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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?

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User Jprusakova
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Explanation:

To determine the confidence level that the newspaper used in their report, we need to use the margin of error and sample size to calculate the standard error and then use a standard normal distribution table to find the corresponding confidence level.

The formula for the margin of error is:

Margin of Error = Critical Value * Standard Error

Where the critical value depends on the level of confidence, and the standard error is the standard deviation of the sample mean, given by:

Standard Error = Standard Deviation / sqrt(Sample Size)

We can estimate the population standard deviation using the sample proportion, given by:

Standard Deviation = sqrt((p * (1-p)) / n)

where p is the proportion of individuals in the sample who are unemployed, which we can estimate using the sample proportion of unemployed individuals.

Since we don't have this information, we can use a conservative estimate of p = 0.5, which will give us the maximum possible standard deviation.

Therefore, the standard error can be estimated as:

Standard Error = sqrt((0.5 * (1-0.5)) / 4000) = 0.0125

Using the margin of error of 1%, we can solve for the critical value:

0.01 = Z * 0.0125

Z = 0.01 / 0.0125 = 0.8

Looking up the corresponding value in a standard normal distribution table, we find that the confidence level is approximately 77.5%.

Therefore, the newspaper most likely used a confidence level of 77.5% in their report.

User Jasminka
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