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

Explanation:

Step1: Identify given values

We know that \( z = 1.96 \) (for 95% confidence level), \( s = 2 \) (standard deviation), and \( n = 35 \) (sample size).

Step2: Substitute into the margin of error formula

The formula for margin of error is \( ME=\frac{z\cdot s}{\sqrt{n}} \). Substituting the values, we get \( ME=\frac{1.96\times2}{\sqrt{35}} \).
First, calculate the numerator: \( 1.96\times2 = 3.92 \).
Then, calculate the denominator: \( \sqrt{35}\approx5.916 \).
Now, divide the numerator by the denominator: \( ME=\frac{3.92}{5.916}\approx0.66 \).

Answer:

0.66