Question

Assume that a sample is used to estimate a population proportion p. Find the 99% confidence ihterval for a

sample of size 188 with 10.1% successes. Enter your answer as a tri-linear inequality using decimals (not
percents) accurate to three decimal places.

Answer 1

The 99% confidence interval for a population proportion with a sample size of 188 and 10.1% successes is (0.044, 0.158) when expressed as a trilinear inequality and rounded to three decimal places.

Step-by-step explanation:

To calculate the 99% confidence interval for the population proportion based on a sample size of 188 with 10.1% successes, we first need to convert the percentage of successes into a decimal. Thus, we have 10.1% successes or 0.101 as our sample proportion (p'). The formula to calculate the confidence interval for a population proportion is as follows:

p' ± Z*sqrt((p'(1-p')/n)),

where Z is the Z-value corresponding to the chosen confidence level (in this case, for 99% confidence interval we use Z = 2.576), p' is the sample proportion, and n is the sample size.

Given:
p' = 0.101
n = 188

We calculate q' as 1 - p' = 1 - 0.101 = 0.899.

Now we compute the standard error (SE):
SE = sqrt((0.101 * 0.899) / 188) = sqrt(0.090799 / 188) = sqrt(0.000483) = 0.02198.

The margin of error (EBP) at the 99% confidence level is:
EBP = Z * SE = 2.576 * 0.02198 = 0.05663.

Finally, the 99% confidence interval is calculated as:
(p' - EBP, p' + EBP) = (0.101 - 0.05663, 0.101 + 0.05663) = (0.04437, 0.15763).

Expressed as a trilinear inequality, we have:
0.044 ≤ p ≤ 0.158 (to three decimal places).