Question

According to a report in USA Today, more and more parents are helping their young adult children buy homes. You would like to construct a 90% confidence interval of the proportion of all young adults in Louisville, Kentucky, who received help from their parents in buying a home. How large a sample should you take so that the margin of error is no more than 0.06?

Answer 1

The value is
`n = 188`

Explanation:

From the question we are told that

The margin of error is
`E = 0.06`

From the question we are told the confidence level is 90% , hence the level of significance is

`\alpha = (100 - 90 ) \%`

=>
`\alpha = 0.10`

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

`Z_{(\alpha )/(2) } =  1.645`

Generally will assume that the sample proportion young adults in Louisville, Kentucky, who received help from their parents in buying a home is
`\^ p = 0.5`

Generally the sample size is mathematically represented as

`n = [\frac{Z_{(\alpha )/(2) }}{E} ]^2 * \^ p (1 - \^ p )`

=>
`n = [(1.645)/(0.06 ) ]^2 * 0.5 (1 - 0.5 )`

=>
`n = 188`