Question

Consider the population of 250,000 voters, 45% of which support Candidate A. A sample of 200 voters will produce a sample proportion p^​. The distribution of p^​ can be approximated with N(p,SE), where SE=√p(1−p)​​/n. The following logical steps lead to the formula for a C% Confidence Interval for p : - C% of sample proportions (of size n ) are expected to fall within ( LL=p−zₐ/₂SE,UL=p+zₐ/₂SE). - Therefore, C% of sample proportions are expected to be within a distance of ME=zₐ/₂​SE from p. - Consequently, p should be within a distance of ME from C% of sample proportions. - C% of (LCl=p−zₐ/₂​SE,UCI=p​+zₐ/₂​SE) should include p. LCI&UCl stand for Lower & Upper Confidence Interval limits. ME stands for Margin of Error. "Should" is used to mean that we expect something to happen; not that it will happen. Use this information to answer the following questions. What is the 95% ME for samples of 200 voters from the above population? Round your answer to four decimal places.

Answer 1

The 95% margin of error for samples of 200 voters from the given population is approximately 0.0616.Therefore, the 95% margin of error for samples of 200 voters is approximately 0.0616, rounded to four decimal places.

Step-by-step explanation:

To find the 95% margin of error (ME) for samples of 200 voters, we will use the formula:

ME = zₐ/₂ * SE

where zₐ/₂ is the z-score corresponding to the desired confidence level (in this case, 95%) and SE is the standard error, which can be calculated using the formula √p(1−p)/n.

Given that 45% of the population supports Candidate A, p = 0.45 and n = 200.

Let's calculate the margin of error:

SE = √(0.45(1−0.45)/200) ≈ 0.0314

Using a z-score table or calculator, the z-score for a 95% confidence level is approximately 1.96.

Plugging these values into the formula:

ME = 1.96 * 0.0314 ≈ 0.0616

Therefore, the 95% margin of error for samples of 200 voters is approximately 0.0616, rounded to four decimal places.