Question

If a 90% confidence interval for the difference between two sample proportions is (0.345, 0.789), what is the 99% confidence interval?

Answer 1

The Upper limit = (p1 - p2) + ME = 0.789 -------------- (1)

The Lower limit = (p1 - p2) - ME = 0.345 -------------- (2)

Adding the 2 equations, we get 2 * (p1 - p2) = 0.789 + 0.345 = 1.134

Therefore p1 - p2 = 0.567

Putting p1 - p2 = 0.567 in equation 1, we get ME = 0.789 - 0.567= 0.222

ME = z critical * SE

For α = 0.10, z critical = 1.645

Therefore SE = ME / 1.645 = 0.222 / 1.645 = 0.1350

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For a 99% CI, the z critical = 2.576

Therefore ME = 2.576 * 0.1350 = 0.3478

The Upper limit = (p1 - p2) + ME = 0.567 + 0.3478 = 0.9148

The Lower limit = (p1 - p2) - ME = 0.567 - 0.3478 = 0.2192

Therefore the 99% CI is ( 0.2192 , 0.9148)

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