Question

When designing a study to determine this population proportion, what is the minimum number of drivers you would need to survey to be 95% confident that the population proportion is estimated to within 0.02

Answer 1

The sample size is
`n =2401`

Explanation:

From the question we are told that

The margin of error is
`E = 0.02`

From the question we are told the confidence level is 95% , hence the level of significance is

`\alpha = (100 - 95 ) \%`

=>
`\alpha = 0.05`

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

`Z_{(\alpha )/(2) } =  1.96`

Generally we will assume the sample proportion to be
`\^ p = 0.5`

Generally the sample size is mathematically represented as

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

=>
`n = [(1.96 )/(0.02) ]^2 *  0.5 (1 - 0.5)`

=>
`n =2401`