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------------------------------------------------ A sample of 40 chipmunks taken from a park in Owensboro has a mean mass of 461.7 g with a standard deviation of 97.9 g. A researcher wants to find the [tex]94\%[/tex] confidence interval of the mean mass of all the chipmunks in the park. What will the margin of error be? Round your answer to 4 decimal places.

[tex]E = \square[/tex]

To find the margin of error for the mean mass of all the chipmunks in the park with a 94% confidence interval, you can follow these steps:

1. Understand the given data:
- Sample size ([tex]\( n \)[/tex]) = 40
- Sample mean = 461.7 grams (not needed for margin of error)
- Sample standard deviation ([tex]\( s \)[/tex]) = 97.9 grams
- Confidence level = 94%

2. Determine the critical value (z-value):
- Since the confidence level is 94%, you need to find the z-value that corresponds to the central 94% of the standard normal distribution.
- This involves looking up the z-value for the cumulative probability of [tex]\( \frac{1 + 0.94}{2} = 0.97 \)[/tex].
- The z-value for 0.97 is approximately 1.8808.

3. Calculate the standard error:
- The standard error (SE) is determined by dividing the sample standard deviation by the square root of the sample size:
[tex]\[
\text{SE} = \frac{\text{sample standard deviation}}{\sqrt{\text{sample size}}} = \frac{97.9}{\sqrt{40}} \approx 15.4793
\][/tex]

4. Calculate the margin of error (E):
- Multiply the z-value by the standard error to find the margin of error:
[tex]\[
E = z \times \text{SE} = 1.8808 \times 15.4793 \approx 29.1135
\][/tex]

Therefore, the margin of error is approximately 29.1135 grams when rounded to four decimal places.