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 $me = \frac{z \bullet s}{\sqrt{n}}$
\bigcirc 0.06
\bigcirc 0.11
\bigcirc 0.34
\bigcirc 0.66

Explanation:

Step1: Identify given values

We have \( z = 1.96 \), \( s = 2 \), and \( n = 35 \).

Step2: Plug into margin of error formula

Using \( ME=\frac{z\cdot s}{\sqrt{n}} \), substitute the values: \( ME=\frac{1.96\times2}{\sqrt{35}} \).
First, calculate \( \sqrt{35}\approx5.916 \).
Then, \( 1.96\times2 = 3.92 \).
Now, divide: \( \frac{3.92}{5.916}\approx0.66 \).

Answer:

0.66