Question

In a sample of 440 adults, 396 had children. Construct a 95% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three places

____ < p < ____

Answer 1

To construct a 95% confidence interval for the true population proportion of adults with children, we can use the following formula:

p ± z*(sqrt(p*(1-p)/n))

where:
- p is the sample proportion (396/440 = 0.9)
- z is the z-score corresponding to the desired confidence level (95% confidence level corresponds to a z-score of 1.96)
- n is the sample size (440)

Plugging in the values, we get:

0.9 ± 1.96*(sqrt(0.9*(1-0.9)/440))

Simplifying this expression, we get:

0.868 < p < 0.932

Therefore, we can conclude that we are 95% confident that the true population proportion of adults with children is between 0.868 and 0.932.