Suppose a certain species bird has an average weight ofx=3.55 grams. Based on previous studies, we can assume that the weights of these birds have a normal distribution with 0=0.37 grams. For a small group of 15 birds, find the margin of error for a 70% confidence interval for the average weights of these birds. how do you work this formula for the given problem
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To find the margin of error for a 70% confidence interval for the average weights of the birds, you can use the following formula:
Margin of error = Z * (standard deviation / √n)
In this formula:
- Z is the z-score corresponding to the desired confidence level. For a 70% confidence interval, the z-score is 1.036 (you can find this value using a standard normal distribution table or a statistical calculator).
- Standard deviation refers to the variability of the weights of the birds, which is given as 0.37 grams.
- n is the sample size, in this case, 15 birds.
Now, let's plug in these values into the formula:
Margin of error = 1.036 * (0.37 / √15)
Calculate the square root of 15:
Margin of error = 1.036 * (0.37 / √15) ≈ 0.1942 grams
Therefore, the margin of error for a 70% confidence interval for the average weights of these birds is approximately 0.1942 grams.