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Assume that a sample is used to estimate a population proportion p. Find the 95% confidence interval for a sample of size 200 with 180 successes. Enter your answer as a tri-linear inequality using decimals (not percentages) accurate to three decimal places.

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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.

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