Question

You measure 32 turtles' weights, and find they have a mean weight of 73 ounces. Assume the population standard deviation is 7.9 ounces. Based on this, what is the maximal margin of error associated with a 95% confidence interval for the true population mean turtle weight. Give your answer as a decimal, to two places

Answer 1

Margin of error = 2.74 ounces

Explanation:

We are given the following in the question:

Sample mean,
`\bar{x}` = 73 ounces

Sample size, n = 32

Alpha, α = 0.05

Population standard deviation, σ = 7.9 ounces

Margin of error:

`z_(critical)* (\sigma)/(√(n))`

`z_(critical)\text{ at}~\alpha_(0.05) = 1.96`

Putting the values, we get,

`1.96* ((7.9)/(√(32)) ) =2.7372 \approx 2.74`

Thus, the maximum margin of error associated with a 95% confidence interval for the true population mean turtle weight is 2.74 ounces