Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 368 with 75 successes

Question

Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 368 with 75 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

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Answer 1

80% confidence interval for a sample proportion expressed as a tri-linear inequality.

x=75

n=368

phat = 75/368 = 0.2038

Margin of error = z(α/2)*sqrt(phat*(1-phat)/n) = z(0.10)*sqrt(0.2038*(1-0.2038)/368) = 1.2816(0.0210) = 0.027

Lower bound = 0.2038 - 0.027 = 0.1768

Upper bound = 0.2038 + 0.027 = 0.2308

( 0.177 < p < 0.231)

Let me know if you have any questions.

Dian