Please advise on solving

Question

Assume that a sample is used to estimate a population proportion p. Find the 95% confidence interval for a sample of size 176 with 16% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

C.I=

Answer 1

The formula for the confidence interval for the population proportion p is p̂ ± z_α/2 √((p̂(1-p̂))/n).

Your sample size n is 176 and your sample proportion p̂ is 0.16.

Since the confidence level is 95%, the significance level α is 5% or 0.05. When drawing a bell shape the confidence interval would be in the center of the bell and they are going to be two tails. Each tail will have an area of 0.025 (0.05÷2). Then with software, you can find which is the critical value z_α/2. Manually you can use a left tail z table and look at 0.0250 in the probabilities and identify its z score. It will be -1.96. You take it as it is positive because in the formula we have ±. Then, the critical values are -1.96 and 1.96.

Applying the formula:

p̂ ± z_α/2 √((p̂(1-p̂))/n)

0.16 ± 1.96 √((0.16(1-0.16))/176)

0.16 ± 1.96 (0.028)

0.16 ± 0.054

(0.106,0.214)

The confidence interval for the population proportion p is (0.106,0.214) which means that we are 95% confident that the population proportion p is between 10.6% and 21.4%. If you are asked for the standard error, well it is √((p̂(1-p̂))/n) = √((0.16(1-0.16))/176) = 0.028. If you are asked for the margin of error, it is z_α/2 √((p̂(1-p̂))/n) = 1.96 √((0.16(1-0.16))/176) = 1.96 (0.028) = 0.054.