Question

A local service agency hires you to estimate the average annual income of all 700 families living in a four square block section. Since time and funds are limited, yu take a random sample of 50 families and want the estimate to have a 92% confidence level. The 50 families had an average income of $35,300. Research shows the population standard deviation for annual income in this four square block section is $1,800

Answer 1

The true estimate of the average annual income of all 700 families living in a four square block section lies within the 92% confidence interval below

`34854.3  <  \mu < 35 745.7`

Explanation:

From the question we are told that

The sample size is n = 50

The sample mean is
`\= x = \$ 35300`

The population standard deviation is
`\sigma = \$ 1800`

From the question we are told the confidence level is 92% , hence the level of significance is

`\alpha = (100 - 92 ) \%`

=>
`\alpha = 0.08`

Generally from the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.751`

Generally the margin of error is mathematically represented as

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

=>
`E = 1.751  *  (1800 )/(√(50) )`

=>
`E =  445.7`

Generally 92% confidence interval is mathematically represented as

`\= x -E <  \mu <  \=x  +E`

=>
`35 300 -445.7 <  \mu < 35 300 + 445.7`

=>
`34854.3  <  \mu < 35 745.7`

So the true estimate of the average annual income of all 700 families living in a four square block section lies within the 92% confidence interval