You want to estimate the mean exam score for students at a university. The confidence level that you use in your calculation is 90%, the standard deviation is 5, and the sample size of the population of students is 500. What is the margin of error? Round your answer to the nearest hundredth.(1 point)
The margin of error can be calculated using the formula:
Margin of Error = Z * (standard deviation / sqrt(n))
Where Z is the critical value for a 90% confidence level, which is 1.645.
So, the margin of error is:
Margin of Error = 1.645 * (5 / sqrt(500)) = 1.45
Therefore, the margin of error is approximately 1.45.
This is wrong.
I apologize for the mistake. Let's recalculate the margin of error using the correct formula.
Margin of Error = Z * (standard deviation / sqrt(n))
Where Z is the critical value for a 90% confidence level, which is approximately 1.645.
Given:
Standard deviation (σ) = 5
Sample size (n) = 500
Confidence level = 90%
Margin of Error = 1.645 * (5 / sqrt(500))
Margin of Error = 1.645 * (5 / 22.36)
Margin of Error = 1.645 * 0.2235
Margin of Error ≈ 0.368
Therefore, the margin of error is approximately 0.37 when rounded to the nearest hundredth.