Question

Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 223 with 148 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

Answer 1

To find the 99.9% confidence interval for a sample of size 223 with 148 successes, we can use the formula CI = p' ± Zα/2 √(p'(1-p')/n), where p' is the sample proportion, Zα/2 is the critical value, and n is the sample size. Plugging in the values, we find the confidence interval is approximately (0.595, 0.731).

Step-by-step explanation:

To find the confidence interval for a population proportion using a sample, we can use the formula:

CI = p' ± Zα/2 √(p'(1-p')/n)

where p' is the sample proportion, Zα/2 is the critical value for the desired confidence level, and n is the sample size.

For a 99.9% confidence level, we can find the critical value by subtracting 0.9999 from 1 and dividing the result by 2, giving us 0.00005. Looking up this value in the Z-table, we find the corresponding critical value is approximately 3.290.

Using the values from the question, p' = 148/223 ≈ 0.663, Zα/2 = 3.290, and n = 223, we can plug these values into the formula to calculate the confidence interval:

CI = 0.663 ± (3.290) √(0.663(1-0.663)/223)

Calculating this expression, we find that the confidence interval is approximately (0.595, 0.731).