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 m e = startfraction z times s over startroot n endroot endfraction. 0.06 0.11 0.34 0.66

Answer 1

The margin of error for the 95% confidence level is 0.66. Then the correct option is D.

What is the margin of error?

The probability or the chances of error while choosing or calculating a sample in a survey is called the margin of error.

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)

The margin of error is given by

$\rm Margin \ of \ error = z \times \dfrac{\sigma}{\sqrt{n}}\\\\Margin \ of \ error = 1.96 \times \dfrac{2}{\sqrt{35}}\\\\Margin \ of \ error = 0.6626 \approx 0.66$

The margin of error for the 95% confidence level is 0.66. Then the correct option is D.

More about the margin of error link is given below.

https://brainly.com/question/6979326