Delinquency Sampling and Provisioning at a Myanmar Microfinance Lender
Microfinance lender Dawn Microfinance has 12,000 active loans. A random pilot sample of 400 loans found 48 delinquent (more than 30 days overdue). The board's policy triggers a provisioning review if the true delinquency rate is above 10%. Average outstanding balance is MMK 1.8 million per loan, and loss given default is estimated at 45%. Management wants a 95% confidence interval for the delinquency rate and a sample size that gives a margin of error of ±2.5 percentage points at the same confidence level. Use z = 1.96. The operations director says that because 12% is above 10%, the board trigger has been met and provisioning should be increased immediately. The finance director is not convinced and asks for a statistical test and the expected loss at the observed rate.
1. Calculate and interpret the 95% confidence interval for the delinquency rate.
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p = 48/400 = 0.12. Standard error = sqrt(p(1-p)/n) = sqrt(0.12 x 0.88 / 400) = sqrt(0.000264) = 0.01625. 95% margin = 1.96 x 0.01625 = 0.03185, or 3.19 percentage points. CI = 0.12 ± 0.03185 = 0.08815 to 0.15185, i.e. about 8.82% to 15.18%. Interpretation: the sample is consistent with a true delinquency rate anywhere in that range. The 10% board trigger lies inside the interval, so the observed 12% alone does not establish that the true rate exceeds 10%.
2. Test at the 5% significance level whether the true delinquency rate exceeds 10%, and state the conclusion for the board trigger.
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Test H0: p ≤ 0.10 against H1: p > 0.10. Use the hypothesised p = 0.10 in the standard error: sqrt(0.10 x 0.90 / 400) = 0.015. z = (0.12 - 0.10) / 0.015 = 1.333. One-sided 5% critical value = 1.645; p-value ≈ 0.091. Since 1.333 < 1.645 and 0.091 > 0.05, do not reject H0. Conclusion: at the 5% level there is not sufficient evidence that the true delinquency rate exceeds 10%. The operations director's immediate trigger is not statistically supported. A cautious board could still monitor the upper confidence limit or use a lower significance level, but that is a policy choice, not proof.
3. Calculate the sample size for a margin of error of ±2.5 percentage points and the expected loss at the observed delinquency rate.
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Use n0 = z^2 p(1-p) / E^2 = 1.96^2 x 0.12 x 0.88 / 0.025^2 = 3.8416 x 0.1056 / 0.000625 = 649.1, so 650 before the finite-population correction. With N = 12,000: n = 650 / (1 + (650 - 1)/12,000) = 650 / 1.0541 = 616.6, so round up to 617 loans. Expected loss at the observed rate: 12,000 x 0.12 = 1,440 delinquent loans; exposure = 1,440 x MMK 1.8m = MMK 2,592m; expected loss = MMK 2,592m x 45% = MMK 1,166.4m. This is an estimate at the observed rate; the confidence interval shows the provisioning range could be materially different.