You measure 30 turtles' weights, and find they have a mean weight of 76 ounces. Assume the population standard deviation is 14.7 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean turtle weight.
To find the maximal margin of error associated with a 90% confidence interval, we need to determine the critical value for a one-tailed interval at a 90% level of confidence.
Since the population standard deviation is known, we can use the formula:
Margin of Error = Critical Value * (Standard Deviation / √(Sample Size))
The critical value for a 90% confidence level can be found using a z-table or calculator, or through the invNorm function in software like Excel.
For a 90% confidence level (one-tailed interval), the critical value is approximately 1.645.
Using the given data:
Mean weight of turtles (x̄) = 76 ounces
Population standard deviation (σ) = 14.7 ounces
Sample size (n) = 30 turtles
Margin of Error = 1.645 * (14.7 / √(30))
Calculating this expression gives a value of approximately 5.78.
Therefore, the maximal margin of error associated with a 90% confidence interval for the true population mean turtle weight is approximately 5.78 ounces.