Reading a District Earnings Survey
A district planning office surveys 9 informal workers in a town to estimate weekly earnings. The sample values, in rupees per week, are: 1,200; 1,500; 1,500; 1,800; 2,000; 2,500; 3,000; 4,500; 9,000. The survey covers only workers registered at a local labour chowk and excludes migrant workers living at construction sites. The officer wants to use the 'average' to argue for a minimum weekly wage of ₹2,200. A workers' group says most workers earn below ₹2,000; an employers' group says the average is ₹3,000 so wages are adequate. The officer also has data showing that workers with more years of schooling tend to have higher earnings, but the sample is small. The office must present a measure of central tendency, a measure of dispersion and a caution about sampling.
1. Compute the mean, median and mode. Which measure best supports the workers' group and which supports the employers' group?
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Sum of earnings = 1,200 + 1,500 + 1,500 + 1,800 + 2,000 + 2,500 + 3,000 + 4,500 + 9,000 = ₹27,000. Mean = 27,000/9 = ₹3,000. With the values ordered, the median is the 5th value = ₹2,000. The mode is ₹1,500 because it appears twice. The employers' group can cite the mean of ₹3,000. The workers' group can cite the median and mode: the middle worker earns ₹2,000, and the most common earning is ₹1,500. The mean is pulled upward by the single ₹9,000 observation, so it is not typical. For a skewed earnings distribution, the median is a better representative measure. Five of the nine workers earn below ₹2,200, so the proposed minimum has support from the median but not from the mean.
2. Calculate the range and the coefficient of variation. What do they imply about using the mean for wage policy?
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Range = highest − lowest = 9,000 − 1,200 = ₹7,800. Deviations from the mean of ₹3,000 are: −1,800; −1,500; −1,500; −1,200; −1,000; −500; 0; 1,500; 6,000. Squared deviations sum to 48,680,000. Population variance = 48,680,000/9 ≈ 5,408,889; standard deviation ≈ ₹2,326. Coefficient of variation = 2,326/3,000 ≈ 0.775, or 77.5%. The dispersion is large relative to the mean. A mean of ₹3,000 therefore hides wide variation, including very low and very high earners. For a minimum-wage argument, the median and the share below the threshold are more informative than the mean. The high CV also warns that a small sample with an outlier can make the mean unstable.
3. Identify the sampling bias and explain how it affects the estimate. How should the office improve the survey?
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The sampling frame is only workers registered at the labour chowk. It excludes migrant workers at construction sites, home-based workers, street vendors and seasonal labourers. Those excluded may have systematically lower or more variable earnings, so the sample is not representative of all informal workers. This is undercoverage or selection bias, not random sampling error. The mean and median may therefore be biased. The office should map all work sites, use stratified random sampling by sector and location, include migrant and home-based workers, increase the sample size, and record hours worked and days employed. It should report the median, the share below ₹2,200 and a confidence interval, and clearly state the limitations before using the data for wage policy.