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 (ME) for the sample mean, you can use the formula:

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

where:

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

Assuming a common confidence level of 95%, the z-value is approximately 1.96.

Given:

  • \( s = 2.3 \) cm
  • \( n = 54 \)

Now, we can compute the margin of error.

  1. Calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{2.3}{\sqrt{54}} \] First, calculate \( \sqrt{54} \): \[ \sqrt{54} \approx 7.348 \] Now compute the SE: \[ SE = \frac{2.3}{7.348} \approx 0.313 \]

  2. Now, calculate the margin of error: \[ ME = 1.96 \times 0.313 \approx 0.613 \]

Finally, round the margin of error to the nearest hundredth: \[ ME \approx 0.61 , \text{cm} \]

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