Question

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

Answer 1

Answer-

The margin of error is 0.66

Solution-

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.

So here,

sample size = n = 35

confidence level (95%) = z = 1.96

mean = μ = 50

standard deviation = s = 2

We know that,

$ME=\dfrac{z\cdot s}{\sqrt{n}}$

Putting the values,

$ME=\dfrac{1.96\times 2}{\sqrt{35}}=0.66$

Therefore, the margin of error is 0.66

Answer 2

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%