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

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.