1 TIMEREMANING $$ 54: 17 $$ 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 $$ M E=\frac{z \bullet s}{\sqrt{n}} $$ 0.06 0.11 0.34 0.66

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To find the margin of error for the population mean, we can use the formula:

$$ ME = \frac{z \times s}{\sqrt{n}} $$

Given:
- $$ z = 1.96 $$ (for a 95% confidence level)
- $$ s = 2 $$ (standard deviation)
- $$ n = 35 $$ (sample size)

Substitute the values into the formula:

$$ ME = \frac{1.96 \times 2}{\sqrt{35}} $$

Now, we can calculate the margin of error.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{1.96\times 2}{\sqrt{35}}$$
- step1: Multiply the numbers:
  $$\frac{3.92}{\sqrt{35}}$$
- step2: Divide the terms:
  $$\frac{98}{25\sqrt{35}}$$
- step3: Multiply by the Conjugate:
  $$\frac{98\sqrt{35}}{25\sqrt{35}\times \sqrt{35}}$$
- step4: Multiply the numbers:
  $$\frac{98\sqrt{35}}{875}$$
- step5: Reduce the fraction:
  $$\frac{14\sqrt{35}}{125}$$
The margin of error for the population mean is approximately 0.66.
The margin of error for the population mean is approximately 0.66.

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