Question

Assume that a sample is used to estimate a population proportion p. Find the 95% confidence interval for a sample of size 243 with 27.2% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

Answer 1

Solution :

Given that,

n = 243

= 27.2% = 0.272

1 - = 1 - 0.272 = 0.728

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 * (((0.272 * 0.728) / 243)

= 0.056

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.272 - 0.056 < p < 0.272 + 0.056

0.216 < p < 0.328

The 95% confidence interval for the population proportion p is : (0.216 , 0.328)