the times it took for 35 loggerhead sea turtle eggs to hatch in a simple random sample are normally…

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 $me=\frac{zcdot s}{sqrt{n}}$. 0.06 0.11 0.34 0.66

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 $me=\frac{zcdot s}{sqrt{n}}$. 0.06 0.11 0.34 0.66

Answer

Explanation:

Step1: Identify the values

$z = 1.96$, $s=2$, $n = 35$

Step2: Substitute values into formula

$ME=\frac{z\cdot s}{\sqrt{n}}=\frac{1.96\times2}{\sqrt{35}}$

Step3: Calculate the denominator

$\sqrt{35}\approx5.916$

Step4: Calculate the numerator

$1.96\times2 = 3.92$

Step5: Calculate the margin - of - error

$ME=\frac{3.92}{5.916}\approx0.66$

Answer:

$0.66$