Confidence interval for a difference in conversion rates

Question

An A/B test runs both variants on 1000 visitors each. Variant A converts 120 visitors; variant B converts 160. Construct a 95% confidence interval for the difference in true conversion rates (B minus A), and say whether it indicates a real difference.

Accepted Answer

Setup. For two independent proportions, the interval for pB - pA is (\hat{p}B - \hat{p}A) \pm z^{}\sqrt{\frac{\hat{p}A(1-\hat{p}A)}{nA} + \frac{\hat{p}B(1-\hat{p}B)}{nB}}. Here \hat{p}A = 0.12, \hat{p}B = 0.16, difference = 0.04, and z^{} = 1.96. Compute the standard error. \sqrt{\frac{0.12 	imes 0.88}{1000} + \frac{0.16 	imes 0.84}{1000}} = \sqrt{0.0001056 + 0.0001344} = \sqrt{0.00024} = 0.0155. Margin of error = 1.96 	imes 0.0155 = 0.0304, so 0.04 \pm 0.0304 = [0.0096,\ 0.0704]. Interpret. The whole interval sits above zero, so at the 5% level variant B's conversion rate is significantly…

Confidence interval for a difference in conversion rates

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