Answer:
Margin of error = 4.21 ounces
Step-by-step explanation:
According to the Question,
- Given That, You measure 25 turtles' weights, and find they have a mean weight of 31 ounces. Assume the population standard deviation is 12.8 ounces
Therefore, Sample mean = 31 ounces , Sample size(n) = 25 , Alpha(α) = 0.10 & Population standard deviation(σ) = 12.8 ounces
- Thus, Margin of error =
× σ / √n (
Putting The Values, We get
1.645 × (12.8 / √25 ) ⇒ 4.2112 ≈ 4.21
Thus, the maximum margin of error associated with a 90% confidence interval for the true population mean turtle weight is 4.21 ounces
