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If a 90% confidence interval for the difference between two sample proportions is (0.043, 0.425), what...

If a 90% confidence interval for the difference between two sample proportions is

(0.043, 0.425), what is the 99% confidence interval?

Math   Statistics-And-Probability   
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Answer #1

The Upper limit = (p1 - p2) + ME = 0.425 -------------- (1)

The Lower limit = (p1 - p2) - ME = 0.043 -------------- (2)

Adding the 2 equations, we get 2 * (p1 - p2) = 0.425 + 0.043 = 0.468

Therefore p1 - p2 = 0.234

Putting p1 - p2 = 0.234 in equation 1, we get ME = 0.425 - 0.234 = 0.191

ME = z critical * SE

For = 0.10, z critical = 1.645

Therefore SE = ME / 1.645 = 0.191 / 1.645 = 0.1161

____________________________

For a 99% CI, the z critical = 2.576

Therefore ME = 2.576 * 0.1161 = 0.299

The Upper limit = (p1 - p2) + ME = 0.234 + 0.299 = 0.533

The Lower limit = (p1 - p2) - ME = 0.234 - 0.299 = -0.065

Therefore the 99% CI is (-0.065, 0.533)

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