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

Answer:

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

Step-by-step explanation:

Given:

Sample size,

n = 500

True proportion,

$\hat {p} = \frac{60}{500}$

  $=0.12$

The standard error will be:

$SE=\sqrt{\frac{\hat p(1-\hat p)}{n} }$

     $=\sqrt{\frac{0.12(1-0.12)}{500} }$

     $=\frac{0.12(0.88)}{500}$

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

= $(0.12-1.96\sqrt{\frac{0.12(0.88)}{500} }, 0.12+1.96\sqrt{\frac{0.12(0.88)}{500} } )$

= $(0.092, 0.148)$