An image analyst classifies an image into three classes, namely, water, vegetation and urban, using a simple classification technique. Using the reference data (ground truth) and classified data, the analyst prepares an error matrix, as shown in the table, and calculates accuracy values (Rounded off to the nearest integer value). Which of the following options is/are CORRECT?
Producer's and User's accuracy values in case of water class are equal
In case of urban class, Producer's accuracy is 20% and User's accuracy is 13%
In case of vegetation class, Producer's accuracy is 20% and User's accuracy is 11%
Show Solution
The Correct Option isA, B
Solution and Explanation
Step 1: Set up the accuracy-assessment definitions. Overall accuracy $=\dfrac{\text{sum of diagonal cells}}{N}$; Producer's accuracy for a class $=\dfrac{\text{diagonal cell}}{\text{column total}}$ (omission error view); User's accuracy for a class $=\dfrac{\text{diagonal cell}}{\text{row total}}$ (commission error view). Step 2: Tabulate all totals. Row totals: 70 (Water), 40 (Vegetation), 110 (Urban). Column totals: 70 (Water), 40 (Vegetation), 110 (Urban). $N=220$. Diagonal values: 48, 16, 84. Step 3: Overall accuracy (option A). \[ OA=\dfrac{48+16+84}{220}\times100=\dfrac{148}{220}\times100=67.27\%\rightarrow67\% \] matches option (A). Step 4: Water class Producer's vs User's accuracy (option B). \[ PA_{water}=\dfrac{48}{70}\times100\approx69\%,\quad UA_{water}=\dfrac{48}{70}\times100\approx69\% \] identical because the water row-sum equals the water column-sum (both 70), confirming (B). Step 5: Urban and vegetation classes (options C and D). \[ PA_{urban}=\dfrac{84}{110}\times100\approx76\%,\ UA_{urban}\approx76\% \] not 20%/13%. \[ PA_{veg}=\dfrac{16}{40}\times100=40\%,\ UA_{veg}=40\% \] not 20%/11%. Both (C) and (D) are wrong. Step 6: Final answer. Only overall accuracy = 67% and equal Producer's/User's accuracy for water hold true.\[ \boxed{\text{Options (A) and (B)}} \]
