Question

A survey of 73 students found that 37% were in favor of raising tuition to pave new parking lots. The standard deviation of the sample proportion is 9.8%. How large a sample (to the nearest person) would be required to reduce this standard deviation to 4.7%?

Answer 1

The value is
`n_2  = 317`

Explanation:

From the question we are told that

The sample size is n = 73

The proportion that in favor of raising the tuition is
`\^(p) =  0.37`

The standard deviation is
`s_1  = 0.098`

The required standard deviation
`s_2  = 0.047`

Generally the requires standard deviation is mathematically represented as

`s_2  = s_1 *  \sqrt{(n_1)/(n_2) }`

=>
`(s_2)/(s_1)  =  \sqrt{(n_1)/(n_2) }`

=>
`(n_1)/(n_2) =[(s_2)/(s_1) ]^2`

=>
`(73)/(n_2) =[(0.047)/(0.098) ]^2`

=>
`(73)/(n_2) =0.2300`

=>
`n_2 =  (73)/(0.2300)`

=>
`n_2  = 317`