how do you go about constructing a 95% confidence interval. you guys have been great with the help links so far.
There is a way to compute that on your calculator.
I do not know the exact details, maybe CaptainBlank or JakeD can explain better.
Those numbers on the table of the normal curve appear as the integration answer to,
$\displaystyle \int_0^t e^{-x^2}dx$
So for example you want to calculate,
from $\displaystyle 0<z<.5$
Then you substitute that into the integral for $\displaystyle t=.5$ and you need to find,
$\displaystyle \int_0^{.5} e^{-x^2}dx$
Problem is that you cannot do that the usual way because there is no closed form anti-derivative. Yet you can approximate by Simson's rule (that is what the calculator does) under the "calculate integral function".
Of course it is not exactly like I said it, the means and standard deviation affect the function you are integration. But that is the basic idea.
Is there a way to calculate the 95% confidence interval for a problem:
say there are 1000 people polled and 600 say they smoke. something like that.
Or.
Say the average marriage lasts 750 days with an sd of 100 (from a poll of 50). how would yyou construct a 90% confidence interval for that?
Why double post the same question?. PH is gonna send his goons after you . I just answered this here.
http://www.mathhelpforum.com/math-he...questions.html
The latter is a confidence interval for the mean. The first is on the other post.
90% CI; $\displaystyle {\sigma}=100$; $\displaystyle \overline{x}=750$; n=50
$\displaystyle E=z\frac{\sigma}{\sqrt{n}}=1.645\frac{100}{\sqrt{5 0}}=23.26$
$\displaystyle 726.74<{\mu}<773.26$
You can say with 90% confidence that the mean marriage time is between 726.74 and 773.26 days.