need help with stats problem

[ercchry]

im having trouble figuring out what n equals and what P(hat) equals

ok so here is the chart dont know how well this will turn out

student | parent |
| neither one both |
never | 141 68 17 |
occasional | 54 44 11 |
regular | 40 51 19 |

(to veiw chart click the quote button, its not lining up any other way)


so its a poll of students and dope linked to parents that smoke or drink

question:

a) for the population of students whose parents do not use drugs or booze, form a 90 percent confidence interval for the population proportion who never use dope.

ok so for this one im pretty sure to find P(hat) its 226/235 and n is 235 but i am douting myself

now the tricky one

b) for the population of students with at least one parent who use drugs or booze, form a 90 percent confidence interval for the population proportion who never use dope.

now i did the same thing as i did in the first one and dont like the numbers im getting

so if someone could help that would be awsome

thanks
eric

[ercchry]

isnt there anyone out there that knows there introductory stats?

[WWJAI]

For the first one, i'm seeing the p=141/235(.60) and n=235 so q=.40
and confidence interval being .547<p<.653

[WWJAI]

And for second one, i'm seeing p=68/163 and n=163
and confidence interval being .354<p<.481

[ercchry]

thankyou very much for the first one i had that but i second guessed myself

[ercchry]

wait i still dont know about b) because it says AT LEAST one not just one

[WWJAI]

ohhh yep..good eye..i didn't notice that..so then p=85/210 and n=210
confidence interval being .349<p<.460

[urbannomad]

why can't u do it on Ti-83?

clicks stats->tests-> #7 and voila input your stuff....and hit calculate....

[ercchry]

umm because i dont have one and also it doesnt help you if you dont know the numbers to use and btw its already handed in.