Final answer:
A 99% confidence interval for the proportion of adult residents who are parents is calculated using a sample proportion of 0.24. The margin of error at this confidence level is 0.058, resulting in an interval of 0.182 to 0.298, with the point estimate being 0.24 ± 0.058.
Step-by-step explanation:
To estimate the proportion (p) of adult residents in a certain county who are parents, we use the sample proportion as a point estimate. In the given sample of 400 adult residents, 96 are parents, resulting in a sample proportion (p') of 0.24 (96/400). To construct a 99% confidence interval for the true proportion, we use the following formula which is based on the normal approximation to the binomial distribution:
Confidence Interval = p' ± Z* ∙ sqrt((p' × (1 - p'))/n)
Where ‘Z*’ is the z-score corresponding to the desired level of confidence and ‘n’ is the sample size. For a 99% confidence level, the z-score is approximately 2.576. Calculating the margin of error (ME) we have:
ME = 2.576 ∙ sqrt((0.24 × 0.76)/400)
ME = 2.576 ∙ sqrt(0.00144)
ME = 0.058
The 99% confidence interval in tri-inequality form is then:
0.182 < p < 0.298
Expressing the same answer using the point estimate and margin of error:
p = 0.24 ± 0.058