College

Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ 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]

Answer :

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.