Question

Assume that a sample is used to estimate a population proportion
*p*. Find the 99% confidence interval for a sample of size
222 with 71 successes. Enter your answer as a tri-linear inequality
using decimals (not percents) accurate to three decimal places.

____< *p* <____

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Answer #1

p = 71 / 222 = 0.32

Z for 99% confidence interval = Z_{0.005} = 2.575

Confidence interval = (p + Z_{0.005} * sqrt(p *
(1 - p) / n))

= (0.32 + 2.575 * sqrt(0.32 * 0.68 / 222))

= (0.32 + 0.081)

= (0.239 < p^ < 0.401)

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