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. Based on the historic data it is known that the population standard deviation is 5 years. The margin of error of a 98 percent CI for the true mean client age is approximately:

Answer 1

2.33

Explanation:

We shall use the z distribution because we know the population standard deviation.

The margin of error formula is given by:

`E=Z_{(\alpha)/(2)}((\sigma)/(√(n)))`

The z-value for 98% confidence interval is 2.326.

The sample size is n=25.

The population standard deviation is is

`\sigma = 5`

We substitute the values into the formula to get:

`E=2.326((5)/(√(25)))`

Evaluate square root:

`E=2.326((5)/(5))`

Simplify:

`E=2.326(1) = 2.326`