# Thread: confidence interval

1. ## confidence interval

39 nuts are gathered. They have an average size of 3.34 cent and a standard deviation of .45 cent. Find the 95% confidence interval for the mean size of all the nuts

2. assuming normality this should be a t with n-1 degrees of freedom,
which most people will approximate with a z anyhow, which is 1.96 for a 95 percent CI.
(by the way, is mr fantasy one of the nuts? It would seem so.)

$\displaystyle \bar X \pm t_{n-1,.025}s/\sqrt{n}$

3. Originally Posted by matheagle
assuming normality this should be a t with n-1 degrees of freedom,
which most people will approximate with a z anyhow, which is 1.96 for a 95 percent CI.
(by the way is mr fantasy one of the nuts?)
Can you direct me where to get the t statistic?

I dont understand where you would use the z value in the formula you gave me

4. Originally Posted by mherr
Can you direct me where to get the t statistic? Mr F says: Use Google eg. 1.3.6.7.2. Upper Critical Values of the Student's-t Distribution

I dont understand where you would use the z value in the formula you gave me Mr F says: You don't know the population sd, only the sample sd. So the Z-statistic is not used. As baldeagle said, the t-statistic is used. But with df = 38, the critical value of z rather than t can be used without too much error.
@baldeagle: *Ahem* .... Space 1999 Eagle Transporter - Where All Eagles Come Home

5. i got the t stat to be 1.94

6. IF we don't have normality, then they want us to use z=1.96 via the central limit theorem. We kind of need to know what the underlying distribution is. Most likley they want the 1.96 percentile from the st normal tables. In any case with n reasonably large (30 according to most books) it doesn't matter which way we go. But one should know where you're sampling from.