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What is the margin of error for a 90% confidence interval to estimate the population proportion with a sample proportion equal to 0.36 and (n=125 )?

(A) 0.06
(B) 0.07
(C) 0.08
(D) 0.09

1 Answer

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Final answer:

The margin of error for a 90% confidence interval to estimate the population proportion with a sample proportion of 0.36 and sample size of 125 is approximately 0.07. This is calculated by using the standard formula for margin of error in a confidence interval for a population proportion. The closest answer choice to the calculated margin of error is option (B) 0.07.

Step-by-step explanation:

The student has asked to calculate the margin of error for a 90% confidence interval to estimate the population proportion when given a sample proportion (p') of 0.36 and a sample size (n) of 125. The formula for the margin of error in this case is:

EBP = Z * sqrt((p' * (1 - p')) / n)

Where:

  • Z is the Z-score corresponding to the desired confidence level, which for a 90% confidence interval is about 1.645.
  • p' is the sample proportion, which is given as 0.36.
  • n is the sample size, which is given as 125.

Plugging in the values we get:

EBP = 1.645 * sqrt((0.36 * (1 - 0.36)) / 125)

EBP = 1.645 * sqrt(0.2304 / 125)

EBP = 1.645 * sqrt(0.0018432)

EBP = 1.645 * 0.0429269

EBP ≈ 0.0706

The closest answer choice to 0.0706 is (B) 0.07. Therefore, the margin of error for the 90% confidence interval for the population proportion is approximately 0.07, which corresponds to option (B).

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