95% confidence interval

chromium · #1 $Old$ June 30th, 2009, 12:08 PM

Hello all! I would really appreciate some help. I kind of used an equation that seemed to work.

If I have 9 observations, specifically 240,216,265,284,229,232,250,225,252, and need to find a 95% confidence interval for the mean $\mu$ assuming that these are observations of a $N(\mu,\sigma^{2})$ distribution, can I do the following?:

( $\bar{x}-t_{\alpha/2}\frac{s}{\sqrt{n}},$ $\bar{x}+t_{0.025,8}\frac{s}{\sqrt{n}})$
*I think $\alpha/2$ becomes 0.025 with 8 degrees of freedom and therefore $t_{\alpha/2}=2.306$

$(243.66667-2.306\frac{21.418449999}{\sqrt{9}}, 243.66667+2.306\frac{21.418449999}{\sqrt{9}})$
(227.20, 260.13)

mr fantastic · #2 $Old$ June 30th, 2009, 05:04 PM

Quote:

Originally Posted by chromium $View Post$

Hello all! I would really appreciate some help. I kind of used an equation that seemed to work.

If I have 9 observations, specifically 240,216,265,284,229,232,250,225,252, and need to find a 95% confidence interval for the mean $\mu$ assuming that these are observations of a $N(\mu,\sigma^{2})$ distribution, can I do the following?:

( $\bar{x}-t_{\alpha/2}\frac{s}{\sqrt{n}},$ $\bar{x}+t_{0.025,8}\frac{s}{\sqrt{n}})$
*I think $\alpha/2$ becomes 0.025 with 8 degrees of freedom and therefore $t_{\alpha/2}=2.306$

$(243.66667-2.306\frac{21.418449999}{\sqrt{9}}, 243.66667+2.306\frac{21.418449999}{\sqrt{9}})$
(227.20, 260.13)

The theory looks OK, I haven't checked the fine detail (where the devil always is). A thread of relevance: constructing a confidence interval please help

matheagle · #3 $Old$ June 30th, 2009, 07:51 PM

Use this table...http://www.danielsoper.com/statcalc/calc10.aspx
or for p-vaulues use...Free p-Value Calculator for the Student t-Test
You can find all of these via the homepage on both of the links I gave you.