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Here is the problem: A sample of 400 has a sample proportion of 0.6. Find the 90 percent confidence interval for the proportion.

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Build a 90 percent confidence interval for the population proportion p.

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Sample proportion.

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Standard error.

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Critical value for 90 percent confidence. The middle 90 percent leaves 5 percent in each tail. Large-sample condition met: np̂ approximately 240 and n(1 minus p̂) approximately 160 are both at least 10.

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Margin of error.

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Interval: estimate plus or minus margin.

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So the answer is 90 percent CI: (0.559709, 0.640291); margin of error 0.0402905.

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We are 90 percent confident that the population proportion p lies between 0.559709 and 0.640291. The confidence describes the method: about 90 percent of intervals built this way from repeated samples would capture the true value.

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The interval is centred on the estimate: (0.559709 plus 0.640291) / 2 equals 0.6.

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Large-sample condition met: np̂ approximately 240 and n(1 minus p̂) approximately 160 are both at least 10.