Question

When 415 junior college students were surveyed, 150 said they have a passport. constructa 95% confidence interval for the proportion of junior college students that have apassport. round to the nearest thousandth?

Answer 1

`0.362-0.045<p<0.362+0.045`

Explanation:

It is given that When 415 junior college students were surveyed, 150 said they have a passport, then sample proportion will be:

Sample proportion=
`(150)/(415)=0.362`

Then,
`ME=1.96\sqrt{\frac{0.362{*}0.638}{415}}`

=
`1.96\sqrt{(0.230)/(415)`

=
`1.96(0.023)`

=
`0.045`

Therefore, at 95% confidence interval, the proportion of junior college students that have a passport is:

`0.362-0.045<p<0.362+0.045`

Answer 2

Solution:
The 95% confident interval will be estimated as follows:

sample proportion: 150/415=0.362

ME=1.96*[0.362*0.638/415]=0.0011
thus
95% CI
0.362-0.0011<p<0.362+0.0011