99% interval for a defect rate

Question

An inspection of 900 items finds 36 defective. Construct a 99% confidence interval for the true defect rate.

Accepted Answer

Setup. Use \hat{p} \pm z^{}\sqrt{\hat{p}(1-\hat{p})/n}. For 99% confidence, z^{} = 2.576. Here \hat{p} = 36/900 = 0.04. Compute. The standard error is \sqrt{\frac{0.04 	imes 0.96}{900}} = \sqrt{0.0000427} = 0.00653. Margin of error = 2.576 	imes 0.00653 = 0.0168, so 0.04 \pm 0.0168 = [0.023,\ 0.057]. Interpret. With 99% confidence the true defect rate is roughly 2% to 6%. There are 36 defective items, above the count of about 10 needed for the normal approximation to hold, so the interval is trustworthy despite the low rate. Takeaway. For rare events, reliability depends on the count of eve…

99% interval for a defect rate

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