Question

An economist is interested in studying the income of consumers in a particular region. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average income of $15,000. What is the width of the 90% confidence interval

Answer 1

The width is
`w = 282.8`

Explanation:

From the question we are told that

The sample size is n = 50

The population standard deviation is
`\sigma = \$ 1000`

The sample size is
`\= x = \$ 15,000`

Given that the confidence level is 90% then the level of significance can be mathematically represented 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_{(0.10 )/(2) } = 1.645`

Generally the margin of error is mathematically represented as

`E = Z_{(0.10)/(2) } * (\sigma )/(√(n) )`

substituting values

`E = 1.645 * (1000 )/(√(50 ))`

=>
`E = 141.42`

The width of the 90% confidence level is mathematically represented as

`w = 2 * E`

substituting values

`w = 2 * 141.42`

`w = 282.8`