Question

In a study with randomly selected participants, it was found that "45.3 % of adults report that they live with one or more chronic conditions". The study also reported a margin of error 3.5 %. Create a 95% confidence interval for the proportion of all U.S. adults living with chronic conditions. don't round.

Answer 1

To create a 95% confidence interval for the proportion of all U.S. adults living with chronic conditions, we can use the following formula:

CI = p ± zsqrt(pq/n)

where:

p is the sample proportion (0.453)

q is the complement of p (1 - p = 0.547)

z is the z-score associated with a 95% confidence level (1.96)

n is the sample size (unknown)

We need to find the sample size (n) in order to calculate the confidence interval. We can do this by using the margin of error formula:

ME = zsqrt(pq/n)

where ME is the margin of error (0.035)

Solving for n, we get:

n = (z^2 * p * q) / ME^2 = (1.96^2 * 0.453 * 0.547) / 0.035^2 = 580.04

Rounding up to the nearest whole number, the sample size is 581.

Now we can substitute the values into the confidence interval formula:

CI = 0.453 ± 1.96sqrt(0.4530.547/581)

CI = 0.453 ± 0.035

The 95% confidence interval for the proportion of all U.S. adults living with chronic conditions is:

CI = (0.418, 0.488)

So we can say with 95% confidence that the true proportion of all U.S. adults living with chronic conditions is between 0.418 and 0.488.