Question

A study was conducted to compare two diagnostic methods say, Method-I and MethodII. The following table shows the classification of the presence of infection on 5-point Likert Scale (none, mild, moderate, severe and extreme): Method-II Method-I None Mild Moderate Severe Extreme None 89 5 16 2 4 Mild 36 3 15 6 2 Moderate 20 4 22 6 1 Severe 14 2 37 18 16 Extreme 4 1 16 23 50 Compute simple and weighted Kappa Statistic(s) to compare both methods. (10) 3.(a) Explain the Kaplan and Meier method of estimation with exampl

13 May 2026
Answer :
Word Count : 648
For simple Kappa Statistic, first construct the contingency table: | Method-I \ Method-II | None | Mild | Moderate | Severe | Extreme | Row Total | | -------------------- | ---- | ---- | -------- | ------ | ------- | --------- | | None | 89 | 5 | 16 | 2 | 4 | 116 | | Mild | 36 | 3 | 15 | 6 | 2 | 62 | | Moderate | 20 | 4 | 22 | 6 | 1 | 53 | | Severe | 14 | 2 | 37 | 18 | 16 | 87 | | Extreme | 4 | 1 | 16 | 23 | 50 | 94 | Column totals are: * None = 163 * Mild = 15 * Moderate = 106 * Severe = 55 * Extreme = 73 Grand total, [ N = 412 ] Observed proportion of agreement: Diagonal frequencies are: [ 89+3+22+18+50 = 182 ] Hence, [ P_o = \frac{182}{412} = 0.4417 ] Expected agreement: [ P_e = \frac{1}{N^2}\sum (Row\ Total \times Column\ Total) ] [ = \frac{1}{412^2} [(116\times163)+(62\times15)+(53\times106)+(87\times55)+(94\times73)] ] [ = \frac{18908+930+5618+4785+6862}{169744} ] [ = \frac{37103}{169744} ] [ P_e = 0.2185 ] Simple Kappa statistic is: \kappa = \frac{P_o-P_e}{1-P_e} [ \kappa = \frac{0.4417-0.2185}{1-0.2185} ] [ \kappa = \frac{0.2232}{0.7815} ] [ \kappa = 0.286 ] Thus, the simple Kappa statistic is: [ \boxed{\kappa = 0.286} ] This indicates fair
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