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The EPA wants to conduct a survey to determine the current approval rating on the government's handling of environmental action. From previous studies, they know that they need to poll 275 people to achieve their desired level of confidence. If they want to keep the same level of confidence but divide the margin of error in third, how many people will they have to poll?

User Darion
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Answer: Assuming that the desired level of confidence and the population size remain constant, the sample size required for a given margin of error is proportional to the square of the standard normal deviate corresponding to the desired level of confidence.

That is, if n is the current sample size and m is the current margin of error, and we want to divide the margin of error by 3 while keeping the same level of confidence, then we need to find the new sample size, n', that satisfies the equation:

n'/n = (zα/2 / z'α/2)²

where zα/2 is the standard normal deviate corresponding to half the desired level of confidence, and z'α/2 is the standard normal deviate corresponding to half the desired level of confidence and one-third the margin of error.

Assuming a desired level of confidence of 95%, we have:

zα/2 = 1.96

z'α/2 = 1.96 / 3 = 0.6533 (rounded to four decimal places)

Substituting these values and the given values of n and m, we get:

n'/275 = (1.96 / 0.6533)²

Solving for n', we get:

n' = 275 * (1.96 / 0.6533)²

Using a calculator, we get:

n' ≈ 887

Therefore, the EPA would need to poll approximately 887 people to achieve the same level of confidence with a margin of error divided by three.

Explanation:

User Dave Michaels
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