Question

# Assume that a sample is used to estimate a population proportion p. Find the 99% confidence...

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.

p = 71 / 222 = 0.32

Z for 99% confidence interval = Z0.005 = 2.575

Confidence interval = (p + Z0.005 * 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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