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A ​90% confidence interval for p is given as ​(0.17​,0.43​). How large was the sample used to construct this​ interval?

n=?

User CoolLife
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Final answer:

To calculate the sample size used to construct a 90% confidence interval for a population proportion, the formula n = (z^2 * p * (1-p))/E^2 is used, where z is the z-score corresponding to the confidence level, p is the estimated proportion, and E is the margin of error. Given a confidence interval of (0.17, 0.43), the estimated proportion is 0.30. Assuming a margin of error of 0.05, the sample size used to construct the interval is approximately 64.

Step-by-step explanation:

To calculate the sample size used to construct a confidence interval for a population proportion, we can use the formula:

n = (z^2 * p * (1-p))/E^2

Where:

  • n is the sample size
  • z is the z-score corresponding to the desired confidence level
  • p is the estimated proportion of the population
  • E is the margin of error

In this case, since the confidence level is 90% and the confidence interval is (0.17, 0.43), we can average the lower and upper bound of the interval to get the estimated proportion, which is (0.17 + 0.43)/2 = 0.30.

Let's assume the margin of error is 0.05. We can find the z-score associated with a 90% confidence level and use the formula to calculate the sample size:

n = (1.645^2 * 0.30 * (1-0.30))/(0.05)^2 ≈ 64

Therefore, the sample size used to construct this confidence interval was approximately 64.

User Steve Samuels
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