}} \) | |||
Mutual information | ||||||
\( {I_{disagreement}}_{+\left(\frac{5}{40}\right){ \log}_2\frac{\left(\frac{5}{40}\right)}{\left(\frac{15}{40}\right)\left(\frac{25}{40}\right)}}^{=\left(\frac{5}{40}\right){ \log}_2\frac{\left(\frac{5}{40}\right)}{\frac{(15)}{40}\left(\frac{25}{40}\right)}} \) | \( {I_{disagreement}}_{+\left(\frac{20}{40}\right){ \log}_2\frac{\left(\frac{20}{40}\right)}{\left(\frac{25}{40}\right)\left(\frac{25}{40}\right)}}^{=\left(\frac{10}{40}\right){ \log}_2\frac{\left(\frac{10}{40}\right)}{\frac{(15)}{40}\left(\frac{15}{40}\right)}} \) | \( {I_{disagreement}}_{+\left(\frac{5}{40}\right){ \log}_2\frac{\left(\frac{5}{40}\right)}{\left(\frac{10}{40}\right)\left(\frac{25}{40}\right)}}^{=\left(\frac{10}{40}\right){ \log}_2\frac{\left(\frac{10}{40}\right)}{\frac{(15)}{40}\left(\frac{30}{40}\right)}} \) | ||||
Step 3 Significance | Mutual information = 0.386-0.227 = 0.159 From (Eq. 6), 2*40*ln2*0.159 = 8.809 Highly significant (with p < 0.001) | Mutual information = -0.227 + 0.386 = 0.159 From (Eq. 6), 2*40*ln2*0.159 = 8.809 Highly significant (with p < 0.001) | Mutual information = 0.098-0.083 = 0.015 From (Eq. 6), 2*40*ln2*0.015 = 0.871 Less significant (with p = 0.35) | |||
Step 4 Local mutual information | I agreement > I disagreement and highly significant mutual information. Thus, agreement | I agreement < I disagreement and highly significant mutual information. Thus, disagreement | I agreement > I disagreement but very low mutual information observed. Thus, inconclusive |