Questions: Suppose another group collects a sample of 48 turtles and finds 24 of them are female. They simulate 30 samples of 48 turtles with a 0.5 probability of each being female and find the mean proportion to be 0.517 . What would be a reasonable estimate for the margin of error? 0.27 0.068 0.034 0.259

Solution Steps

To estimate the margin of error, we first calculate the standard deviation of the sample proportion. The formula for the standard deviation \( \sigma \) of a proportion is given by:

\[ \sigma = \sqrt{\frac{p(1-p)}{n}} \]

where:

• \( p = 0.517 \) (sample proportion)
• \( n = 48 \) (sample size)

Substituting the values:

\[ \sigma = \sqrt{\frac{0.517(1-0.517)}{48}} = \sqrt{\frac{0.517 \times 0.483}{48}} \approx 0.0721 \]

For a 95% confidence level, the Z-score \( Z \) is approximately:

\[ Z = 1.96 \]

The margin of error \( E \) can be calculated using the formula:

\[ E = \frac{Z \times \sigma}{\sqrt{n}} \]

Substituting the values we have:

\[ E = \frac{1.96 \times 0.0721}{\sqrt{48}} \approx 0.0204 \]

Final Answer