Question

Out of 400 people sampled, 152 had children. Based on this, construct a 99% confidence interval for the true population proportion of people with children.

Is it safe to assume that n≤5% of all people with children?
A.No
B.Yes

Answer 1

To construct the 99% confidence interval, calculate the sample proportion and find the z-score, then compute the Margin of Error. The answer for assuming n≤5% of the population is Yes, which allows the use of normal approximation in constructing the interval.

Step-by-step explanation:

To construct a 99% confidence interval for the true population proportion of people with children based on 152 out of 400 people sampled having children, we use the formula for a proportion confidence interval:

± z* ∙ √(p∙(1-p)/n) where p is the sample proportion, n is the sample size, and z* is the z-score corresponding to the 99% confidence level.

First, calculate the sample proportion (p):

p = 152/400 = 0.38

Next, find the z-score for a 99% confidence level, which is approximately 2.576.

Then plug the values into the formula:

Margin of Error = 2.576 ∙ √(0.38∙(1-0.38)/400)

Calculate the Margin of Error and construct the confidence interval:

CI = 0.38 ± (Margin of Error)

Regarding the assumption of n≤5% of the population, the answer is Yes, because typically this assumption allows us to use the normal approximation method for constructing confidence intervals, and with a sample size of 400, it is reasonable to believe that this constitutes less than 5% of all people with children.