Final answer:
To find the 95% confidence interval for a population proportion with a sample of size 200 and 180 successes, we calculate the sample proportion, the standard error, and the margin of error, resulting in a confidence interval from 0.858 to 0.942.
Step-by-step explanation:
The subject of this question is Mathematics, specifically the concept of creating a confidence interval for a population proportion in statistics. Given a sample size of 200 with 180 successes, we first calculate the sample proportion (p') as follows:
p' = x/n = 180/200 = 0.900
Now, calculate q' which is the complement of p':
q' = 1 - p' = 1 - 0.900 = 0.100
Since we are looking for a 95% confidence interval, we use the standard normal distribution (Z-distribution) as the sample size is large and the conditions np' > 5 and nq' > 5 are met. We find the Z value for a 95% confidence interval which is approximately 1.96.
We then calculate the standard error (SE):
SE = sqrt((p' * q') / n) = sqrt((0.900 * 0.100) / 200) = sqrt(0.00045) ≈ 0.0212
The margin of error (ME) is:
ME = Z * SE = 1.96 * 0.0212 ≈ 0.0416
Finally, the 95% confidence interval is given by:
p' - ME < p < p' + ME
0.900 - 0.0416 < p < 0.900 + 0.0416
0.858 < p < 0.942
So, the 95% confidence interval for the population proportion p is 0.858 to 0.942, accurate to three decimal places.