Question

Assume that a sample is used to estimate a population proportion p. Find the 95% confidence interval for a sample of size 238 with 172 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

Answer 1

Answer:

95% confidence interval for a Population proportion

0.6937 ≤ P ≤ 0.7515

Step-by-step explanation:

Explanation:-

Given sample size 'n' = 238

probability of successes  or sample proportion

                     $p = \frac{x}{n} = \frac{172}{238} =0.7226$

95% confidence interval for a Population proportion is determined by

$(p^{-} - Z_{0.05} \sqrt{\frac{p(1-p)}{n} } , p^{-} + Z_{0.05} \sqrt{\frac{p(1-p)}{n} })$

$(0.7226 - 1.96\sqrt{\frac{0.7226(1-0.7226)}{238} } , 0.7226+ 1.96\sqrt{\frac{0.7226(1-0.7226)}{238} })$

(0.7226 -  0.0289 , 0.7226 + 0.0289)

(0.6937 , 0.7515)

Conclusion:-

95% confidence interval for a Population proportion

0.6937 ≤ P ≤ 0.7515