Question

The owner of Limp Pines Resort wanted to know the average age of its clients. A random sample of 25 tourists is taken. It shows a mean age of 46 years with a standard deviation of 5 years. The margin of error of a 98 percent CI for the true mean client age is approximately:

Answer 1

Margin of error is 2.33.

Explanation:

Since we have given that

Sample size = 25

Mean = 46 years

Standard deviation = 5 years

We need to find the margin of error of a 98% confidence interval for the true mean client age.

So, Margin of error is given by

`z* (\sigma)/(√(n))\\\\=2.33* (5)/(√(25))\\\\=2.33* (5)/(5)\\\\=2.33`

Hence, margin of error is 2.33.