90% interval for a difference in email open rates

Question

Email campaign A is opened by 350 of 1000 recipients; campaign B by 390 of 1000. Construct a 90% confidence interval for the difference in open rates (B minus A), and state your conclusion.

Accepted Answer

Setup. Use (\hat{p}B - \hat{p}A) \pm z^{}\sqrt{\hat{p}A(1-\hat{p}A)/nA + \hat{p}B(1-\hat{p}B)/nB}. For 90% confidence, z^{} = 1.645. Here \hat{p}A = 0.35, \hat{p}B = 0.39, difference = 0.04. Compute the standard error. \sqrt{\frac{0.35 	imes 0.65}{1000} + \frac{0.39 	imes 0.61}{1000}} = \sqrt{0.0002275 + 0.0002379} = \sqrt{0.0004654} = 0.0216. Margin of error = 1.645 	imes 0.0216 = 0.0355, so 0.04 \pm 0.0355 = [0.0045,\ 0.0755]. Interpret. The interval stays above zero, so at the 10% two-sided level campaign B opens significantly more often than A. Note that a 95% interval would be wider a…

90% interval for a difference in email open rates

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