A random sample of 54 box turtles was selected from the population in a pasture to determine the average shell diameter. The sample mean is equal to 8.1 cm, with a sample standard deviation of 2.3 cm. What is the sample’s margin of error? Round the answer to the nearest hundredth.(1 point)

margin of error=

1 answer

To calculate the margin of error for the average shell diameter of box turtles, we can use the formula for the margin of error (ME) in the context of a confidence interval. The formula is:

\[ ME = z \cdot \left( \frac{s}{\sqrt{n}} \right) \]

Where:

  • \( z \) is the z-score corresponding to the desired level of confidence.
  • \( s \) is the sample standard deviation.
  • \( n \) is the sample size.

For this example, we need to decide on a confidence level to determine the z-score. A common confidence level is 95%, which corresponds to a z-score of approximately 1.96.

Given:

  • Sample size, \( n = 54 \)
  • Sample mean, \( \bar{x} = 8.1 \) cm (not needed for the margin of error calculation)
  • Sample standard deviation, \( s = 2.3 \) cm

Now we can calculate \( \frac{s}{\sqrt{n}} \):

\[ \frac{s}{\sqrt{n}} = \frac{2.3}{\sqrt{54}} \approx \frac{2.3}{7.348} \approx 0.312 \]

Now we can find the margin of error:

\[ ME = z \cdot \left( \frac{s}{\sqrt{n}} \right) = 1.96 \cdot 0.312 \approx 0.612 \]

Finally, rounding to the nearest hundredth:

\[ \text{Margin of error} \approx 0.61 \]

Thus, the sample’s margin of error is 0.61 cm.