Question

Out of 100 people sampled, 94 had kids. Based on this, construct a 95% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places.

Answer 1

To construct a 95% confidence interval for the proportion of people with kids, use the formula p +/- Z*sqrt((p(1-p))/n), where p is the sample proportion, Z is the Z-score for a 95% confidence level, and n is the sample size.

Step-by-step explanation:

To construct a 95% confidence interval for the true population proportion of people with kids, we need to use the formula:

p ± Z*sqrt((p(1-p))/n)

Where:

• p is the sample proportion (94/100 = 0.94)
• Z is the Z-score for a 95% confidence level (1.96)
• n is the sample size (100)

Plugging in the values:

0.94 ± 1.96*sqrt((0.94(1-0.94))/100)

Simplifying the expression:

0.94 ± 1.96*sqrt(0.0006)

Calculating the square root of 0.0006:

0.94 ± 1.96*0.0245

Calculating the product:

0.94 ± 0.0479

Therefore, the 95% confidence interval for the proportion of people with kids is approximately 0.892 to 0.988.