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)