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. 0.06 0.11 0.34 0.66

Question

Timothy Fernandez

Mathematics

9 July, 09:42

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. 0.06 0.11 0.34 0.66

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Camille Ballard · Answer 1

We are given the following data

n = 35

x = 50

s = 2

CI = 95% (z = 1.96)

The formula for the margin of error is

E = z s / √n

Substituting the given values

E = 1.96 (2) / √35)

E = 0.66 or 66%