Confidence interval for a pass rate

Question

Of 225 candidates who took a certification exam, 180 passed. Construct a 95% confidence interval for the true pass rate.

Accepted Answer

Setup. Use \hat{p} \pm z^{}\sqrt{\hat{p}(1-\hat{p})/n} with z^{} = 1.96. Here \hat{p} = 180/225 = 0.8. Compute. The standard error is \sqrt{\frac{0.8 	imes 0.2}{225}} = \sqrt{0.000711} = 0.0267. Margin of error = 1.96 	imes 0.0267 = 0.0523, so 0.8 \pm 0.0523 = [0.748,\ 0.852]. Interpret. With 95% confidence the true pass rate is about 75% to 85%. Even though \hat{p} = 0.8 leans high, the large sample keeps the interval sensible and safely inside [0,1]. Takeaway. The standard error of a proportion is largest near \hat{p} = 0.5 and shrinks as the rate moves toward 0 or 1, so a rate of 0.8 giv…

Confidence interval for a pass rate

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