The times it took for 35 loggerhead sea turtle eggs to hatch in a simple random sample are normally distributed, with a mean of 50 days and a standard deviation of 2 days. Assuming a 95% confidence level (95% confidence level=z-score of 1.96), what is the margin of error for the population mean? Remember, the margin of error, ME , can be determined using the formula ME is equal to the fraction with numerator z times s and denominator square root of n
Use the margin-of-error formula: ME = z × s ÷ √n. Substituting the values gives ME = 1.96 × 2 ÷ √35 ≈ 3.92 ÷ 5.92 ≈ 0.66 days. Therefore, the margin of error is 0.66, choice A.
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