Assume that a sample is used to estimate a population proportion p. Find the $80 \%$ confidence interval for a sample of size 229 with $19 \%$ successes. a. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. Confidence interval $=$ b. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places. c. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places. \[ p= \] \[ \pm \]

Step 1 ：Given that the sample size is 229 and the sample proportion of successes is 0.19, we want to find the 80% confidence interval for the population proportion.

Step 2 ：The formula for the confidence interval for a proportion is given by \(\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\), where \(\hat{p}\) is the sample proportion, n is the sample size, and Z is the Z-score corresponding to the desired level of confidence.

Step 3 ：In this case, \(\hat{p} = 0.19\), n = 229, and the Z-score for an 80% confidence level is approximately 1.28.

Step 4 ：Substituting these values into the formula, we get \(0.19 \pm 1.28 \sqrt{\frac{0.19(1-0.19)}{229}}\).

Step 5 ：Solving this expression, we find that the 80% confidence interval for the population proportion is approximately (0.157, 0.223).

Step 6 ：\(\boxed{\text{Final Answer: The 80% confidence interval for a sample of size 229 with 19% successes is approximately (0.157, 0.223).}}\)