Question

. A discount brokerage selected a random sample of 64 customers and reviewed the value of their accounts. The mean was $32,000 with a population standard deviation of $8,200. What is a 90% confidence interval for the mean account value of the population of customers

Answer 1

The 90% confidence interval is
`\$ \ 30313.9< \mu < \$ \ 33686.13`

Explanation:

From the question we are told that

The sample size is n = 64

The sample mean is
`\= x = \$ 32, 000`

The standard deviation is
`\sigma= \$ 8, 200`

Given that the confidence interval is 90% then the level of significance is mathematically evaluated as

`\alpha = 100 - 90`

`\alpha = 10 \%`

`\alpha = 0.10`

Next we obtain the critical value of
`( \alpha )/(2)` from the normal distribution table , the value is

`Z_{(\alpha )/(2) } = 1.645`

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } * ( \sigma )/( √(n) )`

=>
`E = 1.645 * ( 8200 )/( √(64) )`

=>
`E = 1686.13`

The 90% confidence interval is mathematically represented as

`\= x - E < \mu < \= x + E`

=>
`32000 - 1689.13 < \mu < 32000 + 1689.13`

=>
`\$ \ 30313.9< \mu < \$ \ 33686.13`