Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 400 adult residents sampled, 296 had kids. Based on this, construct a 95% confidence interval for the proportion p of adult residents who are parents in this county. Express your answer in tri-inequality form. Give your answers as decimals, to three places.

__< p <__ Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

p =__ ±±____

Answer 1

To construct a 95% confidence interval for the proportion of adult residents who are parents in a certain county, we can use the formula p ± Z * √(p(1-p)/n), where p is the sample proportion, Z is the Z-score, and n is the sample size. Given that 296 out of 400 adult residents sampled had kids, the point estimate for p is 0.74. Using a 95% confidence level, the Z-score is approximately 1.96. The confidence interval is 0.708 < p < 0.772.

Step-by-step explanation:

To construct a confidence interval for the proportion p of adult residents who are parents in this county, we can use the formula:

p ± Z * √(p(1-p)/n)

Where p is the sample proportion, Z is the Z-score corresponding to the desired confidence level, and n is the sample size.

Given that out of 400 adult residents sampled, 296 had kids, the point estimate for p is 296/400 = 0.74.

Using a 95% confidence level, the Z-score is approximately 1.96.

Plugging in the values into the formula, we get:

0.74 ± 1.96 * √(0.74(1-0.74)/400)

Simplifying the expression, we find the 95% confidence interval to be:

0.74 - 0.032 < p < 0.74 + 0.032

Expressed as decimals, the confidence interval is:

0.708 < p < 0.772