Question

Assume that a sample is used to estimate a population proportion p. Find the 95% confidence interval for a sample of size 169 with 55 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

Answer 1

To find the 95% confidence interval for a sample of size 169 with 55 successes, we can use the following formula:

Confidence Interval = p-hat ± (Z * sqrt((p-hat*(1-p-hat))/n))

where p-hat is the sample proportion (successes/sample size), Z is the Z-score for a 95% confidence interval (1.96), and n is the sample size.

First, calculate p-hat:
p-hat = 55/169 ≈ 0.325

Next, calculate the margin of error:
Margin of Error = 1.96 * sqrt((0.325*(1-0.325))/169) ≈ 0.075

Finally, find the 95% confidence interval:
Lower Bound = 0.325 - 0.075 ≈ 0.250
Upper Bound = 0.325 + 0.075 ≈ 0.400

Thus, the 95% confidence interval is 0.250 ≤ p ≤ 0.400, expressed as a trilinear inequality with decimals accurate to three decimal places.