. Assume that a sample is used to estimate a population proportion p. Find a mar

Question

. Assume that a sample is used to estimate a population proportion p. Find a margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

99% confidence; n = 2900, x= 580

Explanation / Answer

x > 10 and n - x > 10. Thus, we can use a z-distribution instead of a t-distribution. A 99% confidence interval corresponds to tails of (1 - .99)/2 = .005 on either side. Looking up this value in a z-table, we find the corresponding z-value is 2.575. Thus, we know on the upper end 2.575 = p - p0 / sqrt(p0 * (1-p0) / n) and on the lower end -2.575 = p - p0 / sqrt(p0 * (1-p0) / n). In this case: p0 = 580/2900 = .2 and n = 2900. Plugging in these numbers and solving for p, we find p_upper = .2191 and p_lower = .1809. Thus, our confidence interval is [.1809, .2191]. Notice that the interval is centered about p0, i.e. (.1809 + .2191) / 2 = .2.

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