Question

The current unemployment rate in the country is approximately 8%. A local newspaper reports that in an SRS of 4,000 people in your city, they have a margin of error in their confidence interval of 1%. What is the confidence level that they used in their report?

Answer 1

Answer: The margin of error for a confidence interval depends on the sample size and the level of confidence. For a sample of 4,000 people, a margin of error of 1% implies that the sample proportion is within 1% of the true proportion with a certain level of confidence.

Let's denote the true proportion of unemployed people in the city by p. Then the margin of error can be expressed as:

z*sqrt((p(1-p))/n) = 0.01

where z* is the critical value of the standard normal distribution corresponding to the level of confidence.

Since we don't know the value of p, we'll use the worst-case scenario where p = 0.5, which gives us the maximum possible margin of error. Then we have:

z*sqrt((0.5(1-0.5))/4000) = 0.01

Simplifying and solving for z*, we get:

z* = 2.576

The critical value of 2.576 corresponds to a level of confidence of approximately 99%. Therefore, the confidence level that the newspaper used in their report is 99%.

Explanation: