Use the confidence level and sample data to find the margin of error E. Round your answer to the same number of decimal places as the sample mean unless otherwise noted. College students annual earnings: 99% confidence, n=76, x=3016, o=872
To find the margin of error (E) with a given confidence level, sample size, and sample data, the formula is:
E = z * (o / √n)
Where:
E = Margin of error
z = z-score corresponding to the desired confidence level
o = Standard deviation
n = Sample size
Given:
Confidence level = 99%, which corresponds to a z-score of 2
To find the margin of error (E) with a given confidence level, sample size, and sample data, the formula is:
E = z * (o / √n)
Where:
E = Margin of error
z = z-score corresponding to the desired confidence level
o = Standard deviation
n = Sample size
Given:
Confidence level = 99%, which corresponds to a z-score of 2.576 (obtained from a standard normal distribution table)
Sample size (n) = 76
Sample mean (x) = 3016
Standard deviation (o) = 872
Substituting these values into the formula, we get:
E = 2.576 * (872 / √76)
Calculating the square root of the sample size:
√76 ≈ 8.7178
E = 2.576 * (872 / 8.7178)
E ≈ 256.68
Rounding to the same number of decimal places as the sample mean, the margin of error (E) is approximately 256.68.
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