Question

A sample of 500 nursing applications included 60 from menThe 95% confidence interval of the true proportion of men who applied to the nursing program is

Answer 1

The correct solution is "(0.092, 0.148)".

Explanation:

Given:

Sample size,

n = 500

True proportion,

`\hat {p} = (60)/(500)`

`=0.12`

The standard error will be:

`SE=\sqrt{(\hat p(1-\hat p))/(n) }`

`=\sqrt{(0.12(1-0.12))/(500) }`

`=(0.12(0.88))/(500)`

The confidence interval is "1.96".

hence,

The required confidence interval will be:

=
`(\hat p - 1.96 SE, \hat p + 1.96 SE)`

By substituting the values, we get

=
`(0.12-1.96\sqrt{(0.12(0.88))/(500) }, 0.12+1.96\sqrt{(0.12(0.88))/(500) } )`

=
`(0.092, 0.148)`